Electrostatics Question 506
Question: Two thin wire rings each having a radius R are placed at a distance d apart with their axes coinciding. The charges on the two rings are $ +q $ and $ -q $ . The potential difference between the centres of the two rings is [AIEEE 2005]
Options:
A) Zero
B) $ \frac{Q}{4\pi {\varepsilon _{0}}}[ \frac{1}{R}-\frac{1}{\sqrt{R^{2}+d^{2}}} ] $
C) $ QR/4\pi {\varepsilon _{0}}d^{2} $
D) $ \frac{Q}{2\pi {\varepsilon _{0}}}[ \frac{1}{R}-\frac{1}{\sqrt{R^{2}+d^{2}}} ] $
Show Answer
Answer:
Correct Answer: D
Solution:
Potential at the centre of rings are $ {V _{O _{1}}}=\frac{k.q}{R}+\frac{k(-q)}{\sqrt{R^{2}+d^{2}}} $ , $ {V _{O _{2}}}=\frac{k(-q)}{R}+\frac{kq}{\sqrt{R^{2}+d^{2}}} $
therefore $ {V _{O _{1}}}-{V _{O _{2}}}=2kq[ \frac{1}{R}-\frac{1}{\sqrt{R^{2}+d^{2}}} ] $
$ =\frac{q}{2\pi {\varepsilon _{0}}}[ \frac{1}{R}-\frac{1}{\sqrt{R^{2}+d^{2}}} ] $
