Electrostatics Question 659
Question: Two point charges $ 100\mu C $ and $ 5\mu C $ are placed at points $ A $ and $ B $ respectively with $ AB $ = 40 cm. The work done by external force in displacing the charge 5 $ \mu C $ from $ B $ to $ C $ , where $ BC $ = 30 cm, angle $ ABC $ = $ \pi /2 $ and $ 1/4\pi {\varepsilon _{0}}=9\times 10^{9}Nm^{2}/C^{2} $
Options:
A) $ 9J $
B) $ \frac{81}{20}J $
C) $ \frac{9}{25}J $
D) $ -\frac{9}{4}J $
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Answer:
Correct Answer: D
Solution:
[d] Work done in displacing charge of 5 $ \mu C $ from B to C is $ W=5\times {{10}^{-6}}(V _{C}-V _{B}) $
$ V _{B}=9\times 10^{9}\times \frac{100\times {{10}^{-6}}}{0.4}=\frac{9}{4}\times 10^{6}V $ And $ V _{C}=9\times 10^{9}\times \frac{100\times {{10}^{-6}}}{0.4}=\frac{9}{5}\times 10^{6}V $
So $ W=5\times {{10}^{-6}}\times ( \frac{9}{5}\times 10^{6}-\frac{9}{4}\times 10^{6} )=-\frac{9}{4}J $
