Electrostatics Question 503
Question: To charges $ q _{1} $ and $ q _{2} $ are placed $ 30cm $ apart, . A third charge $ q _{3} $ is moved along the arc of a circle of radius $ 40cm $ from $ C $ to D. The change in the potential energy of the system is $ \frac{q _{3}}{4\pi {\varepsilon _{0}}}k $ , where $ k $ is [CBSE PMT 2005]
Options:
A) $ 8q _{2} $
B) $ 8q _{1} $
C) $ 6q _{2} $
D) $ 6q _{1} $
Show Answer
Answer:
Correct Answer: A
Solution:
Change in potential energy $ (\Delta U)=U _{f}U _{i} $
therefore $ \Delta U=\frac{1}{4\pi {\varepsilon _{0}}}[ ( \frac{q _{1}q _{3}}{0.4}+\frac{q _{2}q _{3}}{0.1} )-( \frac{q _{1}q _{3}}{0.4}+\frac{q _{2}q _{3}}{0.5} ) ] $
therefore $ \Delta U=\frac{1}{4\pi {\varepsilon _{0}}}[8q _{2}q _{3}]=\frac{q _{3}}{4\pi {\varepsilon _{0}}}(8q _{2}) $
$ \therefore k=8q _{2} $
BETA
