A spherical metal shell A of radius and a solid metal sphere B of radius are kept far apart and each is given charge . Now they are connected by a thin metal wire. Then [2011]

Question

A spherical metal shell A of radius $R_{A}$ and a solid metal sphere B of radius $R_{B}$ $(R_{B} < R_{A})$ are kept far apart and each is given charge $' + Q'$ . Now they are connected by a thin metal wire. Then [2011]

Smart Achievers · Accepted Answer

(1, 2, 3, 4)

(1) Electric field inside a spherical metallic shell with charge on the surface is always zero, i.e., $E_{A}^{inside} = 0 .$

(2) When the shells are connected with a thin metal wire, then electric potentials will be equal, i.e., $V_{A} = V_{B} = V .$

$∴$ $\frac{1}{4 π ε_{0}} \frac{Q_{A}}{R_{A}} = \frac{1}{4 π ε_{0}} \frac{Q_{B}}{R_{B}} = V$ $∵$ $R_{A} > R_{B}$ $∴$ $Q_{A} > Q_{B}$

(3) As $\frac{σ_{A}}{σ_{B}} = \frac{\frac{Q_{A}}{4 π R_{A}^{2}}}{\frac{Q_{B}}{4 π R_{B}^{2}}} = \frac{R_{B}^{2}}{R_{A}^{2}} \cdot \frac{Q_{A}}{Q_{B}}$ $= \frac{R_{B}^{2}}{R_{A}^{2}} \cdot \frac{4 π ε_{0} R_{A} V}{4 π ε_{0} R_{B} V} = \frac{R_{B}}{R_{A}}$

(4) $E_{A} = \frac{σ_{A}}{ε_{0}}$ and $E_{B} = \frac{σ_{B}}{ε_{0}}$

$\frac{E_{A}}{E_{B}} = \frac{σ_{A}}{σ_{B}} = \frac{R_{B}}{R_{A}} < 1$

$∴$ $E_{A} < E_{B}$

Solution

Q.
A spherical metal shell A of radius $R_{A}$ and a solid metal sphere B of radius $R_{B}$ $(R_{B} < R_{A})$ are kept far apart and each is given charge $' + Q'$ . Now they are connected by a thin metal wire. Then [2011]

1 $E_{A}^{inside} = 0$

2 $Q_{A} > Q_{B}$

3 $\frac{σ_{A}}{σ_{B}} = \frac{R_{B}}{R_{A}}$

4 $E_{A}^{on surface} < E_{B}^{on surface}$

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Solution

Q. A spherical metal shell A of radius RA and a solid metal sphere B of radius RB(RB<RA) are kept far apart and each is given charge '+Q'. Now they are connected by a thin metal wire. Then [2011]

1 EAinside=0

2 QA>QB

3 σAσB=RBRA