Question

Two thin wire rings each having a radius R are placed at distance d apart with their axes coinciding. The charges on the two rings are +q and -q. The potential difference between the rings are :

• A. $\frac{QR}{4\pi {\epsilon }_0{d}^2}$
• B. $\frac{Q}{2\pi {\epsilon }_0}[\frac{1}{R}-\frac{1}{\sqrt{R^2+d^2}}]$
• C. $\frac{Q}{4\pi {\epsilon }_0}[\frac{1}{R}-\frac{1}{\sqrt{R^2+d^2}}]$
• D. $\text{z}ero$

Answer 1

Answer: (B)

$\frac{Q}{2\pi {\epsilon }_0}[\frac{1}{R}-\frac{1}{\sqrt{R^2+d^2}}]$

$V_1=\frac{Q}{4\pi {\epsilon }_0}[\frac{1}{R}-\frac{1}{\sqrt{R^2+d^2}}]$ and
$V_2=\frac{-Q}{4\pi {\epsilon }_0}[\frac{1}{R}-\frac{1}{\sqrt{R^2+d^2}}]$
$\Delta V=V_1-V_2$

$\Delta V=\frac{Q}{2\pi {\epsilon }_0}[\frac{1}{R}-\frac{1}{\sqrt{R^2+d^2}}]$