Question

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

• A. 1.3 eV
• B. 13 J
• C. 5.8 J
• D. 5.8 eV

Answer 1

Answer: (B)

Given two charges $q_1=12\mu C$,
$q_2=8\mu C$ are at distance
' $d_1{}^{\prime }=10cm$
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 $\left(V_1\right)$ 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 $\left(V_2\right)$ 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 $=V_2-V_1=\frac{1}{4\pi {\epsilon }_0}{q}_1{q}_2\left(\frac{1}{d_2}-\frac{1}{d_1}\right)$
$=9\times 10^9\times 12\times 8\times 10^{-12}\left(\frac{1}{4\times 10^{-2}}-\frac{1}{10\times 10^{-2}}\right)$
$=864\times 10^{-3}\left[\frac{6\times 10^2}{40\times 10^{-2}}\right]$
$=129.6\times 10^1$
$=12.96\approx 13$ Joules