Question

A spherical conductor of radius $4\text{cm}$ carries a charge of $1.11\times {10}^{–10}\text{C}$. At what distance (in $\text{cm}$) from the centre of sphere should a point charge of $4.44\times {10}^{–10}\text{C}$ be placed so that the potential of the conductor becomes $50\text{V}?$ (integer only)
[Assume $V=0\text{at}\infty$]

Answer 1

Given,
$R=4\text{cm};V_{sphere}=50\text{V}{Q}_{sphere}=1.11\times {10}^{–10}\text{C};Q=4.44\times {10}^{–10}\text{C}$



The net potential of the conducting sphere is,

$V_{sphere}=\frac{k\sum Q_{surface}}{R}+\frac{kQ}{x}$

$50=\frac{9\times {10}^9\times 1.11\times {10}^{-10}}{4\times {10}^{-2}}+\frac{9\times {10}^9\times 4.44\times {10}^{-10}}{x}$

$50=25+\frac{4}{x}$

$25=\frac{4}{x}$

$\Rightarrow x=16\text{cm}$