Q: In a region, the potential is represented by V(x, y, z) = 6x − 8xy − 8y + 6yz, where V is in volts and x, y, z are in metres. The electric force experienced by a charge of 2 C situated at point (1, 1, 1) is

4\sqrt{35} N. \vec{E} = -\nabla V = -[(6 - 8y)\hat{i} + (-8x - 8 + 6z)\hat{j} + 6y\,\hat{k}] At (1,1,1): \vec{E} = -(-2\hat{i} - 10\hat{j} + 6\hat{k}), so |E| = \sqrt{4 + 100 + 36} = 2\sqrt{35} N/C F = qE = 2 \times 2\sqrt{35} = 4\sqrt{35} N

Q: Three concentric spherical shells have radii a, b and c (a < b < c) and have surface charge densities σ, −σ and σ respectively. If V_A, V_B and V_C denote the potentials of the three shells, then for c = a + b, we have

V_C = V_B = V_A. Computing each potential from the three charged shells and substituting c = a + b gives V_A = \frac{\sigma}{\varepsilon_0}(a - b + c) = \frac{\sigma}{\varepsilon_0}(2a), and V_B = V_C = \frac{\sigma}{\varepsilon_0}(2a) as well. Hence V_A = V_B = V_C.

Question 1

In a certain region of space with volume 0.2 m³, the electric potential is found to be 5 V throughout. The magnitude of electric field in this region is

Accepted Answer

zero. E = -\frac{dV}{dr}. Since the potential is constant (5 V everywhere), its gradient is zero, so E = 0.

Question 2

Some charge is being given to a conductor, then its potential is

Accepted Answer

same throughout the conductor. A charged conductor is an equipotential body — its surface and interior are all at the same potential.

Question 3

In bringing an electron towards another electron, the electrostatic potential energy of the system

Accepted Answer

increases. U = \frac{1}{4\pi\varepsilon_0}\frac{e^2}{r} (positive for like charges). As r decreases, U increases — work must be done against the repulsion.

Question 4

An electron of mass m and charge e is accelerated from rest through a potential difference of V volt in vacuum. Its final speed will be

Accepted Answer

\sqrt{\frac{2eV}{m}}. eV = \frac{1}{2}mv^2 \Rightarrow v = \sqrt{\frac{2eV}{m}}

Question 5

A hollow metal sphere of radius 10 cm is charged such that the potential on its surface is 80 V. The potential at the centre of the sphere is

Accepted Answer

80 V. Inside a conductor the potential is constant and equal to the surface value. So the potential at the centre is also 80 V.

Question 6

A parallel plate condenser with oil (dielectric constant 2) between the plates has capacitance C. If oil is removed, the capacitance of capacitor becomes

Accepted Answer

\frac{C}{2}. With oil C = K C_0 = 2C_0. Removing it gives C_0 = \frac{C}{K} = \frac{C}{2}

Question 7

Two metal spheres, one of radius R and the other of radius 2R respectively have the same surface charge density σ. They are brought in contact and separated. What will be the new surface charge densities on them?

Accepted Answer

\sigma_1 = \frac{5}{3}\sigma, \sigma_2 = \frac{5}{6}\sigma. Total charge Q_t = 4\pi\sigma R^2 + 16\pi\sigma R^2 = 20\pi\sigma R^2. On contact charges share in ratio 1:2 (potentials equal), so Q_1 = \frac{Q_t}{3}, Q_2 = \frac{2Q_t}{3}.

\sigma_1 = \frac{Q_t/3}{4\pi R^2} = \frac{5}{3}\sigma; \sigma_2 = \frac{2Q_t/3}{4\pi(2R)^2} = \frac{5}{6}\sigma

Question 8

In a region, the potential is represented by V(x, y, z) = 6x − 8xy − 8y + 6yz, where V is in volts and x, y, z are in metres. The electric force experienced by a charge of 2 C situated at point (1, 1, 1) is

Accepted Answer

4\sqrt{35} N. \vec{E} = -
abla V = -[(6 - 8y)\hat{i} + (-8x - 8 + 6z)\hat{j} + 6y\,\hat{k}]

At (1,1,1): \vec{E} = -(-2\hat{i} - 10\hat{j} + 6\hat{k}), so |E| = \sqrt{4 + 100 + 36} = 2\sqrt{35} N/C

F = qE = 2 	imes 2\sqrt{35} = 4\sqrt{35} N

Question 9

Four electric charges +q, +q, −q and −q are placed at the corners of a square of side 2L (the two +q charges on the left corners, the two −q charges on the right corners). The electric potential at point A, mid-way between the two charges +q (i.e. the midpoint of the left side), is

Accepted Answer

\frac{1}{4\pi\varepsilon_0}\frac{2q}{L}\left(1 - \frac{1}{\sqrt{5}}\right). A is at distance L from each +q and \sqrt{(2L)^2 + L^2} = \sqrt{5}L from each −q.

V = 2\cdot\frac{kq}{L} - 2\cdot\frac{kq}{\sqrt{5}L} = \frac{2q}{4\pi\varepsilon_0 L}\left(1 - \frac{1}{\sqrt{5}}\right)

Question 10

Three concentric spherical shells have radii a, b and c (a < b < c) and have surface charge densities σ, −σ and σ respectively. If V_A, V_B and V_C denote the potentials of the three shells, then for c = a + b, we have

Accepted Answer

V_C = V_B = V_A. Computing each potential from the three charged shells and substituting c = a + b gives V_A = \frac{\sigma}{\varepsilon_0}(a - b + c) = \frac{\sigma}{\varepsilon_0}(2a), and V_B = V_C = \frac{\sigma}{\varepsilon_0}(2a) as well. Hence V_A = V_B = V_C.

Question 11

Charges +q and −q are placed at points A and B respectively which are a distance 2L apart. C is the midpoint between A and B. The work done in moving a charge +Q along the semicircle CRD (from C to a point D located a distance L beyond B, i.e. 3L from A) is

Accepted Answer

-\frac{qQ}{6\pi\varepsilon_0 L}. Potential at C (distance L from both): V_C = \frac{kq}{L} - \frac{kq}{L} = 0.

At D (distance 3L from +q, L from −q): V_D = \frac{kq}{3L} - \frac{kq}{L} = -\frac{2kq}{3L}.

W = Q(V_D - V_C) = Q\left(-\frac{2kq}{3L}\right) = -\frac{qQ}{6\pi\varepsilon_0 L}

Question 12

Identical charges (−q) are placed at each of the eight corners of a cube of side b. The electrostatic potential energy of a charge (+q) placed at the centre of the cube will be

Accepted Answer

-\frac{4q^2}{\sqrt{3}\pi\varepsilon_0 b}. Each corner is at half the body-diagonal from the centre: \frac{\sqrt{3}b}{2}.

U = 8 	imes \frac{1}{4\pi\varepsilon_0}\frac{(+q)(-q)}{\frac{\sqrt{3}b}{2}} = -\frac{4q^2}{\sqrt{3}\pi\varepsilon_0 b}

Question 13

A series combination of n_1 capacitors, each of value C_1, is charged by a source of potential difference 4V. When another parallel combination of n_2 capacitors, each of value C_2, is charged by a source of potential difference V, it has the same (total) energy stored in it as the first combination has. The value of C_2, in terms of C_1, is then

Accepted Answer

\frac{16C_1}{n_1 n_2}. Series: U_s = \frac{1}{2}\frac{C_1}{n_1}(4V)^2. Parallel: U_p = \frac{1}{2}(n_2 C_2)V^2.

Setting equal: \frac{16C_1}{n_1} = n_2 C_2 \Rightarrow C_2 = \frac{16C_1}{n_1 n_2}

Question 14

Five 6 µF capacitors are arranged as a Wheatstone bridge between points X and Y — four capacitors in the four arms (X–A–Y and X–C–Y paths) and the fifth as the bridge between the two middle nodes. The effective capacitance between points X and Y is

Accepted Answer

6 µF. The bridge is balanced (\frac{C_{AB}}{C_{BD}} = \frac{C_{AC}}{C_{CD}}), so the middle capacitor carries no charge. Each arm has two 6 µF in series = 3 µF, and the two arms in parallel give 3 + 3 = 6 µF.

Question 15

Twenty seven drops of same size are charged at 220 V each. They combine to form a bigger drop. Calculate the potential of the bigger drop.

Accepted Answer

1980 V. 27 small drops → one big drop: R = 27^{1/3}r = 3r, and total charge Q = 27q.

V' = \frac{KQ}{R} = \frac{K \cdot 27q}{3r} = 9\frac{Kq}{r} = 9V = 9 	imes 220 = 1980 V

Question 16

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

Accepted Answer

\frac{R_2}{R_1}. Connected → same potential: \frac{q_1}{R_1} = \frac{q_2}{R_2}. With \sigma = \frac{q}{4\pi R^2}:

\frac{\sigma_1}{\sigma_2} = \frac{q_1}{q_2}\cdot\frac{R_2^2}{R_1^2} = \frac{R_1}{R_2}\cdot\frac{R_2^2}{R_1^2} = \frac{R_2}{R_1}

Question 17

A short electric dipole has a dipole moment of 16 	imes 10^{-9} C·m. The electric potential due to the dipole at a point at a distance of 0.6 m from the centre of the dipole, situated on a line making an angle of 60° with the dipole axis is (\frac{1}{4\pi\varepsilon_0} = 9 	imes 10^9 N·m²/C²)

Accepted Answer

200 V. V = \frac{1}{4\pi\varepsilon_0}\frac{p\cos	heta}{r^2} = \frac{9 	imes 10^9 	imes 16 	imes 10^{-9} 	imes \frac{1}{2}}{(0.6)^2} = 200 V

Question 18

Four diagrams show regions of equipotentials, each with a 20 V and a 40 V equipotential line (and 10 V, 30 V lines), arranged with different spacings and orientations. A positive charge q is moved from A (on the 20 V equipotential) to B (on the 40 V equipotential) in each diagram. Which statement is correct?

Accepted Answer

In all the four cases, the work done is the same. Work done W = q\Delta V. In every diagram the charge moves from the 20 V line to the 40 V line, so \Delta V = 20 V is the same — hence the work done is the same in all four cases (work is path-independent, depending only on the potential difference).

Question 19

If potential (in volts) in a region is expressed as V(x, y, z) = 6xy − y + 2yz, the electric field (in N/C) at point (1, 1, 0) is

Accepted Answer

-(6\hat{i} + 5\hat{j} + 2\hat{k}). \vec{E} = -
abla V = -[6y\,\hat{i} + (6x - 1 + 2z)\hat{j} + 2y\,\hat{k}]

At (1, 1, 0): \vec{E} = -(6\hat{i} + 5\hat{j} + 2\hat{k}) N/C

Question 20

A conducting sphere of radius R is given a charge Q. The electric potential and the electric field at the centre of the sphere respectively are

Accepted Answer

\frac{Q}{4\pi\varepsilon_0 R} and zero. Charge lies on the surface, so the field inside is zero. The potential is constant inside and equal to its surface value \frac{Q}{4\pi\varepsilon_0 R}.

Electrostatic Potential and Capacitance — NEET Physics PYQ MCQs with Solutions

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