Class 12 Physics

Unit 1 Electrostatics Chapter 2 Electric Field Short Question And Answers

Question 1. An electrostatic field line is a continuous curve. That is, a field line cannot have sudden breaks. Why not?
Answer:

An electrostatic field line is a continuous curve. That is, a field line cannot have sudden breaks.

The tangent drawn at any point of an electrostatic field line indicates the direction of the field at that point Electrostatic field being continuous the field lines (or the lines of force) cannot have any sudden break.

Question 2. Check that the ratio ke²/Gm_em_p is dimensionless. From the table of physical constants, determine the value of this ratio. What does the ratio signify?

Answer:

Dimension of

⇒ \(\left[\frac{k e^2}{G m_e m_p}\right]=\frac{\mathrm{ML}^3 \mathrm{~T}^{-4} \mathrm{I}^{-2} \mathrm{I}^2 \mathrm{~T}^2}{\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2} \cdot \mathrm{M} \cdot \mathrm{M}}=\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^0\)

∴ The ratio is dimensionless

⇒ \(\frac{k e^2}{G m_e m_p}=\frac{\left(9 \times 10^9\right) \times\left(1.6 \times 10^{-19}\right)^2}{\left(6.67 \times 10^{-11}\right) \times\left(9.1 \times 10^{-31}\right)}\)

= 2.27 X 10³¹

The ratio can be written as

⇒ \(=\frac{k e^2}{r^2}: \frac{G m_e m_p}{r^2}\)

⇒ \(\frac{electrostatic force of attraction between a proton and an electron}{gravitational force of attraction between a proton and an electron}\)

Question 3. The track of three charged particles in a uniform electrostatic field, Give the signs of the three charges. Which particle has the highest target mass ratio?

Answer:

The track of three charged particles in a uniform electrostatic field, Give the signs of the three charges.

Particles 1 and 2 are negative charges as these are attracted towards the positive plate while particle 3 is positively charged. The \(\frac{e}{m}\) ratio is highest for particle 3 as its deflection is maximum.

Question 4. Careful measurement of the electric field at the surface of a black box indicates that the net outward flux through the surface of the box is 8.0 x 10³N m².C^-1.

1. What is the net charge inside the box?
2. If the net outward flux through the surface of the box were zero, could you conclude that there were no charges inside the box? Why or why not?

Answer:

1. Flux = \(\frac{q}{\epsilon_0}\)

∴ q = eee0 X flux = 8.854 x 10^-12 x 8 x 10³ C

= 0.07μC

2. It cannot be said that the box has no charge inside it because if it contained an equal amount of positive and negative charges, the flux would have been zero. It can only be said that the net charge inside the box is zero.

Question 5. A point charge of 10μC is placed at a distance 5 cm directly above the center of a square of side 10 cm. What is the magnitude of the electric flux through the square?

Answer:

A point charge of 10μC is placed at a distance 5 cm directly above the center of a square of side 10 cm.

Let the plane ABCD be one of the sides of a cube of side 10 cm. The point P is situated at the center of the cube.

∴ Flux passing through ABCD

⇒ \(\frac{1}{6} \frac{q}{\epsilon_0}=\frac{1}{6} \times \frac{10 \times 10^{-6}}{8.85 \times 10^{-12}} \mathrm{~N} \cdot \mathrm{m}^2 \cdot \mathrm{C}^{-1}\)

= 1.88 X 10⁵ N.m².C^-1

Question 6. A conducting sphere of radius 10 cm has an unknown charge. If the electric field 20 cm from the center of the sphere is 1.5 x 10³ N.C^-1 and points radially inward, what is the net charge on the sphere?
Answer:

A conducting sphere of radius 10 cm has an unknown charge. If the electric field 20 cm from the center of the sphere is 1.5 x 10³ N.C^-1 and points radially inward

Electric field,

⇒ \(E=\frac{1}{4 \pi \epsilon_0} \frac{q}{r^2}\)

∴ \(q=\frac{E r^2}{\left(\frac{1}{4 \pi \epsilon_0}\right)}=\frac{1.5 \times 10^3 \times(0.2)^2}{9 \times 10^9} \mathrm{C}\)

= 6.67nC

Question 7. Two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and of magnitude 17.0 x 10^-22C.m^-2. What is the electric field

1. In the outer region of the first plate?
2. In the outer region of the second plate?
3. Between the plates?

Answer:

Two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and of magnitude 17.0 x 10^-22C.m^-2.

1. \(E_1=-\frac{\sigma_A+\sigma_B}{2 \epsilon_0}=-\frac{\sigma_A-\sigma_A}{2 \epsilon_0}=0\) [∵ \(\sigma_A=-\sigma_B=17.0 \times 10^{-22} \mathrm{C} \cdot \mathrm{m}^{-2}\)]

2. Similarly, E₂ = 0

3. \(E_3=\frac{\sigma_A-\sigma_B}{2 \epsilon_0}\)

= \(\frac{2 \sigma_A}{2 \epsilon_0}\)

= \(\frac{\sigma_A}{\epsilon_0}\)

= \(\frac{17.0 \times 10^{-22}}{8.854 \times 10^{-12}}\)

= 1.92 X 10^-10 N C^-1

Question 8. In a certain region of space, the electric field is along the z direction throughout. The magnitude of the electric field is, however, not constant but increases uniformly along the positive z -direction, at the rate of 105 N.C^-1 per meter. What are the force and torque experienced by a system having a total dipole moment equal to 10^-7 C m in the negative z-direction?
Answer:

In a certain region of space, the electric field is along the z direction throughout. The magnitude of the electric field is, however, not constant but increases uniformly along the positive z -direction, at the rate of 105 N.C^-1 per meter.

Let the dipole be along z -direction

∴ Dipole moment, p_z = -10^-7 C.m

⇒ \(\frac{\partial E}{\partial z}=10^5 \mathrm{~N} \cdot \mathrm{C}^{-1} \cdot \mathrm{m}^{-1}\)

∴ Net force = \(p_x \frac{\partial E}{\partial x}+p_y \frac{\partial E}{\partial y}+p_z \frac{\partial E}{\partial z}\)

= 0 + 0-10^-7 X 10⁵ N

= -10^-2 N

Torque = pEsinθ = pEsin80° = 0

Question 9.

1. A conductor A with a cavity is given a charge Q. Show that the entire charge must appear on the outer surface of the conductor.
2. Another conductor B with charge q is inserted into the cavity keeping B insulated from A. Show that the total charge on the outside die surface of A is (Q + q).
3. A sensitive instrument is to be shielded from the strong electrostatic fields in its environment. Suggest a possible way.

Answer:

1. The charge will reside on the outer surface of the conductor.

2. Charge q inside the cavity would induce a charge -q on the inner side of the cavity and a charge + q would appear on the outer surface making the total charge on this surface equal to (Q + q).

3. The instrument has to be kept inside a closed metallic shell to shield it from the strong electric fields in its environment.

Question 10. A hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the hole is \(\frac{\sigma}{\epsilon_0} \hat{n}\) where h is the unit vector in the outward direction, after is the surface density of charge near the hole.
Answer:

A hollow charged conductor has a tiny hole cut into its surface.

Suppose the hole has been plugged. Then the field inside the conductor is zero and outside it is \(\frac{\sigma}{\epsilon_0} \hat{n}\). This field is essential the sum of two fields.

One due to the plugged hole and the other due to the rest of the conductor. Inside the conductor, these two forces are equal and opposite and the resultant is zero.

But on the outer surface, these two forces are equal and act in the same direction.

Since the resultant is \(\frac{\sigma}{\epsilon_0} \hat{n}\) then each of 0 these forces is equal to \(\frac{\sigma}{2\epsilon_0} \hat{n}\).

Thus the electric field in the hole is equal to \(\frac{\sigma}{\epsilon_0} \hat{n}\).

Question 11. It is now believed that protons and neutrons (which constitute nuclei of ordinary matter) are themselves built out of more elementary units called quarks. A proton and a neutron consist of three quarks each. Two types of quarks, the so-called ‘up’ quark (denoted by’ u ‘) ofcharge + \(\frac{2}{3}\)e and the ‘down’ quark (denoted by ‘ d ‘) of charge –\(\frac{1}{3}\)e, together with electrons build up ordinary matter. Suggest a possible quark composition of a proton and a neutron
Answer:

It is now believed that protons and neutrons (which constitute nuclei of ordinary matter) are themselves built out of more elementary units called quarks. A proton and a neutron consist of three quarks each. Two types of quarks, the so-called ‘up’ quark (denoted by’ u ‘) ofcharge + \(\frac{2}{3}\)e and the ‘down’ quark (denoted by ‘ d ‘) of charge –\(\frac{1}{3}\)e, together with electrons build up ordinary matter.

The charge of a proton +e.

A proton can be made up of 2u ‘s and Id because In dial case the total charge = \(\frac{2 e}{3}+\frac{2 e}{3}-\frac{e}{3}=e\)

∴ A proton can be composed as used.

A neutron has no charge. It can be made up of 2d ‘s and I u because In that case the total charge = \(\frac{2}{3} e-\frac{1}{3} e-\frac{1}{3} e=0\)

∴ A neutron can be composed as udd.

Question 12.

1. Consider an arbitrary electrostatic field configuration. A small test charge is placed at a null point (i.e, where \(\vec{E}\) = 0) of the configuration. Show that the equilibrium of the test charge Is necessarily unstable.
2. Verify this result for the simple configuration of two charges of the same magnitude and sign placed at a certain distance apart.

Answer:

1. Let us assume the free charge to remain in stable equilibrium. Under this condition, the test charge would return to its original position of neutral point when deflected slightly in any direction. This implies that all the field lines near the neutral point are directed towards it.

This proves the existence of inward-bound electric flux associated with the imaginary Gaussian surface surrounding the point But according to Gauss’ theorem, the flux associated with a closed surface in the absence of any enclosed charge is zero. Hence our assumption was not correct, i.e., the free charge should remain in unstable equilibrium.

2. In this case the neutral point is the mid-point of the line joining the two charges. If the free charge is moved slightly from its position along the line connecting the charges, the restoring force will bring the charge back to its original position.

But if it is moved along a direction normal to the line, then it will move further due to the effect of the resultant force. Thus the free charge is not in stable equilibrium.

Question 13. An infinite line charge produces a field of 9 x 10⁴ N.C^-1 at a distance of 2 cm. Calculate the linear charge density.
Answer:

An infinite line charge produces a field of 9 x 10⁴ N.C^-1 at a distance of 2 cm.

The field at a distance r from a system with linear charge density is given by

⇒ \(E=\frac{1}{2 \pi \epsilon_0} \cdot \frac{\lambda}{r} \quad \text { or, } \lambda=4 \pi \epsilon_0 \frac{E r}{2}\)

∴ \(\lambda=\frac{1}{9 \times 10^9} \times \frac{9 \times 10^4 \times 0.02}{2}\)

= 10^-7 C.m^-1

Question 14. A spherical conductor of radius 12 cm has a charge of 1.6 x 10^-7C distributed uniformly on its surface. What is the electric field?

1. Inside the sphere?
2. Just outside the sphere?
3. At a point 18 cm from the center of the sphere?

Answer:

1. The electric field inside the conductor is zero because the charge resides on the outer surface of the conductor.

2. Electric field just outside the sphere

⇒ \(E=\frac{1}{4 \pi \epsilon_0} \cdot \frac{q}{R^2}=9 \times 10^9 \times \frac{1.6 \times 10^{-7}}{(0.12)^2}\)

= 10⁵ N.C^-1

3. Electric field at a distance 18 cm from the centre,

⇒ \(E^{\prime}=\frac{1}{4 \pi \epsilon_0} \frac{q}{r^2}=9 \times 10^9 \times \frac{1.6 \times 10^{-7}}{(0.18)^2}\)

= 4.44 x 10⁴ N.C^-1

Question 15. Two charged conducting spheres of radii a and b are connected to each other by a wire. What is the ratio of electric field at the surfaces of the two spheres?
Answer:

Two charged conducting spheres of radii a and b are connected to each other by a wire.

When two conducting spheres are connected to each other by a wire, a transfer of charge will take place until the two spheres attain the same potential.

i.e., V₁ = V₂

or, E₁a = E₂b

∴ \(\frac{E_1}{E_2}=\frac{b}{a}\)

Question 16. Two large conducting spheres carrying charges Q₁ and Q₂ were brought close to each other. Is the magnitude of the electrostatic force between them exactly given by \(Q_1 Q_2 / 4 \pi \epsilon_0 r^2\), where r is the distance between their centers?
Answer:

No, because Coulomb’s law is applicable only for point charges. For large conductors, if the distance between the conductors becomes very large, then only their size may be neglected. Further, the charges reside only on the surface of the conductor, so the charge distribution is not uniform.

Question 17. A small test charge is released at rest at a point in an electrostatic field configuration. Will it travel along the field line passing through that point?
Answer:

A small test charge is released at rest at a point in an electrostatic field configuration.

The electrostatic force acts along the field line. Hence the test charge will be accelerated along the field line passing through the point. Its velocity (i.e., the motion) will be along the same direction only if the field line is a straight line.

Question 18. Four point charges +q, + q, +q, and -q are placed at the four comers of a square, the length of each side of which is a . Find the magnitude of the intensity of the electric field at the center of the square.
Answer:

Four point charges +q, + q, +q, and -q are placed at the four comers of a square, the length of each side of which is a

The point of intersection O of the diagonals is equidistant from four charges.

∴ \(O A=O B=O C=O D=\frac{a}{\sqrt{2}}\)

Intensities of the electric field at O due to the charges at A and C balance each other, because the two fields are equal and opposite

Intensity 0 due to the charge at \(B=\frac{q}{(a / \sqrt{2})^2} ; \text { along } \overrightarrow{O D}\)

Intensity at 0 due to the charge at \(D=\frac{q}{(a / \sqrt{2})^2}\); along \(\vec{OD}\)

Resultant intensity at O

⇒ \(\frac{q}{(a / \sqrt{2})^2}+\frac{q}{(a / \sqrt{2})^2}=\frac{2 q}{a^2}+\frac{2 q}{a^2}=\frac{4 q}{a^2} ; \text { along } \overrightarrow{O D}\)

Question 19. On the x-y plane, two point charges +q and -q are placed at positions (0, l) and (0, -l) respectively. Find an expression for the intensity of the electric field at a point (0, y) where y >> l. Under what condition does the charging system behave as a dipole and hence express the electric field in terms of the dipole moment of the dipole so formed?
Answer:

On the x-y plane, two point charges +q and -q are placed at positions (0, l) and (0, -l) respectively.

Suppose, the electric field intensity at point C due to the charges at A and B are and E2, respectively

∴ \(E_1=\frac{1}{4 \pi \epsilon_0} \cdot \frac{q}{(y-l)^2} ; \text { along } \overrightarrow{C Y}\)

⇒ \(E_2=\frac{1}{4 \pi \epsilon_0} \cdot \frac{q}{(y+l)^2} ; \text { along } \overrightarrow{C A}\)

As E₁ and E₂ are acting in the opposite directions and E₁ > E₂, the resultant electric field at C is,

E = E₁-E₂

⇒ \(\frac{1}{4 \pi \epsilon_0} \cdot \frac{q}{(y-l)^2}-\frac{1}{4 \pi \epsilon_0} \cdot \frac{q}{(y+l)^2}\)

⇒ \(\frac{q}{4 \pi \epsilon_0}\left[\frac{1}{(y-l)^2}-\frac{1}{(y+l)^2}\right]=\frac{q}{4 \pi \epsilon_0} \cdot \frac{4 y l}{\left(y^2-l^2\right)^2}\)

⇒ \(\frac{1}{4 \pi \epsilon_0} \cdot \frac{4 q y l}{y^4}\) [∵ y >> 1]

⇒ \(-\frac{1}{4 \pi \epsilon_0} \cdot \frac{4 q l}{y^3} \text {; along } \overrightarrow{C Y}\)

If the value of l is small, the combination of the charges behaves as an electric dipole with dipole moment, p = q x 21.

In that case, \(E=\frac{1}{4 \pi \epsilon_0} \cdot \frac{2 p}{y^3} ; \text { along } \overrightarrow{C Y}\)

Question 20. Show graphically how the magnitude of the intensity of the electric field changes with distance r from the center of the shell.
Answer:

The variation of field intensity with distance has been shown by the curved line AB

Question 21. An electric field is expressed as \(\vec{E}=20 \hat{i} \mathrm{~V} \cdot \mathrm{m}^{-1}\). Find the electric flux passing through a surface of area 0.25 m² normal to the x-axis.
Answer:

An electric field is expressed as \(\vec{E}=20 \hat{i} \mathrm{~V} \cdot \mathrm{m}^{-1}\).

The y-z plane is perpendicular to the x-axis.

So, the given area vector is \(\vec{S}=0.25 \hat{i} \mathrm{~m}^2\)

∴ The electric flux across this area,

⇒ \(\phi=\vec{E} \cdot \vec{S}=(20 \hat{i}) \cdot 0.25 \hat{i}\)

= 5V.m

Question 22. State Gauss’s theorem in electrostatics. Find the electric flux through a surface of area 50 m² in x-y plane in the electric field \(\vec{E}=3 \hat{i}+2 \hat{j}+\hat{k} \mathrm{~V} \cdot \mathrm{m}^{-1}\).
Answer:

The unit vector along the normal to the xy plane i.e., z axis is \(\hat{k}\).

So, the area vector = 50 \(\hat{k}\) m²

∴ Electric flux = \(\int \vec{E} \cdot \overrightarrow{d s}=\int(3 \hat{i}+2 \hat{j}+\hat{k}) \cdot \overrightarrow{d s}\)

⇒ \((3 \hat{i}+2 \hat{j}+\hat{k}) \cdot \int \vec{d} s=(3 \hat{i}+2 \hat{j}+\hat{k}) \cdot 50 \hat{k}\)

= 50 V.m

Question 23. If an electric dipole of the moment \(\vec{p}\) is placed along a uniform electric field of intensity \(\vec{E}\), the torque acting on the dipole is

1. \(\vec{\tau}=\vec{p} \times \vec{E}\)
2. \(\vec{\tau}=\vec{p} \cdot \vec{E}\)
3. \(\vec{\tau}=\vec{p}+\vec{E}\)
4. \(\vec{\tau}=0\)

The option 4 is correct.

Question 24. Define the surface density of electric charge. Two large conducting spheres carrying charges Q₁ and Q₂ are brought close to each other. Is the magnitude of the electrostatic force between them exactly given by \(\frac{Q_1 Q_2}{4 \pi \epsilon_0 r^2}\) where r is the distance between their centers?
Answer:

Surface charge density is the amount of electric charge per unit surface area.

Question 25. Define dielectric constant. Two charges ±20 x 10^-6 C , placed 2 mm apart from an electric dipole. Determine the electric field at a point 10 cm away from the center of the dipole on its perpendicular bisector. Given,

⇒ \(\frac{1}{4 \pi \epsilon_0}=9 \times 10^9 \mathrm{~N} \cdot \mathrm{m}^2 \cdot \mathrm{C}^{-2}\)

Answer:

Dipole moment,

p = ql = (20 x 10^-6) x (2 x 10^-3) =4 x 10^-8 C.m

10 cm >> 2 mm; So the distance of the given point is much
greater than the dipole length.

∴ \(E=\frac{1}{4 \pi \epsilon_0} \cdot \frac{p}{r^3}\)

⇒ \(\left(9 \times 10^9\right) \times \frac{\left(4 \times 10^{-8}\right)}{(0.1)^3}\) [10cm = 0.1m]

= 3.6 x 10⁵ V.m^-1

Question 26. Why are electric lines of force, not closed-loop?
Answer:

Electric lines of force start from a positive charge and end on a negative charge. There is no electric field inside a conductor, so electric field lines are not present there. Hence electric lines of force do not form a closed loop.

Question 27. A charge q is placed at the center of a cube of side l. What is the electric flux passing through each face of the cube?
Answer:

A charge q is placed at the center of a cube of side l

Flux passing through the surface of the cube = \(\frac{q}{\epsilon_0}\)

∴ Flux passing through each of the face = \(\frac{1}{6} \frac{q}{\epsilon_0}\)

Question 28. An electric dipole is held in a uniform electric field The dipole is aligned parallel to the field. Find the work done in rotating it through the angle of 180°
Answer:

An electric dipole is held in a uniform electric field The dipole is aligned parallel to the field.

Work done in rotating the dipole through an angle dθ is pEsinθdθ.

where, p = dipole moment,

E = magnitude of the field

∴ Work done in rotating the dipole from 0° (parallel to the field) to 180°

⇒ \(\int_{0^{\circ}}^{180^{\circ}} p E \sin \theta d \theta=p E[-\cos \theta]_0^{180^{\circ}}=2 p E\)

Question 29. A charge q is placed at the center of a cube. What is the electric flux passing through two opposite faces of the cube?
Answer:

A charge q is placed at the center of a cube.

The electric flux passes through two opposite faces of the cube

⇒ \(\frac{1}{6} \frac{q}{\epsilon_0} \times 2=\frac{1}{3} \frac{q}{\epsilon_0}\)

Question 30. A charge q is placed at the center of a cube. What Is the electric flux passing through the cube?
Answer:

A charge q is placed at the center of a cube.

The flux passing through the cube = \(\frac{q}{\epsilon_0}\)

Question 31. Two charges of magnitudes -2Q and +Q points (a, 0) and (4a, 0) respectively. What is the electric flux due to these charges through a sphere of radius 3a with its center at the origin?
Answer:

Two charges of magnitudes -2Q and +Q points (a, 0) and (4a, 0) respectively.

According to Gauss’ theorem,

flux through a surface,

⇒ \(\phi=\frac{\text { charge enclosed by the surface }}{\epsilon_0}\)

Since Q is outside the sphere,

flux, \(\phi=\frac{-2 Q}{\epsilon_0}\)

Question 32. Consider two hollow concentric spheres S₁ and S₂ enclosing charges 2Q and 4Q respectively.

1. Find out the ratio of the electric flux through them,
2. How will the electric flux through S₁ change if a medium of dielectric constant er is introduced in the space inside S₁ in place of air? Deduce the necessary expression.

Answer:

1. According to Gausss’ theorem, flux through any surface

⇒ \(\frac{\text { charge enclosed by the surface }}{\epsilon}\)

∴ \(\frac{\phi_{s_1}}{\phi_{s_2}}=\frac{\frac{2 Q}{\epsilon}}{\frac{(2 Q+4 Q)}{\epsilon}}\)

= \(\frac{1}{3}\)

2. when the medium is filled up with a medium of dielectric constant er then, electric field,

⇒ \(E=\frac{2 Q}{4 \pi \epsilon_0 \epsilon_r r^2}\)

∴ Flux through S₁

⇒ \(\int E \cdot d S=E \cdot 4 \pi r^2=\frac{2 Q}{4 \pi \epsilon_0 \epsilon_r r^2} \times 4 \pi r^2=\frac{2 Q}{\epsilon_0 \epsilon_r}\)

Question 33. A point charge +Q is placed in the vicinity of a conducting surface. Trace the field lines between the charge and the conducting surface.
Answer:

A point charge +Q is placed in the vicinity of a conducting surface.

Question 34. What is the amount of work done in moving a point charge Q around a circular arc of radius ‘r’ at the center of which another point charge ‘q’ is located?
Answer:

The work done in moving a charge along any circular path is zero. So, to move a point charge Q around a circular arc of radius ‘r’ at the center of which another point charge’ is located, no work had to be done.

Question 35. Does the charge given to a metallic sphere depend on whether it is hollow or solid? Give a reason for your ambler.
Answer:

No, the charge given to a metallic sphere does not depend on whether it is hollow or solid, because all the charges given to a metallic sphere will move to the outer surface of the sphere and charges will be distributed uniformly over the surface of a sphere.

Question 36.

1. A point charge q is at a distance of directly above the center of a square of side d, as shown in the figure. Use Gauss’ law to obtain the expression for the electric flux through the square.

2. If the point charge is now moved to a distance d from the center of the square and the side of the square is doubled, explain how the electric flux will be affected.

Answer:

1. Let us imagine the given square is one of the faces of the cube of edge d. Where the point charge q Is at a distance of \(\frac{d}{2}\) above the center of the square. So it is enclosed by the cube. Hence by Gauss’ theorem, electric flux linked with one surface of the cube is \(Q_{\mathrm{sq}}=\frac{q}{6 \epsilon_0}\)

2. If the distance of the point charge from the center of the square is now doubled, the side of the square also gets doubled. So from the same symmetry, the flux through the square will still be \(\frac{q}{6 \epsilon_0}\)

Question 37. Draw a graph to show the variation of E with perpendicular distance r from the line of charge
Answer:

A graph to show the variation of E with perpendicular distance r from the line of charge

Since, \(E=\frac{\lambda}{2 \pi \epsilon_0 r} \quad\)

∴ \(E \propto \frac{1}{r}\)

Therefore, the plot of E versus r will be as shown.

 

Electrostatics Elementary Phenomena Of Electrostatics Introduction

• In about 600 BC, Greek philosopher Thales of Miletus observed that if amber (hard resin from pine tree) is rubbed with flannel acquires the property of attracting small pieces of paper or light bodies towards it.
• In 1600 AD, William Gilbert, a physician to Queen Elizabeth, observed that many other substances behave in the same way. For example, a glass rod rubbed with silk exhibits the same phenomenon. Even when we comb our hair, the comb will show the same property.
• An object which gets such an ability to attract others due to rubbing is called an electrified object and the process is called electrification. Practically, this electrification is due to the transfer of electric charge from one object to another.
• The electricity produced by rubbing is called frictional electricity. This type of electricity remains confined within the body where it originates and cannot move from one place to another. Hence it is also called statical electricity.
• The word ‘electricity’ is derived probably from electron which is the Greek name of amber.

Electric charge:

Gravitational attraction exists between any two particles inside a material. The mass of each particle is the property that is responsible for this attraction.

Read and Learn More Class 12 Physics Notes

On the other hand, the repulsion between two elements or the attraction between an electron and a proton is due to a force known as an electric force. This force originates from a property of the particles known as electric charge. This is a measurable physical quantity.

1. Electric charge is a scalar quantity.
2. Its unit in the CGS system is esu or statcoulomb (state).
3. The unit of charge in SI is coulomb (C).
   1C = 3 x 10⁹ statC
4. The dimension of electric charge is IT

Electrostatics Elementary Phenomena Of Electrostatics Electrification By Rubbing

Experiment: In winter, after combing our hair if we bring the comb near some small pieces of paper, it will attract them. Here, the comb is electrified due to rubbing with hair and it acquires the property of attracting the pieces of paper. Sometimes the comb and the hair are so much electrified during rubbing that we can hear a crackling sound due to tiny sparks.

Example: A metallic chain is often hung from the fuel tank of a motor vehicle and it touches the road below. When the vehicle is in motion, static electricity is developed due to rubbing between the fuel and the inner wall of the tank. The hanging chain helps to move the electric charges to the ground instantly. In the absence of such a chain, the accumulated charges may inflame the fuel in the tank.

Electrostatics

Elementary Phenomena Of Electrostatics Two Kinds Of Electric Charge Positive And Negative

With the help of a simple experiment, it can be proved that electric charges are of two types — positive charge and negative charge.

Experiment: A glass rod rubbed with silk is suspended by a silk thread from a support. An ebonite rod rubbed with fur is suspended from the same support close to the glass rod. The two rods will attract each other.

So two bodies being electrified differently may attract each other. The ebonite rod is now replaced with another glass rod, rubbed with silk, and is suspended in its position. The two rods will repel each other. So two bodies being electrified similarly repel each other.

Inference:

From the above experiment, we can infer that similar electric charges are produced in two similar rods if they are rubbed with similar substances. Since the two rods repel each other, it is proved that similar charges repel each other.

On the other hand, opposite charges may be produced in two different rods if they are rubbed with different substances. Then they attract each other. Here, we can infer that opposite charges attract each other.

The fundamental law of statical electricity:

Opposite or unlike charges attract each other and similar or like charges repel each other. Note that both the charges attract an uncharged body.

American scientist Benjamin Franklin called one of them a positive charge and the other one a negative charge. The reason is that if we give equal amounts of two opposite charges to a body, it will remain unchanged (like 5-5 = 0 ).

The convention is that the charge produced in a glass rod rubbed with silk is positive and the charge produced in an ebonite rod rubbed with flannel is negative.

Electrostatic series:

This is a list where the substances have been arranged in such a way that if any two of them are rubbed together, the one preceding the other in the list acquires a positive charge and the latter acquires a negative charge. This is known as the electrostatic series.

1. Fur
2. Flannel
3. Sealing wax
4. Glass
5. Paper
6. Silk
   
7. Human body
8. Wood
9. Metals
10. Rubber
11. Resin
12. Amber
13. Sulphur
14. Ebonite

From the table, it is clear that a particular substance may be either positively charged or negatively charged if it is rubbed with two different substances. For example, a glass rod becomes positively charged if It is rubbed with silk and negatively charged if it is rubbed with flannel

Repulsion is the Conclusive Test of Electrification:

A charged body attracts an uncharged body as well as an oppositely charged body. But if repulsion takes place between the two bodies, it can be concluded that the experimental body must be charged because repulsion is possible only between two bodies with the same kind of charge. Hence, repulsion is the conclusive test of electrification.

Electrostatics

Elementary Phenomena Of Electrostatics Electronic Theory Of Electricity

Electronic theory accounts for the source of the charge. This is based on the electronic structure of matter. We know that atoms are made up of a positively charged central core called the nucleus, around which negatively charged electrons rotate in various closed orbits. The nucleus contains positively charged protons. The number of electrons in an atom is equal to the number of protons in the nucleus. As a whole, an atom is electrically neutral.

The nucleus of an atom attracts the electrons towards it. Electrons lying in the outermost orbit are rather loosely bound and hence are easy to detach. If one or more electrons are somehow removed from the atom, the atom becomes positively charged. Conversely, if an atom is given one or more electrons, it becomes negatively charged.

Therefore, a positively charged body has a deficiency of electrons in its atoms and a negatively charged body has excess electrons in its This is the electronic theory. Note that electrons alone are responsible for the electrification of a body and not the protons because protons are firmly held in the nucleus.

An atom is composed of three fundamental particles—electron, proton, and neutron. The number of these fundamental particles obviously varies from element to element. The mass of an electron is 9 x 10^-31 kg and the amount of charge in it is 1.6 x 10^-19C. This is the smallest possible amount of charge.

1. The mass of a positively charged body is less than the mass of the same electrically neutral body. In this case, lost mass of the body = number of electrons lost by the body x mass of an electron.
2. The mass of a negatively charged body Is greater than the mass of the same electrically neutral body. In Till’s case, gained mass of the body = number of electrons gained by the body x mass of an electron,

Quantization of charge: The charge carried by a body Is always an Integral multiple of the smallest unit of charge which is the charge of an electron. This is known as the quantization of charge.

That’s why, charge q = ±ne, where, e = charge of an electron and n = 1, 2, 3,… a natural number. Electric charge is a scalar quantity. The net charge of a body is the algebraic sum of all the charges present.

Invariance relative to tiro frame of reference:

The amount of a charge is independent of its state of rest or of motion; and also independent of the state of rest or of motion of the observer. This is expressed by saying that an amount of charge is invariant with respect to the frame of reference. So we can say, q_rest = q_moving

Explanation of Frictional Electricity on the Basis of Electronic Theory:

The electrons are bound to an atom due to the attraction of the positively charged nucleus. This attraction is obviously not equal for the atoms of different substances.

So during rubbing between two bodies, some electrons are transferred from one to the other, producing equal and opposite charges simultaneously. For example, when a glass rod is rubbed with silk, some electrons from the glass migrate to silk.

The force of attraction between the electrons and the nucleus in glass is less than that in silk. Hence the glass rod becomes positively charged with a deficit of electrons and the silk becomes negatively charged with an excess of electrons.

Principle of conservation of electric charge:

It states that the algebraic sum of the positive and the negative charges in an isolated system Is constant. The electronic theory clearly indicates that electric charge can neither be generated nor destroyed. It is only redistributed when electrons migrate from one body to another.

It should be mentioned that the principle had been known, long before the invention of electrons and protons. To date,’ no deviation has been observed for this principle. The principle conservation of charge is a universal law—no physical phenomenon exists for which this law is violated.

Electrostatics

Elementary Phenomena Of Electrostatics Conductor And Insulators

Electric charge cannot flow through all substances with equal ease. According to the ability of charge to move through the materials, they are classified into two major groups

1. Conductor
2. Insulator or non-conductor.

Conductor:

Materials through which charge can move easily are called conductors. If some charge is given to any part of a conductor:

If some charge is given to any part of a conductor, it will spread all over the body of the conductor

Generally, all metals are conductors of varying degrees. Among them silver, copper, and aluminum rank higher. Eaiui, the human body, gas carhop, graphite, mercury, etc., are it conductors. Acid, alkali, and aqueous solutions of salts also conduct electricity. Generally, no material is a perfect conductor.

Insulator or non-conductor:

Materials through which charge cannot move are called insulators or non-conductors:

If any part of an insulator is charged, charges remain confined to that part of the insulator and do not spread all over the body. Dry air, glass, rubber, ebonite, mica, silk, paraffin, bakelite, etc., are insulators. Remember that generally no material can be said to be a perfect insulator.

Besides conductors and insulators, there is a third kind of material called semiconductors, which are neither good conductors nor good insulators. Selenium, germanium, silicon, etc., are semiconductors. They are widely used in electronic circuit elements like transistors, Integrated circuits, etc.

Nowadays a few substances can be made to behave as almost perfect conductors at very low temperatures. These are called superconductors. Aluminum acts as a superconductor at -272°C.

Dry air is a good insulator. But the charge can flow through moist air. Hence experiments on statical electricity cannot be conducted satisfactorily in the rainy season. In the electricity supply system, the transmission wires are joined through porcelain pots, which are non-conductors.

But if the wires were directly connected to the electric post, electricity would have been grounded instantly and there would be a huge transmission loss, and if someone touched the post, there, would be a chance of getting shocked. Rubber, silk, or cotton to reduce the chance of short circuits.

These are called insulated wires. Pure water is a non-conductor, but various types of salts, bases, and acids are dissolved in natural water which makes it a fairly good conductor.

Conductors and Insulators on the basis of electronic theory:

The electrons of the outermost orbits of the atoms of a conductor are very loosely bound to the nucleus and move freely from one atom to another These electrons are called free electrons.

These free electrons carry electricity from one place to another throughout the conductor Metals are good conductors due to the presence of free electrons. In insulators or nonconductors, electrons are tightly bound to the nucleus—they are not free electrons. So electricity cannot flow through them

Electrostatics Elementary Phenomena Of Electrostatics Charging By Conduction

A conductor with a non-conducting handle is called an insulated conductor. Charges given to this conductor cannot go elsewhere and it becomes a charged conductor.

If an uncharged insulated conductor is brought in contact with such a charged insulated conductor, some charges cross over to the uncharged conductor, which becomes charged.

This process of charging a conductor by direct contact with a charged body is known as charging by conduction. It can be explained on the basis of electronic theory.

Explanation of charging by conduction on the basis of free electronic theory:

Suppose a body is negatively charged. So it has excess electrons. When this charged body comes in contact with an uncharged conductor, a few electrons from the charged body move into the conductor. As the conductor now has excess electrons, it becomes negatively charged.

If the body is positively charged initially and is touched with an uncharged conductor, a few electrons of the uncharged conductor move to the charged body due to attraction by the positive charge. As the conductor has a deficit of electrons, it becomes positively charged.

Note that, in both cases, the initial amount of charge in the charged hotly is shared between two bodies due to conduction. As a result, the strength of charge in the first body is reduced to some extent.

 Electrostatics

Elementary Phenomena Of Electrostatics Gold-Leaf Electroscope

An electroscope is an instrument used for detecting the presence of charge and its nature. A gold-leaf electroscope is a widely used instrument.

Description:

A metal rod P passes through an insulating stopper into a vessel fitted with glass panes. Two thin foils of gold (L, L) are hung at the lower end of the metal rod. The upper end of the rod is capped with a metal disc D.

The glass vessel protects the leaves from air current Some fused calcium chloride is placed inside the vessel as a hygroscopic substance. Two tin foils (f, t) are attached to the inner face of the glass walls. These foils which are in contact with the metallic base of the vessel are earthed.

Charging the electroscope by conduction: To charge the gold-leaf electroscope by conduction, a charged body is touched on the disc of the electroscope. A glass rod rubbed with silk becomes positively charged.

If this glass rod is made to touch the disc of the electroscope, a part of the positive charge spreads in the instrument. As a result, the leaves, being positively charged, diverge due to repulsion, and remain so even after the removal of the glass rod.

If the disc is now touched, all the charges of the electroscope will flow’ to die earth through the body of the experimenter, and the leaves will collapse. This is called earthing of a gold-leaf electroscope.

Charging by conduction is however not a good process at all. If the charged both- carries a large amount of charge, then as soon as it is brought in contact with the disc, the divergence of the leaves will be so great that they may get detached from the rod.

Uses:

Detection of charge: The body to be tested is brought slowly near or in contact with the disc of an uncharged electroscope. If the leaves are deflected, both’ are charged otherwise, uncharged.

Determination of the nature of charge: To test the nature of charge on a body, we have to start with a charged electroscope. Suppose, the electroscope is positively charged with its leaves divergent.

The body to be tested is brought slowly to touch the disc. If the divergence of the leaves increases, the body is positively charged. If the divergence decreases, the body is negatively charged. The nature of the charges would be just the opposite if we started initially with a negatively charged electroscope.

An uncharged body may also diminish the divergence to some extent, so it cannot definitely indicate that the charge of the test body is opposite in nature to that on the electroscope. Thus, an increase in the divergence of the leaves of the electroscope provides the surer test for the nature of a charge on a body.

The experimental results are given in the following table:

Obviously, an insulator will have no effect on the divergence of the leaves.

Identification of conductor and Insulator:

The body to be tested is brought slowly to touch the disk of a charged gold leaf electroscope. In this case, the body must be grounded with a conductive wire.

If the test body is a conductor then all the charges of the electroscope will flow to the earth through the conductive wire and the leaves will collapse. If the die body is an insulator then the charge of the electroscope will not be able to flow to the earth.

So there will be no change in the divergence of die leaves of the electroscope. This way, the identification of the conductor and insulator can be done using a gold-leaf electroscope.

Proof Plane:

It is of ten difficult to bring up a strongly charged test body near the electroscope. A portion of the charge from the body can be picked and taken to an electroscope for testing by using what is called a proof plane. A proof plane consists of a small metallic disc mounted with an art-insulating handle.

It is held by the insulating handle and the metal disc is momentarily placed in contact with the charged test body. The disc gets charged by conduction. It is now brought to a charged electroscope for testing the nature of the charge. A proof plane can also be used to detect the presence of charge or the nature of distribution of charge on a body

Electrostatics

Elementary Phenomena Of Electrostatics Electrostatic Induction

When a charged body is brought near, but not in contact with an insulator or insulated uncharged conductor, an opposite charge is produced at the near end and a similar charge at the far end of the insulator or conductor.

The charges so produced disappear as soon as the charged body is removed. This phenomenon is called electrostatic Induction which Is defined as the porary charging of the body by the influence of nearby charges.

Experiment: A positively charged glass rod A brought near the end R of an insulated uncharged conductor RC. Now the following operations are performed.

The disc of a proof plane Is made to touch the end B. Next, the proof plane is brought near an uncharged gold-leaf electroscope without actually touching it.

It is found that the leaves diverge. Hence the end B of the conductor must be charged. The proof plane is now discharged by touching with a hand. If the other end C of the conductor is examined, it will be found that this end is also charged.

But the middle of the conductor BC will be found to have practically no charge. So it may be concluded that induction produces charges only at both ends of a conductor.

The nature of the charges induced can be tested with a charged gold-leaf electroscope. Suppose we take a negatively charged electroscope. Keeping the glass rod A the end B, a proof plane is made to touch the end B.

Now it is taken near the negatively charged gold-leaf electroscope. It is observed that the divergence of the leaves increases which shows that the end B of the conductor is negatively charged.

The gold-leaf electroscope is now charged positively. The proof plane is discharged by touching with a hand. The end C of the conductor is touched with the proof plane. It is brought near the gold-leaf electroscope and it is found that the leaves diverge further. So the end C is positively charged, i.e., it has a charge similar to that of the glass rod.

In this experiment if, instead of the glass rod A, an ebonite rod rubbed with flannel be taken, it will be found that the end B has a positive charge and the end C has a negative charge. In the middle of BC, there is no charge.

Now if the glass rod or the ebonite rod is removed from the vicinity of the conductor BC, it will be found that no charge exists either at the end B or at the end C. So if the inducing body is removed, the charges of the conductor disappear. It is proved that under the influence of a charged body an uncharged conductor

1. Becomes temporarily charged,
2. Developsunlike charge at the near end and like charge at the far end, and
3. The induced charges disappear when the charged body is moved away from the conductor.

Explanation of Electrostatic Induction from Electronic Theory:

Electrostatic Induction In Conductor:

Electrostatic induction in a conductor can be easily explained by the electronic theory. Every conductor has a large number of free electrons which can flow from one atom to another within the conductor.

In the first experiment, due to the presence of the positive charge in rod A, some free electrons in the conductor BC are attracted to end B and cause an abundance of electrons at that end.

Hence the end B becomes negatively charged. On the other hand, a deficit of electrons by the same number has occurred at the end C and so it becomes equally positively charged.

In the second experiment, the negative charges of the ebonite rod repel the free electrons from the end B to the end C causing an excess of electrons at C and a deficiency of electrons at the end B. Thus the end C becomes negatively charged and the end B positively charged.

It may be noted that induction only changes the arrangement of the free electrons in the conductor. The total number of electrons in the conductor remains the same. So when the charged body is removed from the vicinity of the conductor, the electrons are redistributed uniformly and the conductor becomes uncharged.

Electrostatic Induction In insulator: Insulators have no free electrons. Electrostatic induction takes place in the insulators due to the polarisation of atoms placed in an electric field.

If a positively charged body is placed near an insulator, and the closest atom of the insulator finds itself in the field of the positive charge. This attracts the electrons of the atom and repels the nucleus.

So a deformation takes place in the structure of the atom. As a result, the nucleus of the atom is no longer symmetrical with respect to the electrons.

A slight relative displacement occurs between the positive and the negative charges inside the atom. This happens for all the atoms of the insulator. This is called the polarisation of atoms.

When a positively charged body is brought near an insulator, the negative charge of each atom is pulled toward the body while the positive charge is repelled in the opposite direction.

Since both types of charges exist in equal amounts inside an insulator, they neutralize each other. Hence, the inside of an insulator is electrically neutral. Just the opposite phenomenon will happen if a negatively charged body is brought near an insulator.

Inducing Charge and Induced Charge Free Charge and Bound Charge:

Inducing and Induced charge: The charge responsible for creating induction, is called an inducing charge and the charge created due to induction is called an induced charge.

In section 1.8, the positive charge on the glass rod A or the negative charge on the ebonite rod is inducing charge and the charge developed in the conductor BC is induced charge.

From and bound charges: The induced charge at the end of a conductor near the inducing charge is opposite in nature to that of the inducing charge.

Hence the charges induced at the near end of the conductor remain immobile due to the electrostatic force of attraction and cannot escape by conduction. These fixed charges at the near end of the induced conductor are called bound charges.

The charges induced at the far end of the conductor are of the same kind as the inducing charge and hence a force of repulsion exists between these two.

So these charges can immediately flow to the earth by conduction if the conductor is touched by hand. Hence these charges at the far end of the induced conductor are called free charges.

In section 1.8, charges developed due to induction at the end B of the conductor BC are called bound charges and those developed at the tend C are called free charges.

A few facts about induction:

• From the discussions in the previous sections regarding electrostatic induction the following conclusions can be made
• Two kinds of charges, positive and negative are produced simultaneously due to induction.
• Unlike charge is induced at the near end of the conductor and like charge at the far end.
• Positive and negative charges are induced in equal amounts.
• It is a temporary phenomenon. The induced opposite charges neutralize, each other as soon as the inducing body is removed.

Induction Precedes Attraction:

• When a charged body is brought near an uncharged body, the uncharged body is attracted towards the charged body. This is due to electrostatic induction.
• When achargedbodyisbroughtnearan an uncharged body, unlike charges are induced at the near end and like charges at the far end of the uncharged body. The force of attraction or repulsion between two charges varies inversely to the square of the distance between them.
• As the unlike-induced charge is nearer to the inducing charge than the like-induced charge, the attraction between the unlike charges predominates over the repulsion between the like charges.
• So a resultant force of attraction acts on the uncharged body and the body as a whole is attracted towards the charged body. Soinductionprecedes attraction. Effectively, induction is the cause and attraction is its effect.

Electrostatics

Elementary Phenomena Of Electrostatics Charge Always Resides On The Outer Surface Of A Conductor

• If one end of an insulated rod is charged, the charge is confined to that end only. But when any part of a conductor is charged, the charge distributes itself over the whole surface.
• No charge is found to exist on the inside of a solid conductor or on the inner surface of a hollow conductor. According to the property of a conductor, a charge may flow easily through it.
• Like charges repel each other and try to move away from each other as far as possible. So, they distribute themselves on the outer surface of the conductor, where their mutual distance becomes maximum.
• Under special circumstances, charges may reside on the inner surface of a hollow conductor. A charged body is kept inside a hollow sphere in such a way that it does not touch the hollow sphere.
• In this case, unlike charges are induced on the inner surface of the hollow sphere and like charges on the outer surface. Without removal, the charged body of the hollow sphere is touched by hand.
• The free(charges on the outer surface of the hollow sphere move to the earth but the bound charges on the inner charge surface of the sphere exist there.
• As long as the inducing charge remains inside the hollow sphere, the bound charge also remains on the inner surface. When the charged body is removed, charges shift to the outer surface of the sphere.
• Electric Screen: Any arrangement, that can keep any space free from external electrical influences, is called an electric screen, Charge always resides on the outer surface of a conductor. Electric screens are based on this property.

A gold-leaf electroscope enclosed in a wire-gauge cage C is placed on an insulated base A. Now if a charged body is brought near to or in contact with the cage, no effect is produced on the electroscope because the charge resides on the outer surface of the cage, not inside. So the region enclosed by the cage is free from external electrical influences.

The space inside a closed metallic box is free from electrical influences for the same reason. The valves of a radio are placed in metallic cases to shield it from external electrical influences. Sensitive electrical instruments are always kept within electric screens.

Electrostatics

Elementary Phenomena Of Electrostatics Distribution Of Charge On The Surface Of A Conductor Surface Density Of Charge

Distribution of charge on a conducting surface: Although the charge on a conductor distributes itself all over the surface, it should not however be concluded that the distribution is always uniform all over the surface.

The distribution of charge depends on the shape of the conductor. The greater the curvature at any point, the greater will be the accumulation of charge at that point. Distributions of charge on charged conductors of different shapes are shown by dotted lines B.

The boundary of each conductor is shown by the line A. The density of charge in each case is roughly represented by the distance of the dotted line B from the boundary line A of each conductor.

Surface density of charge: The surface density of charge at a point on a charged conductor is the amount of charge per unit area of the surface of the conductor surrounding the point. The surface density of charge is generally denoted by the symbol cr.

If Q is the charge distributed uniformly over the surface of area A of a spherical conductor having radius r, the surface density of charge is given by,

⇒ \(\sigma=\frac{Q}{A}=\frac{Q}{4 \pi r^2} \text { or, } \sigma \propto \frac{1}{r^2}\)

So, the surface density of charge reduces with the increase of the radius of the object concerned and vice versa. Hence, at sharp bends or pointed portions of a conductor, the surface density of charge will be greatest. So a conductor having different curvatures at different points has different surface densities of charge at those points.

Unit of σ:

Dimension of σ: [σ] = L^-2TI

Electrostatics

Elementary Phenomena Of Electrostatics Numerical Examples

Example 1. A hollow spherical conductor of radius 2 cm is charged with 62.8 states. Determine the surface density of charge on the inner and outer surfaces of the conductor. If the sphere is solid, what will be the values of the above quantities?
Solution:

No charge resides on the inner surface of a hollow conductor. So surface density of charge on the internal surface of the hollow sphere is zero.

⇒ \(\sigma=\frac{Q}{4 \pi r^2}\) [Q = 62.8 statC; r = 2 cm]

⇒ \(\frac{62.8}{4 \pi(2)^2}=\frac{62.8}{16 \pi}\)

= 1.249 StaC cm^-2

If the sphere is a solid one, it has no interned surface. The surface area of a hollow sphere and that of a solid sphere of the same radius are equal. So the surface density of charge on the external surface of the solid sphere will be the same as that of the hollow sphere.

Example 2. 27 drops of water, each of radius 3 mm and having equal charge are combined to form a large drop. Find the ratio of the surface density of charge on the large drop to that on each small drop.
Solution:

Suppose, the charge on each small drop of water is q. So the charge in the combined drop will be 27q. In the first case, the surface density of charge,

⇒ \(\sigma_1=\frac{q}{4 \pi r^2}=\frac{q}{4 \pi(0.3)^2}\) [here, r = 3mm = 0.3 cm]

If R is the radius of the large drop, we have,

⇒ \(\frac{4}{3} \pi R^3=27 \times \frac{4}{3} \pi(0.3)^3\)

or, R = 0.9 cm.

In the second case,

⇒ \(\sigma_2=\frac{27 q}{4 \pi R^2}=\frac{27 q}{4 \pi(0.9)^2}\)

∴ \(\frac{\sigma_2}{\sigma_1}=\frac{27 q}{4 \pi(0.9)^2} \times \frac{4 \pi(0.3)^2}{q}\)

= \(\frac{27 \times 0.09}{0.81}\)

= \(\frac{3}{1}\)

σ₂ :  σ₁ = 3:1

Example 3. A hollow spherical conductor of radius 2 cm is electrified with 20 states. Determine the surface density of charge on the external surface of the conductor
Solution:

Surface density of charge of a spherical conductor,

⇒ \(\sigma=\frac{Q}{4 \pi r^2}\) [Q – charge on the surface of the sphere; r = radius of the sphere]

Here, Q = 20 statC; r = 2 cm

∴ \(\sigma=\frac{20}{4 \pi(2)^2}\)

= 0.398 statC.cm^-2

 Electrostatics

Elementary Phenomena Of Electrostatics Action Of A Pointed Conductor

We know that if a conductor with a sharp point is charged, the surface density of charge at the pointed part becomes very high. It will then induce unlike charges of the air molecules and dust particles in the vicinity which will be attracted towards it.

Due to this attraction, they come in contact with the sharp end, and their unlike charges get neutralized.

Only the particles having charges similar to the sharp end retain their charge and are therefore repelled strongly by the pointed end.

As a result of both these processes (attraction and repulsion), the conductor gradually loses its charge through the pointed end. This is known as the discharging action of points. So a conductor should be round and without any sharp end to retain its charge for a long time.

Lightning Conductor or Lightning Arrester:

• In 1752 AD, Benjamin Franklin experimentally proved the existence of charges in the clouds and in the atmosphere.
• Scientists found that cosmic rays, ultraviolet rays, and the rays emitted from radioactive substances on earth, charge the air particles and water drops of the clouds.
• Moreover, due to mutual friction of the clouds, the water drops in it become charged. Both types of charges, positive and negative, are produced.
• A flash of lightning is nothing but a discharge of electricity along an air-tracking the sky.
• Such discharges are possible because of the enormous difference in potential that may exist between a charged cloud and the earth or between two oppositely charged clouds.
• The air, in the path of the lightning, is heated up due to the discharge and expands suddenly.
• This sudden expansion highly reduces the pressure in this area and as a result, the surrounding air rushes there with a tremendous force.
• The report of the thunder is due to these sudden expansions and contractions of the air.
• A highly charged cloud usually causes electrostatic induction on Earth.
• So the potential difference between cloud and ground may become high enough to start an electric discharge.
• This is known as lightning. The sound heard just after lightning is called a thunderclap.
• During lightning, a high current is set up from the cloud to the ground.
• This current follows the least resistive path. For this, thunder generally strikes on trill buildings or trees.
• To protect highrise buildings from thunderbolts, lightning conductor is used. It consists of a long and thick copper strip.
• The upper end of it is designed with shaip points and the lower end is fixed to a metal plate buried deep m the ground.
• The lightning conductor protects the building using the discharging action of sharp points of the conductor.
• When a charged cloud passes above the points of the lightning conductor, it induces, unlike charge on the sharp points.
• The like charge induced on the other end passes to the earth. The discharging action of the sharp points partially neutralizes the charge of the cloud.
• Hence the possibility of lightning is reduced markedly.
• For this reason, the lightning conductor is also called a copper plate lightning arrester.
• Sometimes even after the disground charging action of the sharp points, a discharge may take place between the charged cloud and the building.
• Then the thick copper strip provides the path of least resistance for the charge to flow to the earth without damaging the building.
• For this reason, copper strips with sharp points are used as lightning conductors.

Necessary qualities of a good lightning conductor:

A good lightning conductor should have the following properties:

• The copper strip should not melt due to heat evolved during lightning discharge.
• The upper part of the copper strip should be provided with a sufficient number of sharp points.
• No discontinuity should exist in the copper strip. The lower end of it should be buried deep in the ground.

Safe shelters and Unsafe places during lightning:

Buildings on metal frames and houses fitted with lightning conductors are the safest places during lightning. Staying in a car with a metal frame with windows closed also offers adequate protection, if it is connected to the earth.

On the other hand, stray tall trees, telegraph and telephone posts, high fences of mud, etc. are unsafe places at the time of lightning.

It may be noted that, although lightning and thunder take place simultaneously, the sound of thunder reaches us much later, because the velocity of sound is much less than that of light. So, if a man hears the sound of thunder, there will be no chance of his being struck by lightning.

Electrostatics

Elementary Phenomena Of Electrostatics Ncert Text Book Questions With Answer Hint

Question 1. Why can one ignore the quantization of electric charge when dealing with macroscopic i.e., large-scale charges?
Answer:

The charge at the macroscopic level is so large compared to the charge of an electron that quantization of charge has no practical importance at this level. The charge of an electron is 1.6 x 10^-19C which means that a small charge of 1μC has about 10¹³ electrons presentinit. For such a large number of electrons, there is no significance of quantization, and should be treated as continuous.

Question 2. A polythene piece rubbed with wool is found to have a negative charge of 3 x 10^-7 C.

1. Estimate the number of electrons transferred (from which to which?)
Answer:

q = ne

∴ N = \(\frac{q}{e}\)

or, \(n=\frac{-3 \times 10^{-7}}{-1.6 \times 10^{-19}}\)

= 1.875 x 10¹²

Here, q = -3 x 10^-7C

e = -1.6 x 10^-19C

So, 1.875 x 10¹² electrons are transferred from wool to polythene.

2. Is there a transfer of mass from wool to polythene?
Answer:

Yes, some electrons are transferred from wool to polythene but the mass of electrons transferred is infinitesimally small, the transfer of mass may be neglected.

Electrostatics

Elementary Phenomena Of Electrostatics Conclusion

• Due to rubbing, an object which gets the ability to attract others is called an electrified object and the process is known as electrification.
• The principle of conservation of electric charge states that the total charge in an isolated system remains constant.
• If electrons are added to anatomy, it will be negatively charged and if electrons are removed from an atom, it becomes positively charged.
• This is the electronic theory of electricity.
• According to the ability of charge to move through the materials, they are classified into two major groups:
  1. Conductor
  2. Non-conductor orinsulator.
• Materials through which charge flows easily are called conductors. Generally, all metals are conductors.
• Materials through which charge is not found easily are called non-conductors or insulators.
• Dry air, glass, rubber, etc., are examples of conductors.
• Any conductor with a non-conducting base is called an insulated conductor.
• Friction produces simultaneously equal and opposite charges in two bodies.
• Electrostatic induction is defined as the charging of a body temporarily by the influence of nearby charges.
• The charge of a body that induces a charge on a conductor Is known as an inducing charge and the charge on the conductor is called Induced charge.
• The induced charge disappears as soon as the charged body is removed.
• The tire opposite charge Induced at the nearer end of the conductor due to electrostatic induction is called the bound charge while the similar charge induced at the far end is called the free charge.
• Equal and opposite charges are developed simultaneously in the same body due to induction.
• Induction precedes attraction. Charge always resides on the outer surface of a charged conductor.
• The arrangement that shields or screens a space from external electrical Influences is called electrical shielding or screening.
• The surface density of charge at a given point on a conductor is the amount of charge per unit area surrounding the point on the tire surface of the conductor.
• The greater the die curvature at any point of a conductor, the greater will be die accumulation of charge at that point.
• A lightning conductor or lightning arrester is a device used to protect a tall building from a thunderbolt.

 Electrostatics  Elementary Phenomena Of Electrostatics Useful Relations For Solving Numerical Problems

• If Q is the charge distributed uniformly on a surface of area A, then the surface density of charge is given by,
  ⇒ \(\sigma=\frac{Q}{A}\)
• For a spherical conductor,
  ⇒ \(\sigma=\frac{Q}{4 \pi r^2}\) (where r is the radius of the spherical conductor)

Unit 1 Electrostatics Chapter 1 Elementary Phenomena Of Electrostatics Exercise Multiple Choice Question And Answer

Question 1. A positively charged glass rod attracts a light-hanging body and thereafter repels it. The hanging body initially was

1. Positively charged
2. Earth-connected
3. Negatively charged
4. Uncharged

Answer: 3. Negatively charged

Question 2. Five balls marked by numbers from 1 to 5 are hung by different threads. It is seen that the pairs of balls (1,2), (2,4), and (4,1) attract each other. Again the pairs (2, 3) and (4, 5) repel each other. So the ball marked by 1 is

1. Positively charged
2. Negatively charged
3. Uncharged
4. Made by metal

Answer: 3. Uncharged

Question 3. The charges of the clouds responsible for lightning are produced due to

1. Conversion of raindrops into elections
2. The electric field of the earth
3. Creation of ions by the sun
4. Friction among the water drops.

Answer: 4. Friction among the water drops

Question 4. 106 electrons are given to a pith ball. The charge of the ball will be

1. 1.6 x 10^-13 C
2. 1.6 X 10^-25 C
3. 1.6 x 10^-19 C
4. None of the above

Answer: 3. 1.6 x 10^-19C

Question 5. A glass rod rubbed with silk becomes positively charged because

1. Protons are added to the glass rod
2. Protons are removed from the glass rod
3. Electrons are added to the glass rod
4. Electrons are removed from the glass rod

Answer: 4. Electrons are removed from the glass rod

Question 6. Two identical metallic spheres of the same mass are taken. Positive Q C charge is developed on one and an equal amount of negative charge is developed on the other. After charging

1. The two spheres will have equal mass
2. The sphere charged negatively will have a greater mass
3. The sphere charged positively will have a greater mass
4. The sphere charged negatively will have a smaller mass

Answer: 2. The sphere charged negatively will have a greater mass

Question 7. If a body is charged by rubbing, its weight

1. Does not change
2. Increases a little
3. Decreases a little
4. May increase or decrease a little

Answer: 4. May increase or decrease a little

Question 8. If the charge of an electron is 1.6 x 10^-19 C, which one of the following cannot be the charge of a body?

1. 3.2 x 10^-10 C
2. 4.8 x 10^-10 C
3. 5.6 x 10^-19 C
4. 1.6 x 10^-19 C

Answer: 3. 5.6 x 10^-19 C

Question 9. Electrostatic induction can be brought about

1. In conductors only
2. In insulators only
3. In bad conductors only
4. In both conductors and insulators

Answer: 4. In both conductors and insulators

Question 10. Two charged spheres attract each other with a force. They are touched by each other and thereafter brought back to their initial positions, the two spheres

1. Attract each other with a smaller force
2. Attract each other with a greater force
3. Repel each other with a smaller force
4. Repel each other with a greater force

Answer: 3. Repel each other with a smaller force

Question 11. The charge of a conductor resides on its outer surface of This statement is correct

1. In all cases
2. In the case of solid and hollow conductors
3. In the case of only spherical conductors
4. In the case of the conductors having no pointed ends

Answer: 1. In all cases

Question 12. With respect to a hollow sphere, a solid metallic sphere of the same radius will retain

1. More charge
2. Less charge
3. An equal amount of charge
4. None of the above

Answer: 3. Equal amount of charge

Question 13. The unit of surface density of charge in SI is

1. C
2. C m^-1
3. C.m^-2
4. C.m^-3

Answer: 3. C.m^-2

Question 14. The diameter of a hollow conducting sphere is 2 cm. The sphere has a 12.56 unit charge, the surface density of charge in its inner surface in CGS units

1. 0
2. 1
3. 12.56
4. 6.28

Answer: 1. 0

Unit 1 Electrostatics Chapter 1 Elementary Phenomena Of Electrostatics Very ShortAnswer Type Questions

Question 1. If a glass rod is rubbed with silk, what kind of charge is produced on the rod?
Answer: Positive

Question 2. If an ebonite rod is rubbed with flannel, what kind of charge is produced on the rod?
Answer: Negative

Question 3. A, B, and C are three charged bodies. If A and B repel each other and A attracts C, what will be the nature of the force acting between B and C?
Answer: Attractive

Question 4. Two point charges qx and q2 are such that qxq2< 0. What is the nature of the force acting between the two
Answer: Attractive

Question 5. What is the number of electrons in a 1C charge?
Answer: 6.25 x 10¹⁸

Question 6. If a body has a 1.5 x 107 number of excess electrons, what is the charge on the body?
Answer: -2.4 x 10^-12C

Question 7. If 1022 electrons are transferred from a metallic sphere, what will be the charge of the sphere?
Answer: 1600 C positive charge

Question 8. Write the name of a good conductor.
Answer: Silver

Question 9. Write the name of an insulator.
Answer: Glass

Question 10. What kind of charges are produced at the near end and at the far end of a conductor due to induction?
Answer: Opposite, similar

Question 11. Which is the appropriate season for performing experiments on statical electricity?
Answer: Winter

Question 12. Where does the charge of a conductor reside?
Answer: Outer surface

Question 13. What should be the shape of a conductor to retain its charge?
Answer: Spherical

Question 14. If a cubical conductor is charged, where will the surface density of charge be maximum?
Answer: At each of the vertices of the cube

Question 15. Is it safe or unsafe to remain inside a car at the time of
lightning?
Answer: Safe

Question 16. Why are two metal plates placed on the two inner sides of a gold-leaf electroscope?
Answer: To increase the divergence of the gold leaves

Unit 1 Electrostatics Chapter 1 Elementary Phenomena Of Electrostatics Fill In The Blanks

1. Repulsion is the surer test of electrification.

2. In SI, the amount of charge of an electron is 1.6 x 10^-19C

3. The charge on an electron is the smallest magnitude of charge in nature.

4. Dry air is a good insulator

5. Diamonds, ebonite, and bakelite are non-conductors of electricity.

6. If complete induction takes place, the amount of the inducing charge and that of the induced charge become equal.

7. In the case of electrostatic induction there should be a gap between the charged body and the uncharged body.

8. Induction precedes attraction.

9. The greater the curvature of a region of a conductor, the greater the accumulation of charge at that region.

10. In the case of a spherical conductor, the surface density of charge is equal everywhere.

11. Charge resides only on the outer surface of a conductor.

12. Charging of a gold-leaf electroscope by the process of conduction is not a good process.

13. If the charge of an experimental body and that of a gold-leaf electroscope are of the same nature, the divergence of the leaves of the gold-leaf electroscope will increase

14. To determine the nature of the charge of a body it is brought slowly to a charged electroscope from a distance.

Unit 1 Electrostatics Chapter 1 Elementary Phenomena Of Electrostatics Problem Set – 1

Question 1. A body has a -80μC charge. What is the number of excess electrons in the body?
Answer: 5 X 10¹⁴

Question 2. A conductor is clinked with 14.4 x 10¹⁹C. Determine the number of the deficit of electrons. In it, What will he the change In men of the conductor?
Answer: 9, 81 x 10^-13 kg mass (decrease)

Question 3. How much charge Is to be given to a sphere of radius 30cm so that its surface density of charge will be \(\frac{2}{\pi}\) CGS unit?
Answer: 7200 CGS unit

Question 4. The surface area of a body Is 25 cm2 and Its surface density of charge Is 5 CGS units. What Is the total charge on it?
Answer: 125 CGS unit

Question 5. What will be the surface density of charge of a sphere of radius 4 cm, If it is given 182 esu of charge?
Answer: 0.9 esu.cm^-2

Question 6. Two spheres of radii 4 cm and 8 cm have the same amount of charge. Determine the ratio of their surface densities of charge.
Answer: 4:1

Question 7. The diameter of a sphere is 2 cm. The sphere is hollow and conducting and is given a charge of 6.16 units. Determine the surface density of charge of the sphere on its outer and inner surfaces in the CGS unit.
Answer: 0.49 CGS unit, 0

Question 8. 64 equally charged water droplets, each of radius 4 mm, are combined to form a large water drop. Determine the ratio of the surface densities of charges in the two cases.
Answer: 4:1

Question 9. The ratio of the radii of two spheres is 5: 2 and that of their charges is 5:3. Determine the ratio of the surface densities of charge of the two spheres.
Answer: 4:15

Unit 1 Electrostatics Chapter 1 Elementary Phenomena Of Electrostatics Assertion-reason type

Direction: These questions have statement I and statement II. Of the four choices given below, choose the one that best describes the two statements.

1. Statement 1 Is true, statement 2 is true; statement 2 is a correct explanation for statement 1
2. Statement 1 is true, and statement 2 is true; statement 2 Is not a correct explanation for statement 1.
3. Statement 2 is true, statement 2 is false.
4. Statement 1 is false, and statement 2 is true

Question 1.

Statement 1: If there exists attraction between two bodies, both of them may not be charged.

Statement 2: A charged body can attract a neutral body.

Answer: 1. Statement 1 Is true, statement 2 is true; statement 2 is a correct explanation for statement 1

Question 2.

Statement 1: The Mass of a body decreases slightly when it is negatively charged.

Statement 2: Charging is due to the transfer of electrons

Answer:  4. Statement 1 is false, statement 2 Is true

Question 3.

Statement 1: Total charge on a body is the algebraic sum of charges located at different points of the body.

Statement 2: Electric charge is additive in nature.

Answer: 1. Statement 1 Is true, statement 2 is true; statement 2 is a correct explanation for statement 1

Question 4.

Statement 1: The tires of an aircraft are slightly conducting.

Statement 2: If a conductor is grounded, the extra charge induced on the conductor will flow to the ground.

Answer: 1. Statement 1 Is true, statement 2 is true; statement 2 is a correct explanation for statement 1

Question 5.

Statement 1: The top of a lightning conductor of a high building has sharp pointed ends.

Statement 2: The surface density of charge at sharp points is very high, resulting in the setting up of an electric wind.

Answer: 1. Statement 1 Is true, statement 2 is true; statement 2 is a correct explanation for statement 1

Unit 1 Electrostatics Chapter 1 Elementary Phenomena Of Electrostatics Multiple Choice Question And Answers

In this type of question, more than one option is correct.

Question 1. Two identical metallic spheres are given charges +q and -q respectively. Now

1. Both spheres have equal masses
2. The positively charged sphere has a mass smaller than that of the negatively charged sphere
3. The negatively charged sphere has a mass smaller than that of the positively charged sphere
4. The change in the masses depends on the magnitude of the charge transfer

Answer:

2. The positively charged sphere has a mass smaller than that of the negatively charged sphere

4. The change in the masses depends on the magnitude of charge transfer

Question 2. A spherical conductor A lies inside a hollow spherical conductor B. Charge Q₁ and Q₂ are given to A and B respectively.

1. Charge Q₁ will appear on the outer surface of A
2. Charge -Q₁ will appear on the inner surface of B
3. Charge Q₂ will appear on the outer surface of B
4. Charge Q₁ + Q₂ will appear on the outer surface of B

Answer:

1. Charge Q₁ will appear on the outer surface of A

2. Charge -Q₁ will appear on the inner surface of B

4. Charge Q₁ + Q₂ will appear on the outer surface of B

Question 3. The minimum quantity of charge available in nature is

1. 1C
2. 4.8 x 10^-13 C
3. 1.6 X 10^-1C
4. 4.8 X 10^-10 esu

Answer:

3. 1.6 X 10^-1C

4. 4.8 X 10^-10 esu

Question 4. A, B, and C are three concentric metallic shells. Shell A is the innermost and shell C is the outermost. A is given some charge.

1. The inner surfaces of B and C will have the same charge
2. The inner surfaces of B and C will have the same surface density of charge
3. The outer surfaces of A, B, and C will have the same charge
4. The outer surfaces of A, B, and C will have the same surface density of charge

Answer:

1. The inner surfaces of B and C will have the same charge

3. The outer surfaces of A, B, and C will have the same charge

Unit 1 Electrostatics Chapter 1 Elementary Phenomena Of Electrostatics Match The Columns

Answer: 1-B, 2-A, 3 D, 4-C

Answer: 1-D, 2-A, 3-B, 4-C

Answer: 1-C, 2-B, 3-E, 4-A

Comprehension type:

Read the following passage carefully and answer the questions at the end of it.

Question 1. A glass rod when rubbed- with silk acquires ‘a. change of +3.2 x 10^-7C.

1. The amount of charge on the silk is

1. 0
2. -3.2 x 10^-7C
3. +1.6 x 10^-7C
4. Not possible to calculate

Answer: 2. +1.6 x 10^-7C

2. Transfer of mass from glass rod to silk is

1. 9 x 10^-19 kg
2. 0
3. 18 x 10^-19 kg
4. None of the above

Answer: 3. 18 x 10^-19 kg

Question 2. A hollow spherical conductor of radius 3 cm is charged with a charge of 36πC.

1. The surface density of charge on the inner surface of the following conductor is

1. 1 C .cm^-2
2. 0
3. 104 C.cm^-2
4. ∞

Answer: 2. 0

2. The surface density of charge on ‘the outer surface of the hollow conductor is

1. 0
2. 1 C.m^-2
3. 104C.m^-2
4. ∞

Answer: 3. 104C. m^-2

3. If the hollow sphere is a solid one, the surface density of charge on its outer surface is

1. 0
2. 1C m^-2
3. ∞
4. 104 C.m^-2

Answer: 4. 104 C.m^-2

Integer answer type:

In this type, the answer to each of the questions is a single-digit integer ranging from 0 to 9:

Question 1. An insulated spherical conductor of radius lm is charged with a positive charge of 8πC. What is the surface density of charge on, the surface of the conductor in C.m^-2?
Answer: 2

Question 2. The surface area of a body is 10 cm? and its surface density of charge is 0.4 unit/cm². What is the total charge on it?
Answer: 4

Question 3. A body has a -0.8 x 10^-18C, charge. What is the number of excess electrons in the body?
Answer: 5

Question 4. A cube of the side has a 72 πC charge. What is the average surface density of charge of the cube in μC.cm^-2?
Answer: 3

Unit 1 Electrostatics Chapter 1 Elementary Phenomena Of Electrostatics Examination Archive With Solutions

Question 1. The number of electrons in 2 C of charge is

1. 12.5 X 10^-18
2. 12.5 X 10(-19)
3. 12.5 X 10¹⁸
4. 12.5 X 10¹⁹

Answer: 3.

Charge of one electron

= 1.6 X 10^-19 C

∴Number of electrons in’ 2 C charge

⇒ \(\frac{2}{1.6 \times 10^{-19}}\)

= 12.5 X 10¹⁸

The option 3 is correct.

Unit 1 Electrostatics Chapter 1 Elementary Phenomena Of Electrostatics Long Questions and Answers

Question 1. When an insulated charged spherical conductor is brought near a light, a small spherical conductor suspended with a silk fiber,

1. It quickly comes and sticks to the charged sphere.
2. Instantly move away and remain stationary in a deflected position to explain the phenomena.

Answer:

When the insulated charged spherical conductor is brought near the small spherical conductor suspended with a silk fiber, unlike charge is induced on the near end of the small sphere and like charge on the far end.

As the unlike induced charge is nearer to the inducing charge, the small conductor suspended by the silk fiber is attracted and comes in contact with the insulated charged sphere.

As soon as the small sphere comes in contact with the insulated charged sphere, the former gets the same charge on it. Due to mutual repulsion of like charges, the small sphere instantly moves away from the insulated charged sphere.

Up to a distance such that the tension in the fiber, weight of the suspended ball, and force given by the charged sphere keep the suspended ball in equilibrium

Question 2. State whether attraction can occur between two same kind of charges.
Answer:

We know the same charges repel each other. But, if a highly charged object is brought near to a little charged object quickly, even though it has the same charge, attraction may happen. This happens due to electrostatic induction.

When a highly charged object is brought near to a little charged object, an opposite charge may be induced in the latter. If the quantity of induced charge exceeds the former amount of charge present, attraction instead of repulsion may happen between them

Question 3. If a charged ebonite rod is made to touch the disc of a gold-leaf electroscope, the leaves diverge. Then the rod is removed from the disc and it is found that the divergence of the leaves decreases a little—explain.
Answer:

Suppose, the ebonite rod is charged positively. Ebonite is a non-conductor of electricity. So when the rod touches the disc of a gold-leaf electroscope, the charges of both of them existing at the place of contact spread on the disc and the leaves of the electroscope. So the leaves spread apart

Now the charges on the other parts of the ebonite rod do not move to the electroscope. So these charges induce a negative charge on the disc and a positive charge on the leaves. For this reason, the divergence of the leaves increases further.

Divergence of the leaves depends on both induction and conduction of charges. When the ebonite rod is removed, induction will be absent. So then the leaves will have positive charges due to conduction only. So, the divergence of the leaves decreases a little.

Question 4. Why are gold leaves used in a gold-leaf electroscope?
Answer:

The leaves of the gold-leaf electroscope should be thin and light. They diverge due to the repulsion of similar charges on them. The divergence of the leaves is more if they are light. Large divergence makes observation easier.

Now gold is a ductile metal, that can be beaten to produce very thin leaves. Moreover, gold does not react with air. For these reasons, gold leaves are used in an electroscope.

Question 5. Where will the surface charge density be maximum in a charged cubical conductor?
Answer:

The surface density of charge will be maximum at each of the cubical conductors.

Question 6. Why Is the metal box of the gold-leaf electroscope
earthed?
Answer:

During experiments with a gold-leaf electroscope, free charges exist on the outer surfaces of the tin (metallic) plates used in it. These free charges repel like charges on the leaves and decrease their divergence.

Because of the earthing of the electroscope, these free charges move to the ground. Hence the sensitivity of the device increases.

Question 7. Why is a drying agent kept inside a gold-leaf electroscope?
Answer:

Though dry air is a good insulator, moist air conducts electricity moderately. If a charged body finds itself in moist air, charges will leak and ultimately get discharged.

Hence, due to the presence of moist air inside a gold-leaf electroscope, the divergence of the leaves will gradually decrease. To avoid this unwanted condition, drying agents (hygroscopic substances) are kept inside a gold-leaf electroscope.

Question 8. Why should not a strongly charged body be brought very close to a gold-leaf electroscope?
Answer:

If a strongly charged body is brought very close to a gold-leaf electroscope, the amount of charge induced on the leaves of the electroscope will be very large. Then the divergence of the leaves may be large enough to detach them from the rod.

Question 9. Can an alternating static charge at one end of an isolated conductor, developed by an alternating current, be detected by a gold-leaf electroscope?
Answer:

Yes, a gold-leaf electroscope can defect alternating static charges. The divergence of the gold leaves in this electroscope depends only on the magnitude of charge in a test body, not on the nature of the charge, i.e., not on whether the charge is positive or negative.

The average of this magnitude of charge, over a complete period of the alternating source, has a finite value—this average is never zero. So, when the test body is brought near the disc of a gold-leaf electroscope, this average non-zero value of charge will be the input on the electroscope. The leaves will diverge accordingly

Question 10. Why it is not possible to electrify a metal rod by rubbing it while holding it with a bare hand?
Answer:

A metal rod as well as a human body is a good conductor of electricity. 5a by holding a metal rod with a bare hand if it is rubbed. the charge acquired due to friction flows through the body of the experimenter. That’s why the rod remains uncharged.

Question 11. How much is one safe while taking shelter in a vehicle for protection during lightning?
Answer:

The steel frame but not die tires of the topped vehicle can protect one from lightning in the condition that the fellow is not touching metal parts another hide. The lightning charge flows around die outside of there, creating a partial electric screen and protecting the inside

Question 12. How can charge be fully transferred from one spherical conductor to another?
Answer:

If a charged spherical conductor is placed inside a relatively bigger uncharged spherical conductor and a wire is connected between the two, then a charge will be completely transferred from the small spherical conductor to the bigger spherical conductor as the charge resides on die outer surface

Question 13. Why it is not safe to stand under a tree during lightning?
Answer:

Electricity seeks the path of least resistance. Due to the moisture inside, a tree is a much better conductor than its surroundings. The result, a tree provides a preferred path for lightning to easily reach the ground. This is why, one should not stand under a tree during lightning.

Question 14. A spherical shell of charge +Q has an outer radius r₂ and an Inner radius r₁. If a charge +q is placed at the center of the shell then what are the values of surface charge density of inner and outer surfaces?
Answer:

Here charge +Q resides on the outer surface. As the +q charge is placed at the center, a charge -q is induced on the inner surface and a charge +q is induced on the outer surface.

Hence, the total charge on the outer surface =Q+ q; and the surface charge density of the outer surface,

⇒ \(\sigma_2=\frac{Q+q}{4 \pi r_2^2}\)

The surface charge density of the inner surface,

⇒ \(\sigma_1=\frac{q}{4 \pi r_1^2}\)

Question 15. Charge of 3.2 X 10^-7 C Is obtained by rubbing a piece
Answer:

Since polythene has a negative charge, electrons are transferred from flannel to polythene.

Number of electrons transferred

= \(\frac{\text { total charge transferred }}{\text { charge of an electron }}\)

⇒ \(\frac{3.2 \times 10^{-7}}{1.6 \times 10^{-19}}\)

= \(2 \times 10^{12}\)

Here some mass is transferred from flannel to pohthene.

Mass transferred = number of electrons transferred x mass of an electron

= 2 x 10^-12 X 9.1 x 10^-31

= 1.82 x 10^-18 kg.

Question 16. Why does a spherical conductor retain its charge for a longer time in comparison with conductors of any other shape?
Answer:

Any conductor has a higher charge density at its sharp ‘ edge. Such an edge discharges faster than a spherical surface. For a spherical conductor, the distribution of charge over its surface is uniform.

As it has no sharp edge, it loses its charge very slowly due to the discharging action. So a spherical conductor can retain its charge for a longer time.

Question 17. Does a solid metallic sphere retain more charge than a hollow sphere of the same diameter?
Answer:

A solid sphere does not retain more charge than a hollow sphere of the same diameter, because charge resides on the outer surface of a conductor, and the outer surface areas of the two spheres are equal.

Question 18. When a conductor is charged, its charge resides on its outer surface. What is the reason?
Answer:

According to the property of a conductor, a charge may flow easily through it. Like charges repel each other and repulsion causes them to move as far apart as possible. An equilibrium is reached only when they move to the outer surface of the conductor.

Question 19. What is the minimum amount of charge acquired by a charged body? The charge of a body 5.6 X 10^-14C justifies it.
Answer:

The minimum charge acquired by a charged body is ±e = ±4.8 X 10^-10 esu of charge = +1.6 X 10^-19C. All charges present in nature are always an integral multiple of an electronic charge e.

The charge of a body is 5.6 X 10^-14 C this statement means that it has a deficit of electrons, and hence is positively charged.

The number of electron deficit

⇒ \(\frac{5.6 \times 10^{-14}}{1.6 \times 10^{-19}}\)

= 350000

Question 20. Can charge reside on the inner surface of a hollow conductor? Explain.
Answer:

A charged body is kept inside a hollow conductor in such a way that it does not touch the inside of the conductor. In this case, unlike charges are induced on the inner surface of the hollow conductor and like charges on its outer surface. It is clearly a special arrangement in which the inner surface of a hollow conductor is charged.

Question 21. An isolated metallic conductor is positively charged. Did its mass increase, decrease or remain the same? How will the mass of the conductor change if it is negatively charged?
Answer:

Since the conductor is positively charged, electrons have been removed from it So the mass of the conductor has decreased. Electrons are to be added to the conductor to charge it negatively. So the mass of the conductor will increase.

Question 22. A sensitive instrument is influenced by a strong electric field. Suggest a possible way to prevent the influence.
Answer:

The instrument should be enclosed in a cage made of metal wire. The cage would act as an electric screen; external electric fields would have no influence inside

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