Electrostatics Question 235
Question: Two identical thin rings each of radius R meters are coaxially placed at a distance R meters apart. If Q1 coulomb and Q2 coulomb are respectively the charges uniformly spread on the two rings, the work done in moving a charge qfrom the centre of one ring to that of other is [MP PMT 1999; AMU (Engg.) 1999]
Options:
A) Zero
B) $ \frac{q(Q _{1}-Q _{2})(\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} $
Show Answer
Answer:
Correct Answer: B
Solution:
$ W=q({V _{O _{2}}}-{V _{O _{1}}}) $
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}} $
therefore $ {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}} $
