JEE Physics Electrostatic Potential and Capacitance MCQs Set 11

Answer: (d) \( 3\times10^{-2} \) C

Explanation:
1. The total charge \( = (-1+6)\times10^{-2} = 5\times10^{-2} \) C.
2. The spheres reach the same potential, so \( \dfrac{q_1}{r_1} = \dfrac{q_2}{r_2} \) and the charge is shared in the ratio of the radii 2 : 3.
3. The bigger sphere has \( \dfrac35\times5\times10^{-2} = 3\times10^{-2} \) C.

Teacher's Note:
1. Joined conductors share charge in proportion to their radii (equal potential), not equally.

Answer: (d) all of the above

Explanation:
1. The potential decreases along the direction of the field, so it is highest at A and lowest at C.
2. B and D lie on the same equipotential line (perpendicular to the field), so \( V_B = V_D \).
3. All three statements are correct.

Teacher's Note:
1. Field lines point from high to low potential, and equipotential surfaces are perpendicular to the field.

Answer: (b) \( \dfrac{Q}{4\pi\varepsilon_0R} \) and zero

Explanation:
1. Inside a conductor the electric field is zero.
2. The potential is constant throughout and equal to that on the surface: \( V = \dfrac{Q}{4\pi\varepsilon_0R} \).

Teacher's Note:
1. Zero field does not mean zero potential. It means the potential is constant.

Answer: (b) \( \dfrac{2Q}{4\pi\varepsilon_0R}+\dfrac{q}{4\pi\varepsilon_0R} \)

Explanation:
1. The point charge Q gives \( \dfrac{Q}{4\pi\varepsilon_0(R/2)} = \dfrac{2Q}{4\pi\varepsilon_0R} \) at P.
2. The shell, with an induced charge -Q on its inner surface and q + Q on its outer surface, gives a potential at every interior point equal to its surface potential \( \dfrac{q}{4\pi\varepsilon_0R} \).
3. Total \( V = \dfrac{2Q}{4\pi\varepsilon_0R}+\dfrac{q}{4\pi\varepsilon_0R} \).

Teacher's Note:
1. Use superposition: the point charge, plus the net effect of the shell charges (-Q inside and q + Q outside, which combine to q).

Answer: (a) 120 V/m

Explanation:
1. Outside: \( E = \dfrac{kQ}{r^2} \Rightarrow kQ = 100\times(0.2)^2 = 4 \).
2. Inside a uniformly charged sphere: \( E = \dfrac{kQr}{R^3} = \dfrac{4\times0.03}{(0.1)^3} = 120 \) V/m.

Teacher's Note:
1. Inside a uniformly charged sphere the field grows linearly with r. Outside it falls as 1/r².

Answer: (a) zero

Explanation:
1. A is at the same distance from B and from C.
2. \( V_A = \dfrac{k q}{AB} - \dfrac{kq}{AC} = 0 \).

Teacher's Note:
1. Every point on the perpendicular bisector of a dipole's axis is at zero potential.

Answer: (d) V

Explanation:
1. The potential difference between the sphere and the shell depends only on the charge Q of the inner sphere: \( \dfrac{Q}{4\pi\varepsilon_0}\left(\dfrac1a-\dfrac1b\right) \).
2. Giving the shell extra charge raises its potential and that of everything inside by the same amount, so the difference is unchanged: V.

Teacher's Note:
1. Adding charge to the outer shell shifts all potentials inside it equally, so potential differences inside do not change.

Answer: (b) 2 V

Explanation:
1. \( \int_{\infty}^{0}-\vec E\cdot d\vec\ell = V(0) - V(\infty) = V_{centre} \).
2. For any charge distribution on a ring, the potential at the centre is \( \dfrac{kQ}{r} = \dfrac{9\times10^9\times1.11\times10^{-10}}{0.5} \approx 2 \) V.

Teacher's Note:
1. The potential at the centre of a ring does not depend on how the charge is distributed, since every element is at distance r.

Answer: (c) \( \dfrac{\sigma}{\varepsilon_0}(R+r) \)

Explanation:
1. The potential of a shell of radius a and surface density σ at its centre is \( \dfrac{\sigma4\pi a^2}{4\pi\varepsilon_0a} = \dfrac{\sigma a}{\varepsilon_0} \).
2. Adding for both shells: \( V = \dfrac{\sigma}{\varepsilon_0}(R+r) \).

Teacher's Note:
1. The potential is a scalar sum. Each shell contributes its own surface value at the centre.

Answer: (c) \( r_1/r_2 \)

Explanation:
1. At the surface, \( E = \dfrac{V}{r} \), so \( V = Er \).
2. With the same E, \( V_1 : V_2 = r_1 : r_2 \).

Teacher's Note:
1. For a charged conducting sphere, E = V/R at the surface, which links field and potential directly.

Answer: (b) 5 volt/metre

Explanation:
1. At x = 13 m the point is on the last segment, from (12 m, 30 V) to (16 m, 10 V).
2. \( E = -\dfrac{dV}{dx} = -\dfrac{10-30}{16-12} = +5 \) V/m.

Teacher's Note:
1. The field is the negative slope of the V-x graph. A falling graph means a positive field.

Answer: (c) \( -q\dfrac{r_1}{r_2} \)

Explanation:
1. The grounded inner sphere is at zero potential: \( \dfrac{kq'}{r_1} + \dfrac{kq}{r_2} = 0 \).
2. \( q' = -q\dfrac{r_1}{r_2} \).

Teacher's Note:
1. Earthing fixes the potential at zero, not the charge. Use V = 0 to find the induced charge.

Answer: (a) \( \phi_1\dfrac{r_1}{r_2} \)

Explanation:
1. The charge on A is \( Q = 4\pi\varepsilon_0r_1\phi_1 \).
2. After connection all the charge goes to the outer shell, and the potential of both is \( \dfrac{Q}{4\pi\varepsilon_0r_2} = \phi_1\dfrac{r_1}{r_2} \).

Teacher's Note:
1. When a charged inner conductor is connected to an outer shell, its charge moves to the outer surface.

Answer: (b) \( -\dfrac{qQ}{6\pi\varepsilon_0L} \)

Explanation:
1. \( V_C = \dfrac{kq}{L}-\dfrac{kq}{L} = 0 \).
2. \( V_D = \dfrac{kq}{3L}-\dfrac{kq}{L} = -\dfrac{2kq}{3L} \).
3. \( W = Q(V_D-V_C) = -\dfrac{2kqQ}{3L} = -\dfrac{qQ}{6\pi\varepsilon_0L} \).

Teacher's Note:
1. Work done depends only on the end points. The path (semicircle) is irrelevant for a conservative field.

Answer: (a) zero along each path

Explanation:
1. All points on the circle are at the same distance from Q, so they are at the same potential.
2. \( W = q(V_B-V_A) = 0 \) for every path.

Teacher's Note:
1. Moving a charge between points of the same potential needs no net work, whatever the path.

Answer: (c) \( \dfrac{E^2q^2t^2}{2m} \)

Explanation:
1. Acceleration \( a = \dfrac{qE}{m} \), so \( v = \dfrac{qEt}{m} \).
2. \( KE = \tfrac12mv^2 = \dfrac{q^2E^2t^2}{2m} \).

Teacher's Note:
1. The particle starts from rest, so v = at. Substitute into the kinetic energy formula.

Answer: (b) \( qEY \)

Explanation:
1. The force is \( F = qE \) along the field, and the displacement along the field is Y.
2. \( W = F\,Y = qEY \).

Teacher's Note:
1. Work = force × displacement along the force.

Answer: (c) \( \dfrac{q(Q_1-Q_2)(\sqrt2-1)}{\sqrt2\cdot4\pi\varepsilon_0R} \)

Explanation:
1. The distance from the centre of one ring to a point on the other is \( \sqrt2R \).
2. \( V_1 = k\left(\dfrac{Q_1}{R}+\dfrac{Q_2}{\sqrt2R}\right) \) and \( V_2 = k\left(\dfrac{Q_2}{R}+\dfrac{Q_1}{\sqrt2R}\right) \).
3. \( W = q(V_1-V_2) = \dfrac{kq(Q_1-Q_2)}{R}\left(1-\dfrac{1}{\sqrt2}\right) = \dfrac{q(Q_1-Q_2)(\sqrt2-1)}{\sqrt2\cdot4\pi\varepsilon_0R} \).

Teacher's Note:
1. The field does positive work when a charge moves from higher to lower potential: W = q(V initial − V final).

Answer: (c) 8q2

Explanation:
1. The distance from q1 stays 40 cm on the arc, so its contribution to the energy does not change.
2. Distance from q2: at C it is \( \sqrt{30^2+40^2} = 50 \) cm, at D it is \( 40-30 = 10 \) cm.
3. \( \Delta U = \dfrac{q_3q_2}{4\pi\varepsilon_0}\left(\dfrac{1}{0.1}-\dfrac{1}{0.5}\right) = \dfrac{q_3}{4\pi\varepsilon_0}(8q_2) \).

Teacher's Note:
1. Only the charge whose distance changes affects the change in potential energy.

Answer: (d) \( \sqrt{\dfrac{2kQq}{mR}} \)

Explanation:
1. The potential energy at the centre is \( \dfrac{kQq}{R} \), and at infinity it is zero.
2. \( \tfrac12mv^2 = \dfrac{kQq}{R} \Rightarrow v = \sqrt{\dfrac{2kQq}{mR}} \).

Teacher's Note:
1. Use energy conservation: the initial potential energy becomes kinetic energy at infinity.