A) 0.11 J
B) 11 J
C) 0.01 J
D) 1.1 J
Correct Answer: C
Solution :
Initial PE of the three charges, \[{{U}_{i}}=k\frac{{{q}_{1}}{{q}_{2}}+{{q}_{2}}{{q}_{3}}+{{q}_{1}}{{q}_{3}}}{r}\] \[=9\times {{10}^{9}}\left[ \frac{\begin{align} & 1\times 2\times {{10}^{-12}}+2\times 3\times {{10}^{-12}} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+1\times 3\times {{10}^{-12}} \\ \end{align}}{1} \right]\] \[=99\times {{10}^{-3}}J\] Final PE of the system \[{{U}_{f}}=9\times {{10}^{9}}\left[ \frac{\begin{align} & 1\times 2\times {{10}^{-12}}+2\times 3\times {{10}^{-12}} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+1\times 3\times {{10}^{-12}} \\ \end{align}}{0.5} \right]\] \[=\frac{99\times {{10}^{-3}}}{0.5}=198\times {{10}^{-3}}J\] \[W={{U}_{f}}-{{U}_{i}}=(198-99)\times {{10}^{-3}}\] \[=99\times {{10}^{-3}}J\] \[=0.99=0.01\,J\]
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