Electrostatics Question 796
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 center of one ring to that of other is
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:
[b] Work done $ W _{21}=( V _{1}-V _{2} )q $
$ V=\frac{1}{4\pi {\in _{0}}}[ \frac{Q _{1}}{R}+\frac{Q _{2}}{\sqrt{2}R} ] $ and $ V _{2}=\frac{1}{4\pi {\in _{0}}}[ \frac{Q _{2}}{R}+\frac{Q _{1}}{\sqrt{2}R} ] $
$ \text{Thus,}{{\text{W}} _{21}}=\frac{q( Q _{1}-Q _{2} )( \sqrt{2}-1 )}{\sqrt{2}.4\pi {\in _{0}}R}. $
