Question

Two positive point charges are of $12C$ and $8C$ are $10cm$ apart from each other. The work done in bringing them $4cm$ closer is

• A. $5.8J$
• B. $13eV$
• C. $5.8eV$
• D. $13J$

Answer 1

The correct option is D $13J$

Given two charges $q_1=12\mu C$, $q_2=8\mu C$ are at distance '$d_1$'$=10cm$

We have to find the work done in bringing them $d_2=4cm$ closer.

For this, we find the charge in the potential energy of the system and then take the difference.

Now $d_1=10cm=10\times {10}^{-2}={10}^{-1}m$

$d_2=4cm=4\times {10}^{-2}m$

So, potential ($V_1$) when $q_1$ and $q_2$ are distance $d_1$ apart,

$V_1=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{d_1}$

Potential ($V_2$) when $q_1$ and $q_2$ are distance $d_2$ apart,

$V_2=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{d_2}$

Work done $=V_2-V_1=\frac{1}{4\pi {\epsilon }_0}{q}_1{q}_2(\frac{1}{d_2}-\frac{1}{d_1})$

$=9\times {10}^9\times 12\times 8\times {10}^{-12}(\frac{1}{4\times {10}^{-2}}-\frac{1}{10\times {10}^{-2}})$

$=864\times {10}^{-3}\left[\frac{6\times {10}^2}{40\times {10}^{-2}}\right]$

$=129.6\times {10}^1$

$=12.96\approx 13Joules$