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} $



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