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.

• A. $\frac{q}{4\pi\varepsilon_{0}}\left\lbrack \frac{1}{R} - \frac{1}{\sqrt{R^{2} + d^{2}}} \right\rbrack$
• B. Zero (“kwU;)
• C. $\frac{q}{2\pi\varepsilon_{0}}\left\lbrack \frac{1}{R} - \frac{1}{\sqrt{R^{2} + d^{2}}} \right\rbrack$
• D. $\frac{qR}{4\pi\varepsilon_{0}d^{2}}$

Answer 1

Answer: (C)

$\frac{q}{2\pi\varepsilon_{0}}\left\lbrack \frac{1}{R} - \frac{1}{\sqrt{R^{2} + d^{2}}} \right\rbrack$

Answer 2

: $V_{1} = \frac{q}{4\pi\varepsilon_{0}R} - \frac{q}{4\pi\varepsilon_{0}\sqrt{R^{2} + d^{2}}}$

$V_{2} = - \frac{q}{4\pi\varepsilon_{0}R} + \frac{q}{4\pi\varepsilon_{0}\sqrt{R^{2} + d^{2}}}$

$\therefore V_{1} - V_{2} = \frac{2q}{4\pi\varepsilon_{0}}\left\lbrack \frac{1}{R} - \frac{1}{\sqrt{R^{2} + d^{2}}} \right\rbrack$

$V_{1} - V_{2} = \frac{q}{2\pi\varepsilon_{0}}\left\lbrack \frac{1}{R} - \frac{1}{\sqrt{R^{2} + d^{2}}} \right\rbrack$