WBCHSE Class 12 Physics Electric Potential Short Answer Questions

Unit 1 Electrostatics Chapter 3 Electric Potential Short Answer Questions

Question 1. A charge of 8 mC is located at the origin. Calculate the work done in taking a small charge of -2 x 10-9 C from a point. P(0, 0, 3 cm) to a point Q(0, 4 cm,0), via a point R (0, 6 cm, 9 cm).

Class 12 Physics Unit 1 Electrostatics Chapter 3 Electric Potential Question 1 initial and final position ofthe particle

Answer:

A charge of 8 mC is located at the origin.

Work done depends only on the linear distance between the initial and final position of the particle.

∴ Work done

⇒ \(=\frac{q_1 q_2}{4 \pi \epsilon_0}\left(\frac{1}{r_2}-\frac{1}{r_1}\right)\)

⇒ \(9 \times 10^9 \times\left(8 \times 10^{-3}\right) \times\left(-2 \times 10^{-9}\right) \times\left(\frac{1}{0.04}-\frac{1}{0.03}\right)\)

= 1.2 J

Question 2. In a hydrogen atom, the electron and proton are bound at a distance of about 0.53 A.

  1. Estimate the potential energy of the system in eV, taking the zero of the potential energy at an infinite separation of the electron from the proton.
  2. What is the minimum work to free the electron, given that its kinetic energy in the orbit is half the magnitude of potential energy obtained in (a)?
  3. What are the answers to (a) and (b) above if the zero of the potential energy is taken at 1.06 A separation?

Answer:

In a hydrogen atom, the electron and proton are bound at a distance of about 0.53 A.

1. Potential energy

⇒ \(E_p=\frac{1}{4 \pi \epsilon_0} \cdot \frac{q_1 q_2}{r}\)

⇒ \(\frac{\left(9 \times 10^9\right) \times\left(-1.6 \times 10^{-19}\right) \times\left(1.6 \times 10^{-19}\right)}{0.53 \times 10^{-10} \times 1.6 \times 10^{-19}} \mathrm{eV}\)

= -27.17 eV

2. Kinetic energy of the electron,

⇒ \(E_K=\frac{1}{2} E_P=13.585 \mathrm{eV}\)[Ek is always positive]

∴ The total energy of the electron,

E = Ep + EK = (-27.17 + 13.585)eV

=-13.585 eV

Minimum work done to free an electron from the atom

= [0-(-13.585)]eV

= 13.585 eV

Potential energy at a separation of 1.06 A

⇒\(E_P=\frac{\left(9 \times 10^9\right) \times\left(-1.6 \times 10^{-19}\right)\left(1.6 \times 10^{-19}\right)}{1.06 \times 10^{-10} \times 1.6 \times 10^{-19}} \mathrm{eV}\)

= -13.585 eV

and kinetic energy, \(E_K=\frac{E_P}{2}\)

= 6.792 eV

∴ The total energy of the electron

= [6.792 + (-13.585)]eV

=-6.792 eV

∴ Minimum work done to free the electron from the atom

= [0-(-6.792)]eV

= 6.792 eV

Question 3. Two charges -q and +q are located at points (0, 0, -a) and (0, 0, a) respectively.

  1. What is the electrostatic potential at the points (0, 0, z) and (x, y, 0)?
  2. Obtain the dependence of potential on the distance f of a point from the origin when r >> a.
  3. How much work is done in moving a small test charge from the point (5, 0, 0) to (-7, 0, 0) along the x-axis? Does the answer change if the path of the test charge between the same points is not along the x-axis?

Answer:

Two charges -q and +q are located at points (0, 0, -a) and (0, 0, a) respectively.

1. The point (0, 0, z) lies on the axis of the dipole.

∴ Potential on a point on the axis of the dipole,

⇒ \(V= \pm \frac{1}{4 \pi \epsilon_0} \cdot \frac{p}{x^2-a^2}\)

where p = dipole moment = q.2a

and x²-a² = z²-a²

∴ \(V= \pm \frac{1}{4 \pi \epsilon_0} \cdot \frac{2 q a}{z^2-a^2}\)

The point (x,y,0) is on the line perpendicular to m, x the axis of the dipole, the potential at that point is zero.

2. Let P be the point at a distance r from the dipole. OP makes an angle θ with the axis of the dipole. The length of the dipole = 2a.

r2 = BP == CP = r – a cosθ

r1 = AP == DP = r + a cosθ

Potential at P,

⇒ \(V=V_1+V_2=\frac{q}{4 \pi \epsilon_0}\left(\frac{1}{r_2}-\frac{1}{r_1}\right)\)

⇒ \(\frac{q}{4 \pi \epsilon_0}\left(\frac{1}{r-a \cos \theta}-\frac{1}{r+a \cos \theta}\right)\)

Class 12 Physics Unit 1 Electrostatics Chapter 3 Electric Potential Question 3 The length of the dipole

∴ \(V=\frac{q}{4 \pi \epsilon_0} \frac{2 a \cos \theta}{r^2-a^2 \cos ^2 \theta}\)

⇒ \(\frac{2 a q \cos \theta}{4 \pi \epsilon_0}\left\{\frac{1}{r^2}\left(1-\frac{a^2}{r^2} \cos ^2 \theta\right)^{-1}\right\}\)

⇒ \(\approx \frac{p \cos \theta}{4 \pi \epsilon_0 r^2}\) ∵\( \frac{a^2}{r^2} \ll 1\)

∴ \(V \propto \frac{1}{r^2}\)

3. The work done is zero because the potential along the x-axis is zero.

The work done is always zero between these two points because work done does not depend on the path taken, it only depends on the position of the endpoints.

Question 4. A charge array is known as an electric quadrupole. For a point on the axis of the quadrupole, obtain the dependence of potential on r for \(\frac{r}{a}\) >> 1, and contrast your results with that due to an electric dipole and an electric monopole (i,e., single charge).

Class 12 Physics Unit 1 Electrostatics Chapter 3 Electric Potential Question 4 electric depole

Answer:

A charge array is known as an electric quadrupole.

Potential at P,

⇒ \(V_1=\frac{1}{4 \pi \epsilon_0}\left[\frac{q}{r-a}-\frac{2 q}{r}+\frac{q}{r+a}\right]\)

⇒ \(\frac{q}{4 \pi \epsilon_0} \cdot \frac{2 a^2}{r\left(r^2-a^2\right)} \approx \frac{q}{4 \pi \epsilon_0} \cdot \frac{2 a^2}{r^3}\)

∴ \(V_1 \propto \frac{1}{r^3}\)

Potential due to dipole,

⇒ \(V_2=\frac{1}{4 \pi \epsilon_0} \cdot \frac{q}{r^2} \quad\)

∴ \(V_2 \propto \frac{1}{r^2}\)

Potential due to monopole,

⇒ \(V_3=\frac{1}{4 \pi \epsilon_0} \cdot \frac{q}{r} \quad\)

∴ \(V_3 \propto \frac{1}{r}\)

Question 5. If Coulomb’s law involved \(\frac{1}{r^3}\) dependence (instead of \(\frac{1}{r^2}\) ), would Gauss’ law be still true?
Answer:

Gauss’ law and Coulomb’s law depend on each other. So if Coulomb’s law changes its form, Gauss’ law would also change its present form.

Question 6. Describe schematically the equipotential surfaces corresponding to

  1. A constant electric field in the z-direction,
  2. A field that uniformly increases in magnitude but remains in a constant (say, z ) direction,
  3. A single positive charge at the origin and
  4. A uniform grid consisting of long equally spaced parallel charged wires in a plane.

Answer:

1.  A plane parallel to xy-plane.

2. A plane parallel to the xy-plane. These planes will come closer as the field intensity increases.

3. Concentric spheres with the charge at the centre.

4. Nearer the grid the equipotential surfaces will change their shape periodically slowly becoming planer and parallel to the grid at far-off distances from the grid.

Question 7. The top of the atmosphere is at about, 440 kV with respect to the surface of the earth, corresponding, to an electric field that decreases with altitude. Near the surface of the earth, the field is about 100 V.m-1. Why then we do not get an electric shock as we step out of our house into the open? (Assume the house to be a still cage so there is no field inside.)
Answer:

The top of the atmosphere is at about, 440 kV with respect to the surface of the earth, corresponding, to an electric field that decreases with altitude. Near the surface of the earth, the field is about 100 V.m-1.

Our body and the earth’s surface become equipotential i.e., there is no potential difference between the earth and our body. Hence no current flows through our body and therefore we do not experience a shock

Question 8. Write down the name of the physical quantity whose unit is joule/coulomb.
Answer: Electric potential

Question 9. Determine how much work is to be done to move a 10 C positive charge 1 m along the y-axis in a uniform electric field ⇒ \(\vec{E}=5(\hat{i}+\hat{j}) \mathrm{V} \cdot \mathrm{m}^{-1}\)
Answer:

Force acting on a charge q placed in a uniform electric field \(\vec{E}\) is given by,

⇒ \(\vec{F}=q \vec{E}=10 \times 5(\hat{i}+\hat{j}) \mathrm{N}\)

Again, displacement along y-axis, \(\vec{s}=1 \hat{j} \mathrm{~m}\)

∴ \(\vec{F} \cdot \vec{s}=50(\hat{i}+\hat{j}) \cdot \hat{j}\)

Question 10. The electric field strength at a point in an electric field is zero. Is the electric potential also zero at that point? Answer with reason.
Answer:

The electric field strength at a point in an electric field is zero.

If the intensity of the electric field be E and the potential be V, thenrthear&atitinjbetween them is, E = \(\frac{dV}{dx}\)

So, if E = 0 at any point, we have \(\frac{dV}{dx}\) = 0

or, V = constant

Thus, the potential has a constant value, not necessarily zero, around that point.

Question 11. The electric potential at a point (x, y, z) is given by V = -x2y- xz3 + 4 . Find the intensity of electric field \(\vec{E}\) at that point
Answer:

The electric potential at a point (x, y, z) is given by V = -x2y- xz3 + 4 .

⇒ \(E_x=-\frac{d V}{d x}=2 x y+z^3, E_y=-\frac{d V}{d y}=x^2\)

⇒ \(E_z=-\frac{d V}{d z}=3 x z^2\)

∴ \(\vec{E}=E_x \hat{i}+E_y \hat{j}+E_z \hat{k}\)

= \(\left(2 x y+z^3\right) \hat{i}+x^2 \hat{j}+3 x z^2 k\)

Question 12. A test charge q is moved without acceleration along the path from A to B and from D to C in electric field E.

Class 12 Physics Unit 1 Electrostatics Chapter 3 Electric Potential A test charge q is moved without acceleration

1. Calculate the potential difference between A and C

2. At which point (of the two) is the electric potential more and why?

Answer:

1. \(V_A-V_C=-\int_2^6 \vec{E} \cdot d \vec{x}=-\int_2^6 E d x \cos 0^{\circ}=-\int_2^6 E d x\)

⇒ \(-E \int_2^6 d x=-4 E\)

∴ \(V_C-V_A=4 E\)

2. In the direction of the electric field potential decreases. So, point C is a a higher potential.

Question 13. Draw the equipotential surfaces due to an electric dipole. Locate the points where the potential due to the dipole is zero
Answer:

Some equipotential surfaces of a dipole

Class 12 Physics Unit 1 Electrostatics Chapter 3 Electric Potential the equipotential surfaces

All the points on the equatorial plane of a dipole, being equidistant from +q and -q, have zero potential.

Question 14. Obtain an expression for the work done to dissociate the system of three charges placed at the vertices of an equilateral triangle of side a as shown below.
Answer:

Work done to keep the system bound Is

⇒ \(W=\frac{1}{4 \pi \epsilon_0}\left[\frac{q_1 q_2}{a}+\frac{q_2 q_3}{a}+\frac{q_1 q_3}{a}\right]\)

⇒ \(\frac{1}{4 \pi \epsilon_0}\left[\frac{q(-4 q)}{a}+\frac{(-4 q) 2 q}{a}+\frac{q(2 q)}{a}\right]\)

⇒ \(\frac{1}{4 \pi \epsilon_0}\left[\frac{-4 q^2}{a}+\frac{8 q^2}{a}+\frac{2 q^2}{a}\right]\)

⇒ \(\frac{q^2}{4 \pi \epsilon_0 a}(-4-8+2)=\frac{-10 \cdot q^2}{4 \pi \epsilon_0 a}\)

Therefore, the work done to dissociate the system is,

⇒ \(W_d=-W=\frac{10 q^2}{4 \pi \epsilon_0 a}\)

Question 15. An infinitely large thin plane sheet has a uniform surface charge density +σ. Obtain the expression for the amount of work done in bringing a point charge q from infinity to a point, distant r, in front of the charged plane sheet.
Answer:

An infinitely large thin plane sheet has a uniform surface charge density +σ.

Electric field due to the infinitely large plane sheet = \(\frac{\sigma}{2 \epsilon_0}\)

Work done to bring a point charge q from infinity to a point at a distance r =q(Vr-Vco)

⇒ \(-q \int_{\infty}^r E d r=-q \int_{\infty}^r \frac{\sigma}{2 \epsilon_0} d r=\infty\)

Question Four point charges Q, q, Q and q are placed at the comers of a square of side.

Class 12 Physics Unit 1 Electrostatics Chapter 3 Electric Potential placed at the comers

Find the

1. Resultant electric force on a charge Q, and

2. Potential energy of this system

Answer:

The electric force on charge Q due to charge q

Class 12 Physics Unit 1 Electrostatics Chapter 3 Electric Potential Electric force on charge

⇒ \(F_q=\frac{1}{4 \pi \epsilon_0} \times \frac{q Q}{a^2}\)

The electric force on charge Q due to the other charge Q

⇒ \(F_Q=\frac{1}{4 \pi \epsilon_0} \times \frac{Q^2}{(a \sqrt{2})^2}=\frac{1}{4 \pi \epsilon_0} \frac{Q^2}{2 a^2}\)

The net force on charge Q

⇒ \(F_{\text {net }}=F_Q+\sqrt{F_q^2+F_q^2}=F_Q+F_q \sqrt{2}\)

⇒ \(\frac{1}{4 \pi \epsilon_0} \times \frac{Q^2}{2 a^2}+\frac{1}{4 \pi \epsilon_0} \times \frac{q Q}{a^2} \sqrt{2}\)

⇒ \(\frac{Q^2}{4 \pi \epsilon_0 a^2}\left[\frac{Q}{2}+\sqrt{2} q\right]\) along diagonal

2. Potential energy of the system,

⇒ \(\dot{U}=U_{q Q}+U_{Q q}+U_{q Q}+U_{Q q}+U_{q q}+U_{Q Q}\)

⇒ \(4 U_{q Q}+U_{q q}+U_{Q Q}\)

⇒ \(\frac{4 q Q}{4 \pi \epsilon_0 a}+\frac{q^2}{4 \pi \epsilon_0(\sqrt{2} a)}+\frac{Q^2}{4 \pi \epsilon_0(\sqrt{2} a)}\)

⇒ \(\frac{1}{4 \pi \epsilon_0 a}\left[4 q Q+\frac{q^2}{\sqrt{2}}+\frac{Q^2}{\sqrt{2}}\right]\)

Question 16.

1. Three point charges q, -4q, and 2q are placed at the vertices of an equilateral triangle ABC of side. Obtain the expression for the magnitude of the resultant electric force acting on the charge q.

2. Hind out this amount of work done to separate the charges at Infinite distance.

Class 12 Physics Unit 1 Electrostatics Chapter 3 Electric Potential separate the charges at infinite distance

Answer:

1.

⇒ \(F_{A B}=\frac{1}{4 \pi \epsilon_0} \frac{q(4 q)}{l^2}=\frac{1}{4 \pi \epsilon_0} \frac{4 q^2}{l^2}\)

⇒ \(F_{A C}=\frac{1}{4 \pi \epsilon_0} \frac{q(2 q)}{l^2}=\frac{1}{4 \pi \epsilon_0} \frac{2 q^2}{l^2}\)

Class 12 Physics Unit 1 Electrostatics Chapter 3 Electric Potential equilateral triangle ABC

The angle between forces \(\vec{F}_{A B} \text { and } \vec{F}_{A C}\) is 120°.

The magnitude of the resultant force,

⇒ \(F=\sqrt{F_{A B}^2+F_{A C}^2+2 F_{A B} F_{A C} \cos 120^{\circ}}\)

⇒ \(\frac{1}{4 \pi \epsilon_0}\left(\frac{q^2}{l^2}\right) \sqrt{(4)^2+(2)^2+2 \times 4 \times 2 \times\left(\frac{-1}{2}\right)}\)

⇒ \(\frac{1}{4 \pi \epsilon_0} \frac{q^2}{l^2}(2 \sqrt{3})\)

2. Amount of work done = change in potential energy of the system

= Uf – Ui

= 0 – (UAB + UBC + UCA)

⇒ \(\frac{-1}{4 \pi \epsilon_0 l}[q(-4 q)+(-4 q)(2 q)+(q)(2 q)]\)

⇒ \(\frac{10 q^2}{4 \pi \epsilon_0 l}\)

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