WBCHSE Class 12 Physics Electric Potential  Long Question and Answers

Unit 1 Electrostatics Chapter 3 Electric Potential Long Question and Answers

Question 1. If the intensity at a point in an electric field is zero, will the electric potential be also zero at that point? If the electric potential be zero at a point, will the Intensity be also zero at that point?
Answer:

If the intensity of the electric field is E and the potential is V, then the relation between 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. Now, for example, we consider an electric dipole (q,-q) placed at AB along the x-axis.

The variation of the potential V with the dis- Answer: We know that charges of a conductor reside on its outer stance x is shown in the figure. At the surface.

Placing the smaller conductor inside the bigger one and mid-point O, clearly, V = 0; but the connecting them with a wire, we can prepare a single conductor.

The slope of the V-x curve is not zero, i.e., \(\frac{dV}{dx}\) not = 0. As E = – \(\frac{dV}{dx}\), we Thus charge from the smaller conductor will then flow to the can conclude that E not = 0 at O.

Therefore, if the potential is zero at a point, but it is not a constant in the immediate neighbourhood of that point, the electric field intensity is non-zero there.

In particular, the potential due to a dipole at any point on the perpendicular bisector is zero, but the field at that point is certainly non-zero being directed opposite to the direction of the dipole moment.

Question 2. Two hollow conductors are charged positively. The potential of the smaller conductor is 50 V and that of the larger conductor is 100 V. How are these two conductors to be placed so that when connected by a wire, charges will flow from the smaller conductor to the larger one?
Answer:

Two hollow conductors are charged positively. The potential of the smaller conductor is 50 V and that of the larger conductor is 100 V.

We know that the charges of a conductor reside on its outer surface. Placing the smaller conductor inside the bigger one and connecting them with a wire, we can prepare a single conductor. Thus charges from the smaller conductor will then flow to the outer surface of the larger one.

Question 3. Explain the variations of potential and field Intensity with distance due to a hollow charged spherical conductor (radius a and charge q ), both inside and outside the sphere, with a graph.
Answer:

Electric potential: Potential at all points inside a hollow charged spherical conductor and on its surface are equal. Here this potential is, V = ⇒ \(\frac{q}{a}\). So, from the centre to the surface of the tire conductor, the potential is constant and it has been shown by the line AB, parallel to the x-axis.

If the distance of an external point from the centre of the conductor be r, then electric potential at that point is given by, So as r increases, V decreases. This decrease has been shown by the curved line BC. The inverse variation is illustrated by the moderate slope of the curve.

Electric intensity: As the potential at all points inside a hollow charged spherical conductor is constant, intensity at those points, i.e., inside the hollow spherical conductor is zero.

As the radius of the conductor is an intensity at a point on its surface, \(E=\frac{q}{a^2}\). If r is the distance of an external point from the centre of the sphere, the intensity at that point will be \(E=\frac{q}{a^2}\)

So, intensity on the surface of the conductor is maximum and it decreases as we move away from the conductor. This variation has been shown by the curved line AB.

The inverse square variation has been shown by the steeper slope, as compared to the V-r curve.

Question 4. The potential difference between two conductors is very large. What will happen under the following three conditions?

1. The conductors are connected by a metallic wire,
2. Both positive and negative ions are present in the air medium in between the conductors,
3. The conductors are placed in a vacuum.

Answer:

1. Charge will continue to flow till the potentials of the two conductors become equal.

2. Positive ions will move to the conductor having lower potential and negative ions will move to the conductor having higher potential.

3. Charges on the conductors will not move and the potential difference between them will be maintained.

Question 5. Give an Example of a line In the electric field of an electric dipole along which a positive charge may be moved without any work being done.
Answer:

There are numerous equipotential surfaces in the electric field of a dipole. Any movement of a charge, on any of these surfaces, would involve no work. For Example, the plane perpendicular to the dipole and passing through its mid-point is an equipotential, actually zero-potential surface. So if a positive charge is moved along any line on this plane, no work is to be done.

Question 6. Can two different equipotential surfaces intersect each other?
Answer:

If two different equipotential surfaces intersected each other, the potential would have two different values at the point of intersection. Moreover, there would be two normals on the two surfaces at the point of intersection. That would mean two elec-/ trie fields in two different directions at the same point. These situations are absurd. So two different equipotential surfaces cannot intersect each other.

So, the intensity on the surface of the conductor is maximum and it decreases as we move away from the conductor. This variation has been shown by the curved line AB. The inverse square variation has been shown by the steeper slope, as compared to the V-r curve.

Question 7. Two pairs of planes A, B and A’, B’ are kept in two uniform electric fields. The potential dx difference between the planes In the two cases are equal. Which pair of planes Is situated in a stronger electric field?

Answer:

Since E = – \(\frac{dV}{dx}\); therefore, dx = – \(\frac{dV}{E}\)

So it is seen that dx will be smaller if E is stronger.

Therefore, planes A and B are in a stronger electric field.

Question 8. A positively charged and a negatively charged body are connected to the earth separately. What will be their potential before and after the connection?
Answer:

A positively charged and a negatively charged body are connected to the earth separately.

Before connection with the earth, the positively charged body will have positive potential and the negatively charged body will have negative potential. But after connection with the earth both the bodies will have zero potential.

Question 9. When the body Is connected with the earth, electrons are found to move from the earth to the body. What Is Your Idea about the nature of the charge of the body?
Answer:

When the body Is connected with the earth, electrons are found to move from the earth to the body.

Since electrons are moving from the earth to the body, Its potential Is positive. As the body has positive potential, it Is positively charged.

Question 10. In a vacuum, equal charges of the same nature are placed at the vertices of an equilateral triangle. What will be the electric field Intensity and potential at the centroid of the triangle
Answer:

In a vacuum, equal charges of the same nature are placed at the vertices of an equilateral triangle.

Let three charges, each equal to +q, be placed at the vertices of the equilateral triangle. The length of the side of the triangle is r.

We know that the centroid of the triangle G divides the median of the n triangle In (the ratio 2:1).

Height of the equilateral triangle

⇒ \(\frac{\sqrt{3}}{2} r\)

So, the potential at G Is,

⇒ \(V=\frac{q}{4 \pi \epsilon_0\left(\frac{2}{3} \cdot \frac{\sqrt{3} r}{2}\right)} \times 3\)

= \(\frac{3 q}{4 \pi \epsilon_0 \cdot \frac{r}{\sqrt{3}}}\)

= \(\frac{1}{4 \pi \epsilon_0} \frac{3 \sqrt{3} q}{r}\)

If E be the electric field intensity at G due to each charge +q (HI and if they are resolved into two perpendicular components, we see that the two components, ‘Ecos30° are equal and opposite balance each other.

Again 2sin30° \(\left(=\frac{2 E}{2}=E\right)\) and F being equal and opposite, balance each other. So the electric field Intensity at the centroid of the angle is zero.

Question 11. Draw three equipotential surfaces corresponding to a field that uniformly Increases In magnitude but remains constant along the z-direction. How are these surfaces different from those of u constant electric field along the z-direction?
Answer:

The separation between the equipotential surfaces will gradually decrease along the direction of the Increment of the electric field (in this case z-axis).

In the case of a constant uniform electric field, the equipotential surfaces are equally spaced.

Question 12. A unit positive charge is moving along PQR in an electric field of intensity E. What is the potential difference between the points P and R?

Answer:

A unit positive charge is moving along PQR in an electric field of intensity E.

The work done to move a test charge from one point to another is Independent of the path followed.

⇒ \(V_{P R}=W_{P R}=-\int \vec{E} \cdot d \vec{l}=-\int_P^R E d l \cos 180^{\circ}\)

or, V_VR = Er

i.e., the potential difference is r times of the electric field intensity.

Question 13. The variation of electric potential V for two charges Q₁ and Q₂ (where r Is the distance from a point charge).

1. What are the natures of the charges Q₁ and Q₂?
2. Which of the two charges Is greater In magnitude? Justify

Answer:

1. Electric potential due to a point charge Q is given by

⇒ \(V=\frac{1}{4 \pi \epsilon_0} Q\)

V is positive if Q is positive and V is negative if Q is negative.

Therefore, Q₁ is a positive charge and Q₂ is a negative charge.

2. Since V₂ > V₁, so charge Q₂ Is greater In magnitude as compared to charge Q₁

Question 14. Two electric charges q and -2q ore placed 6 m apart on a horizontal plane. Find the locus of any point on this plane where the potential has a value zero.
Answer:

Two electric charges q and -2q ore placed 6 m apart on a horizontal plane.

The potential at P due to the charges q and -2q,

⇒ \(V=\frac{1}{4 \pi \epsilon_0}\left(\frac{q}{r_1}-\frac{2 q}{r_2}\right)\)

But V = 0 (given)

∴ \(\frac{1}{r_1}=\frac{2}{r_2} \quad\)

or, \(r_2=2 r_1 \quad\)

or, \(r_2^2=4 r_1^2\)

or, y² + (6-x)² = 4(x²+y²)

or, y² + 36 + x²- I2x = 4x² + 4y²

or, 3x² + 12x + 3y² = 36

or, x² + 4x+ y² = 12

or, (x + 2)² + y² = 42

or, [x-(-2)]² + (y-0)² = 42

Therefore, the locus of point P is a circle whose centre is at (-2, 0) and whose radius is 4 m.

Question 15. A charge Q Is uniformly distributed over a long rod YA AB of length L as shown In the figure. Determine the electric potential at the point O lying at a distance L from the end A.

Answer:

A charge Q Is uniformly distributed over a long rod YA AB of length L as shown In the figure.

Potential at O,

⇒ \(V=\frac{1}{4 \pi \epsilon_0} \int_0^L \frac{d q}{L+x}\)

⇒ \(\left.\frac{1}{4 \pi \epsilon_0} \int_0^L \frac{\lambda d x}{L+x} \text { [linear charge density, } \lambda=\frac{d q}{d x}=\frac{Q}{L}\right]\)

⇒ \(\frac{\lambda}{4 \pi \epsilon_0} \int_0^L \frac{d x}{L+x}\)

= \(\frac{1}{4 \pi \epsilon_0} \cdot \frac{Q}{L} \ln \frac{L+L}{L}\)

= \(\frac{Q \ln 2}{4 \pi \epsilon_0 L}\)

Question 16. A conducting of radius a, thickness l(<<a) has a potential V. Now the bubble transforms into a droplet. Find the potential on the surface of the droplet. Answer t Let q be the charge on the bubble. The potential on the surface of the bubble, V =
Answer:

A conducting of radius a, thickness l(<<a) has a potential V. Now the bubble transforms into a droplet.

Let q be the charge on the bubble

The potential on the surface of the bubble, \(V=\frac{1}{4 \pi c_0} \frac{q}{a}\)

∴ \(q=\frac{V a}{k}\left[\text { where } k=\frac{1}{4 \pi \epsilon_0}\right]\)

Let R be the radius of the droplet.

By comparing their volumes, we get

⇒ \(\left(4 \pi a^2\right) t=\frac{4}{3} \pi R^3\)

∴ \(R=\left(3 a^2 t\right)^{1 / 3}\)

Now potential on the surface of the droplet becomes

⇒ \(V_1=\frac{k q}{R}\)

or, \(V_1=k \frac{\frac{V a}{k}}{\left(3 a^2 t\right)^{1 / 3}}\) [from equation (1)]

⇒ \(V_1=V\left(\frac{a}{3 t}\right)^{1 / 3}\)

Question 17. Some equipotential surfaces. What can you say about the magnitude and the direction of the electric field A 30 cm Intensity?

Answer:

The electric field is always perpendicular to the equipotential surface. Also, potential decreases along the direction of electric field intensity.

The electric field makes an angle of 120° with the X-axis.

The magnitude of the electric field along X -axis,

⇒ \(E \cos 120^{\circ}=-\frac{(20-10)}{(20-10) \times 10^{-2}}\)

or, \(-E \cdot \frac{1}{2}=-\frac{10}{0.10}\)

E = 200 V/m

The direction of the electric field is radially outward.

Here, rV = constant = 6 V.m hypotenuse,

Thus potential at any distance r from the centre,

V(r) = \(\frac{6}{r}\)

Hence,\(E=-\frac{d V}{d r}=\frac{6}{r^2} \mathrm{~V} / \mathrm{m}\)

Question 18. The radii of two concentric metal spheres are a and b (b>a). The outer sphere Is charged with a charge q. If the Inner sphere is connected to the earth, what will be its charge?
Answer:

The radii of two concentric metal spheres are a and b (b>a). The outer sphere Is charged with a charge q.

Let the charge of the inner sphere be Q.

Given that the charge of the outer sphere is q.

Radii of the inner and outer spheres are a and b respectively.

As the inner sphere is earthed, the potential is zero.

Thus tire net potential is

⇒ \(\frac{1}{4 \pi \epsilon_0} \frac{Q}{a}+\frac{1}{4 \pi \epsilon_0} \frac{q^{\circ}}{b}=0\) [∵ \(\frac{1}{4 \pi \epsilon_0}\) is constant, so it is not equal to zero.]

∴ \(\frac{Q}{a}=-\frac{q}{b}\)

or, \(Q=-q \frac{a}{b}\)

Therefore charge of the inner sphere is \(-q \frac{a}{b}\).

Question 19. +Q, +q, +q charge θ arc placed on the vertices of an Isosceles right-angled triangle. If a the electric potential energy of the system of charges Is zero, what will +q be the value of Q?

Answer:

Electric potential energy due to Q.and q through the hypotenuse,

⇒ \(U_1=\frac{1}{4 \pi \epsilon_0}\left[\frac{Q \cdot q}{\sqrt{2 a}}\right]\)

Similarly, potential energy due to Q and q through the perpendicular

⇒ \(U_2=\frac{1}{4 \pi \epsilon_0}\left[\frac{Q \cdot q}{a}\right]\)

and potential energy due to q and q through the base,

⇒ \(U_3=\frac{1}{4 \pi \epsilon_0}\left[\frac{q \cdot q}{a}\right]\)

Thus the net potential energy,

⇒ \(U=U_1+U_2+U_3\)

= \(\frac{1}{4 \pi \epsilon_0}\left[\frac{Q q}{\sqrt{2} a}+\frac{Q q}{a}+\frac{q q}{a}\right]=0\)

or, \(\frac{Q q}{\sqrt{2} a}[1+\sqrt{2}]=-\frac{q q}{a}\)

or, \(Q=-\frac{\sqrt{2} q}{1+\sqrt{2}}\)

∴ \(Q=-\frac{2 q}{\sqrt{2}+2}\)

Question 20. Discuss the variation of electric potential due to positive and negative point charges with distance from a charge.
Answer:

Electric potential at a distance r from a charge +q ,

⇒ \(V=\frac{+q}{4 \pi \epsilon_0 r}\)

Electric potential at a distance r from a charge -q, \(V=\frac{-q}{4 \pi \epsilon_0 r}\)

The variation of V with r. From the nature of the curves it can be said that the electric potential gradually decreases as the distance from charge +q increases and it gradually increases as the distance from charge -q increases.