Electrostatics Question 377
Question: Two point charges $ 100\mu C $ and $ 5\mu C $ are placed at points $ A $ and $ B $ respectively with $ AB=40cm $ . The work done by external force in displacing the charge $ 5\mu C $ from $ B $ to $ C $ , where $ BC=30cm $ , angle $ ABC=\frac{\pi }{2} $ and $ \frac{1}{4\pi {\varepsilon _{0}}}=9\times 10^{9}Nm^{2}/C^{2} $ [MP PMT 1997]
Options:
A) $ 9J $
B) $ \frac{81}{20}J $
C) $ \frac{9}{25}J $
D) $ -\frac{9}{4}J $
Show Answer
Answer:
Correct Answer: D
Solution:
Work done in displacing charge of 5 m C from B to C is $ W=5\times {{10}^{-6}}(V _{C}-V _{B}) $
where $ 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}}}{0.5}=\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 $
BETA
