Electrostatics - NEET Physics Questions
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Electrostatics

Practice Questions:
Question 41: easy

Two point charges exert a force \(F_0\) on each other when placed in vacuum. Now the charges are increased to four times, separation between them is doubled and the system is placed is an insulating medium. Now they experience the same force. What should be the dielectric constant of the medium?

1. 3
2. 4
3. 2
4. 5
View Answer

Initial force in vacuum \(F_0 = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r^2}\) . New force in medium \(F' = \frac{1}{4\pi\epsilon_0 K} \frac{(4q_1)(4q_2)}{(2r)^2} = \frac{16}{4K} \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r^2}\) . Since \(F' = F_0\), we have \(frac{4}{K} = 1\), so \(K = 4\).

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Question 42: easy

A regular polygon has \(n\) sides each of length \(l\). Each corner of the polygon is at a distance \(r\) from the centre. Identical charges each equal to \(q\) are placed at all the corners except one. The magnitude of electric field at the centre of the polygon is

1. \(\frac{kq(n-1)}{r^2}\)
2. \(\frac{2kq}{r^2}\)
3. \(\frac{kqn}{r^2}\)
4. \(\frac{kq}{r^2}\)
View Answer

A completely symmetric distribution of \(n\) charges has a net field of zero at the center. Removing one charge is equivalent to superimposing a charge of \(-q\) at the empty corner on an otherwise full polygon, which produces a field of magnitude \(\frac{kq}{r^2}\).

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Question 43: easy

An electric dipole of dipole moment \(\vec{P}\) is initially aligned along a uniform electric field \(\vec{E}\). The work done in rotating it by \(90^\circ\) will be

1. \(-PE\)
2. Zero
3. \(PE\)
4. \(\frac{PE}{2}\)
View Answer

The work done in rotating a dipole in a uniform electric field is given by \(W = PE(cos\theta_1 - cos\theta_2)\). Since it is rotated from \(\theta_1 = 0^\circ\) to \(\theta_2 = 90^\circ\), \(W = PE(cos 0^\circ - cos 90^\circ) = PE\).

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Question 44: easy

Column A contains some charge distribution and column B contains corresponding electric flux or field. Match the columns and choose the correct option.

Column A:
A. Charge outside a closed gaussian surface
B. Charge \(q\) inside a closed gaussian surface
C. Infinite plane sheet of charge
D. Field outside a charged conducting sphere

Column B:
(P) \(\oint \vec{E} \cdot d\vec{A} = \frac{q}{epsilon_0}\)
(Q) \(E = \frac{\sigma}{2\epsilon_0}\)
(R) \(E = \frac{KQ}{r^2}\)
(S) Net flux is zero

 

1. A(S), B(P), C(R), D(Q)
2. A(S), B(P), C(Q), D(R)
3. A(P), B(R), C(Q), D(S)
4. A(P), B(Q), C(S), D(R)
View Answer

A charge outside a closed surface contributes zero net flux (A-S). Gauss's law states that for an enclosed charge \(q\), the net flux is \(\frac{q}{\epsilon_0}\) (B-P). The electric field due to an infinite plane sheet is \(E = \frac{\sigma}{2\epsilon_0}\) (C-Q). For a charged conducting sphere, the external field is \(E = \frac{KQ}{r^2}\) (D-R).

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Question 45: easy

Two charged spherical conductors of radius \(R_1\) and \(R_2\) are connected by a wire. Then the ratio of final surface charge densities of the spheres \(\sigma_1 / \sigma_2\) is

1. \(\frac{R_1^2}{R_2^2}\)
2. \(\frac{R_1}{R_2}\)
3. \(\frac{R_2}{R_1}\)
4. \(\sqrt{\frac{R_1}{R_2}}\)
View Answer

When connected, their electric potentials become equal, i.e., \(V_1 = V_2\). Since \(V = \frac{\sigma R}{\varepsilon_0}\), we get \(\sigma_1 R_1 = \sigma_2 R_2\), which gives \(\frac{\sigma_1}{\sigma_2} = \frac{R_2}{R_1}\).

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Question 46: easy

Twenty seven drops of same size are charged at \(220\text{ V}\) each. They combine to form a bigger drop. Calculate the potential of the bigger drop.

1. 1980 V
2. 660 V
3. 1320 V
4. 1520 V
View Answer

By volume conservation, \(R = N^{1/3} r = 27^{1/3} r = 3r\). The total charge of the combined drop is \(Q = 27q\). The potential of the bigger drop is \(V' = \frac{kQ}{R} = \frac{k(27q)}{3r} = 9 V = 9 \times 220 = 1980\text{ V}\).

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Question 47: easy

Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).

Assertion (A): When a neutral body is positively charged, its mass decreases.

Reason (R): A body acquires positive charge when it loses electrons.

In the light of the above statements, the correct option is

1. Both (A) and (R) are true and (R) is the correct explanation of (A)
2. Both (A) and (R) are true but (R) is not the correct explanation of (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer

Charging a body positively involves removing electrons. Since electrons have mass, removing them decreases the body's mass. Thus, both statements are true and Reason is the correct explanation.

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Question 48: easy

Four electric charges of \(8\ mu\text{C}\), \(5\ mu\text{C}\), \(-3\ mu\text{C}\), \(-10\ mu\text{C}\) are placed at the corners of a square of side \(\sqrt{2}\text{ m}\). The potential at the centre of the square is

1. \(9 \times 10^3 V\)
2. Zero
3. \(1.8 \times 10^3 V\)
4. \(2.7 \times 10^3 V\)
View Answer

The total electric potential at the centre is \(V = \sum \frac{kq_i}{r}\). Since the distance \(r\) from each corner to the centre is equal and the sum of charges \(\sum q_i = 8 + 5 - 3 - 10 = 0\), the net potential is zero.

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Question 49: easy

Consider the following statements:

(A) In a region of uniform electric field, the net charge contained in a volume is zero.

(B) Gauss law gives incorrect result for uniform electric field.

The correct option(s) is/are:

1. Only (A)
2. Only (B)
3. Both (A) and (B)
4. Neither (A) nor (B)
View Answer

In a uniform electric field, net flux through any closed surface is zero, which means net enclosed charge is zero. Gauss's law is universally correct, so statement (B) is false.

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Question 50: easy

If some positive charge is given to a solid conductor, then its potential is

1. Minimum at surface
2. Zero at centre
3. Same throughout the conductor
4. Maximum somewhere outside the surface
View Answer

The electric field inside a conductor is zero, so no work is done in moving a charge inside it. Consequently, the potential is constant and same throughout the conductor.

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