A spherical metal shell A of radius $\mathrm{R}_{\mathrm{A}}$ and a solid metal sphere B of radius $\mathrm{R}_{\mathrm{B}}\left(<\mathrm{R}_{\mathrm{A}}\right)$ are kept far apart and each is given charge '+Q'. Now they are connected by a thin metal wire. Then

(A) $E_{A}^{\text {inside }}=0$; (B) $Q_{A}>Q_{B}$; (C) $\frac{\sigma_{A}}{\sigma_{B}}=\frac{R_{B}}{R_{A}}$; (D) $E_{A}^{\text {on surface }}<E_{B}^{\text {on surface }}$

JEE Advanced · 2011 · 10 Apr 2011 · Paper 1 · Electrostatics

WhatsApp Email

A spherical metal shell A of radius

R_{A}

and a solid metal sphere B of radius

R_{B} (< R_{A})

are kept far apart and each is given charge '+Q'. Now they are connected by a thin metal wire. Then

(A) $E_{A}^{inside} = 0$ ✓ Correct
(B) $Q_{A} > Q_{B}$ ✓ Correct
(C) $\frac{σ _{A}}{σ _{B}} = \frac{R _{B}}{R _{A}}$ ✓ Correct
(D) $E_{A}^{on surface} < E_{B}^{on surface}$ ✓ Correct

Answer: (A), (B), (C), (D)

Solution

When connected by a wire, both conductors reach the same potential:

\frac{k Q _{A}}{R _{A}} = \frac{k Q _{B}}{R _{B}}

This gives

Q_{A} R_{B} = Q_{B} R_{A}

, and since

Q_{A} + Q_{B} = 2 Q

Q_{A} = \frac{2 Q R _{A}}{R _{A} + R _{B}}, Q_{B} = \frac{2 Q R _{B}}{R _{A} + R _{B}}

Since

R_{A} > R_{B}

, we have

Q_{A} > Q_{B}

(option B).

For surface charge densities:

σ_{A} = \frac{Q _{A}}{4 π R _{A}^{2}}

σ_{B} = \frac{Q _{B}}{4 π R _{B}^{2}}

Using

Q_{A} R_{B} = Q_{B} R_{A}

\frac{σ _{A}}{σ _{B}} = \frac{Q _{A} R _{B}^{2}}{Q _{B} R _{A}^{2}} = \frac{R _{B}}{R _{A}}

(option C).

Electric field on surface:

E_{A} = \frac{k Q _{A}}{R _{A}^{2}}

E_{B} = \frac{k Q _{B}}{R _{B}^{2}}

. Since both are at same potential,

E_{A} R_{A} = E_{B} R_{B}

, giving

E_{A} < E_{B}

(option D).

Inside a conductor, electric field is always zero (option A).

JEE Advanced · 2011 · 10 Apr 2011 · Paper 1 · One or more correct options

Practise more: all JEE Advanced PYQs on Electrostatics →

JEE Advanced · 2011 · 10 Apr 2011 · Paper 1 · Electrostatics

Solution

More JEE Advanced questions on Electrostatics