Question

Work performed when a point charge $2\times {10}^{-8}$ is transformed from infinity to a point at a distance of $1cm$ from the surface of the ball with a radius of $1cm$ and a surface charge density $\sigma ={10}^{-8}C/c{m}^2$ is:

• A. $1.1\times {10}^{-4}J$
• B. $11\times {10}^{-4}J$
• C. $0.11\times {10}^{-4}J$
• D. $113\times {10}^{-4}J$

Answer 1

Answer: (B)

$11\times {10}^{-4}J$

The potential due to ball at distance r is $V=\frac{kQ}{r}=\frac{k4\pi R^2\sigma }{r}=(9\times {10}^9)\frac{4\pi (0.01{)}^2\times {10}^{-8}\times {10}^4}{(0.01+0.01)}=5.65\times {10}^{4}V$

Work done is the potential energy so, $W=qV=(2\times {10}^{-8})(5.65\times {10}^4)=11\times {10}^{-4}J$