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. $\frac{qR}{4\,\pi\varepsilon_{0}d^{2}}$
• B. $\frac{q}{2\pi\varepsilon_{0}}\left[\frac{1}{R}-\frac{1}{\sqrt{R^{2}+d^{2}}}\right]$
• C. zero
• D. $\frac{q}{4\pi\varepsilon_{0}}\left[\frac{1}{R}-\frac{1}{\sqrt{R^{2}+d^{2}}}\right]$

Answer 1

Answer: (B)

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

Answer 2

$V_A$ = potential due to charge $+ q$ on ring $A +$ potential due to charge $- q$ on ring $B$ $= \frac{1}{4\,\pi\varepsilon_{0}}\left(\frac{q}{R}-\frac{q}{d_{1}}\right), d_{1}=\sqrt{R^{2}+d^{2}}$ $=\frac{1}{4\,\pi \varepsilon _{0}} \left(\frac{q}{R}-\frac{q}{\sqrt{R^{2}+d^{2}}}\right)\,...\left(i\right)$ Similarly, $V_{B}=\frac{1}{4\,\pi \varepsilon _{0}}\left(-\frac{q}{R}+\frac{q}{\sqrt{R^{2}+d^{2}}}\right)$ Potential difference $V_{A}-V_{B}$ $=\frac{1}{4\,\pi \varepsilon _{0}}\left(\frac{q}{R}-\frac{q}{\sqrt{R^{2}+d^{2}}}\right) -\frac{1}{4\,\pi \varepsilon _{0}} \left(-\frac{q}{R}+\frac{q}{\sqrt{R^{2}+d^{2}}}\right)$ $=\frac{1}{4\,\pi \varepsilon _{0}} \frac{q}{R}+\frac{1}{4\,\pi \varepsilon _{0}} \frac{q}{\sqrt{R^{2}+d^{2}}}-\frac{1}{4\,\pi \varepsilon _{0}} \frac{q}{\sqrt{R^{2}+d^{2}}}$ $=\frac{1}{2\,\pi\varepsilon _{0}}\left(\frac{q}{R}-\frac{q}{\sqrt{R^{2}+d^{2}}}\right)$