Question

Two thin rings each of radius R are placed at a distance 'd' apart. The charges on the rings are +q and -q. The potential difference between their centres will be-

• A. $\frac{qR}{4\pi\varepsilon_{0}d^{2}}$
• B. $\frac { \mathrm { q } } { 2 \pi \varepsilon _ { 0 } }$ $\left\lbrack \frac{1}{R} - \frac{1}{\sqrt{R^{2} + d^{2}}} \right\rbrack$
• C. Zero
• D. $\frac { \mathrm { q } } { 4 \pi \varepsilon _ { 0 } }$ $\left\lbrack \frac{1}{R} - \frac{1}{\sqrt{R^{2} + d^{2}}} \right\rbrack$

Answer 1

Answer: (B)

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

Answer 2

V_A = $\frac { \mathrm { kq } } { \mathrm { R } }$ - $\frac { \mathrm { kq } } { \sqrt { \mathrm { R } ^ { 2 } + \mathrm { d } ^ { 2 } } }$

\f0 V_A - V_B = -

\f0 V_A - V_B = $\frac { \mathrm { q } } { 2 \pi \varepsilon _ { 0 } }$ $\left[ \frac { 1 } { \mathrm { R } } - \frac { 1 } { \sqrt { \mathrm { R } ^ { 2 } + \mathrm { d } ^ { 2 } } } \right]$