Electrostatics Question 358
Question: Four equal charges $ Q $ are placed at the four corners of a square of each side is $ ‘a’ $ . Work done in removing a charge Q from its centre to infinity is [AIIMS 1995]
Options:
A) 0
B) $ \frac{\sqrt{2}Q^{2}}{4\pi {\varepsilon _{0}}a} $
C) $ \frac{\sqrt{2}Q^{2}}{\pi {\varepsilon _{0}}a} $
D) $ \frac{Q^{2}}{2\pi {\varepsilon _{0}}a} $
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Answer:
Correct Answer: C
Solution:
Potential at centre O of the square $ V _{O}=4( \frac{Q}{4\pi {\varepsilon _{0}}(a/\sqrt{2})} ) $
Work done in shifting (- Q) charge from centre to infinity $ W=-Q({V _{\infty }}-V _{O})=QV _{0} $
$ =\frac{4\sqrt{2}Q^{2}}{4\pi {\varepsilon _{0}}a} $
$ =\frac{\sqrt{2}Q^{2}}{\pi {\varepsilon _{0}}a} $