Electrostatics Question 427
Question: In the rectangle, shown below, the two corners have charges $ q _{1}=-5\mu C $ and $ q _{2}=+2.0\mu C $ . The work done in moving a charge $ +3.0\mu C $ from $ B $ to $ A $ is (take $ 1/4\pi {\varepsilon _{0}}=10^{10}N\text{-}m^{2}/C^{2} $ ) [AMU 2001]
Options:
A) 2.8 J
B) 3.5 J
C) 4.5 J
D) 5.5 J
Show Answer
Answer:
Correct Answer: A
Solution:
Work done $ W=3\times {{10}^{-6}}(V _{A}-V _{B}); $
where $ V _{A}=10^{10}[ \frac{(-5\times {{10}^{-6}})}{15\times {{10}^{-2}}}+\frac{2\times {{10}^{-6}}}{5\times {{10}^{-2}}} ]=\frac{1}{15}\times 10^{6}volt $
and $ V _{B}=10^{10}[ \frac{(2\times {{10}^{-6}})}{15\times {{10}^{-2}}}-\frac{5\times {{10}^{-6}}}{5\times {{10}^{-2}}} ]=-\frac{13}{15}\times 10^{6}volt $
$ W=3\times {{10}^{-6}}[ \frac{1}{15}\times 10^{6}-( -\frac{13}{15}\times 10^{6} ) ] $ = 2.8 J
