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Q.
Two positive point charges of $ 12\,\mu \,C $ and $ 8\,\mu \,C $ are $10 \,cm$ apart. The work done in bringing them $4 \,cm$ closer, is
AMUAMU 2000
Solution:
Given two charges $q _{1}=12 \,\mu C$,
$ q _{2}=8 \,\mu C$ are at distance
' $d_{1}{ }'=10 \,cm$
We have to find the work done in bringing them $d _{2}=4 \,cm$ closer.
For this, we find the charge in the potential energy of the system and then take the difference.
Now $d_{1}=10\, cm =10 \times 10^{-2}=10^{-1} m$
$d _{2}=4 \,cm =4 \times 10^{-2} m$
So, potential $\left( V _{1}\right)$ when $q _{1}$ and $q _{2}$ are distance $d _{1}$ apart,
$V _{1}=\frac{1}{4 \pi \varepsilon_{0}} \frac{ q _{1} q _{2}}{ d _{1}}$
Potential $\left( V _{2}\right)$ when $q _{1}$ and $q _{2}$ are distance $d _{2}$ apart,
$V _{2}=\frac{1}{4 \pi \varepsilon_{0}} \frac{ q _{1} q _{2}}{ d _{2}}$
Work done $= V _{2}- V _{1}=\frac{1}{4 \pi \varepsilon_{0}} q _{1} q _{2}\left(\frac{1}{ d _{2}}-\frac{1}{ d _{1}}\right)$
$=9 \times 10^{9} \times 12 \times 8 \times 10^{-12}\left(\frac{1}{4 \times 10^{-2}}-\frac{1}{10 \times 10^{-2}}\right)$
$=864 \times 10^{-3}\left[\frac{6 \times 10^{2}}{40 \times 10^{-2}}\right]$
$=129.6 \times 10^{1}$
$=12.96 \approx 13$ Joules