JEE Main 2021 · Physics · Electrostatics · medium

A solid conducting sphere of radius 10 cm carries a charge of +4 nC and is surrounded by a concentric thin conducting shell of radius 20 cm carrying…

Numerical answer

A solid conducting sphere of radius

10\ \mathrm{cm}

carries a charge of

+4\ \mathrm{nC}

and is surrounded by a concentric thin conducting shell of radius

20\ \mathrm{cm}

carrying

-2\ \mathrm{nC}

. The two are then joined by a thin wire. Taking

\dfrac{1}{4\pi\varepsilon_0}=9\times10^9\ \mathrm{N\,m^2C^{-2}}

, the decrease in potential of the inner sphere, in volts, is

Show answer and solution

Answer: 180

Initially

V_{in}=9\times10^9\left(\dfrac{4\times10^{-9}}{0.1}+\dfrac{-2\times10^{-9}}{0.2}\right)=9(40-10)=270\ \mathrm{V}

. After joining, the charges rearrange so the whole net charge

+2\ \mathrm{nC}

sits on the outer shell and both are at the same potential:

V=\dfrac{9\times10^9\times2\times10^{-9}}{0.2}=90\ \mathrm{V}

. The decrease is

270-90=180\ \mathrm{V}

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