Electrostatics Question 660
Question: Two identical thin rings each of radius $ R $ meters are coaxially placed at a distance R meters apart. If $ Q _{1} $ coulomb and $ Q _{2} $ coulomb are respectively the charges uniformly spread on the two rings, the work done in moving a charge$ q $ from the centre of one ring to that of other is
Options:
A) Zero
B) $ \frac{q(Q _{2}-Q _{1})(\sqrt{2}-1)}{\sqrt{2}.4\pi {\varepsilon _{0}}R} $
C) $ \frac{q\sqrt{2}(Q _{1}+Q _{2})}{4\pi {\varepsilon _{0}}R} $
D) $ \frac{q(Q _{1}+Q _{2})(\sqrt{2}+1)}{\sqrt{2}.4\pi {\varepsilon _{0}}R} $
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Answer:
Correct Answer: B
Solution:
[b] $ W=q(V _{O2}-V _{O1}) $
Where $ {V _{O _{1}}}=\frac{Q _{1}}{4\pi {\varepsilon _{0}}R}+\frac{Q _{2}}{4\pi {\varepsilon _{0}}R\sqrt{2}} $
And $ {V _{O _{2}}}=\frac{Q _{2}}{4\pi {\varepsilon _{0}}R}+\frac{Q _{1}}{4\pi {\varepsilon _{0}}R\sqrt{2}} $
$ \Rightarrow {V _{O _{2}}}-{V _{O _{1}}}=\frac{(Q _{2}-Q _{1})}{4\pi {\varepsilon _{0}}R}[ 1-\frac{1}{\sqrt{2}} ] $ So, $ W=\frac{q.(Q _{2}-Q _{1})}{4\pi {\varepsilon _{0}}R}\frac{(\sqrt{2}-1)}{\sqrt{2}} $
