Electrostatics

Practice Questions:

Question 151: 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 152: 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 153: 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 154: 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 155: 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 156: 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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Question 157: easy

Identify the incorrect statement among the following

1. Electrostatic field at the surface of a charged conductor is proportional to the surface charge density

2. There is no net charge at any point inside conductor when a charge is placed outside it

3. Inside charged or neutral conductor, electrostatic field is zero

4. The electrostatic field at surface of charged conductor must be tangential to surface at any point when placed in external electric field

View Answer

Under electrostatic conditions, the electric field at the surface of a charged conductor must be perpendicular (normal) to the surface at every point. It cannot be tangential, otherwise charges would flow along the surface.

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Question 158: easy

The number of electrons that should be removed from a metal coin such that coin acquires a positive charge of \(10^{-10}\text{ C}\) is

1. \(1.6 \times 10^{-19}\)

2. \(6.25 \times 10^{9}\)

3. \(6.25 \times 10^{8}\)

4. \(1.6 \times 10^{8}\)

View Answer

Using quantization of charge, \(q = ne ⇒ n = \frac{q}{e} = \frac{10^{-10}}{1.6 \times 10^{-19}} = 6.25 \times 10^8\).

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Question 159: easy

Sixty four identical drops of water having equal charge combine to form a bigger drop. The factor by which potential of bigger drop change in comparison to a small drop is

1. 64

2. 32

3. 16

4. 8

View Answer

When \(N = 64\) drops combine, the new radius is \(R = N^{1/3}r = 4r\). The new charge is \(Q = Nq = 64q\). The potential of the bigger drop is \(V' = \frac{kQ}{R} = \frac{k(64q)}{4r} = 16 \left(\frac{kq}{r}\right) = 16V\).

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Question 160: easy

Match column I and Column II.

**Column I**

(A) Coulomb’s law
(B) Surface charge density
(C) Quantisation of charge
(D) Electric flux

**Column II**
(P) Charge/Area
(Q) \(\oint \vec{E} \cdot d\vec{s}\)
(R) \(q = ne\)
(S) Force is inversely proportional to square of distance

1. A → P, B → Q, C → R, D → S

2. A → S, B → P, C → R, D → Q

3. A → S, B → R, C → P, D → Q

4. A → S, B → P, C → Q, D → R

View Answer

A matches with S (inverse square law), B with P (charge/area), C with R (\(q=ne\)), and D with Q (flux).

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