Question

A charge of $2C$ is moved from a point $2m$ away from a charge of $1C$ to a point $1m$ away from that charge. The work done is:

• A. ${10}^{9}J$
• B. ${10}^{6}J$
• C. $9\times {10}^{9}J$
• D. ${10}^{10}J$

Answer 1

Answer: (C)

$9\times {10}^{9}J$

Voltage due to $1C$ at a distance $r$ is $\frac{k\times 1}{r}$.

Total change in potential in moving $2C$ charge from $2m$ away to $1m$ away:
$V_2-V_1=\frac{k\times 1}{1}-\frac{k\times 1}{2}$
or, $V_2-V_1=\frac{k\times 1}{2}=\Delta V$

So, work done $=\Delta VQ$$=\frac{k\times 1}{2}\times 2=9\times {10}^{9}J$