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(B) The electric potential at a distance $x$ from the center of a charged ring of radius $R$ along its axis is $V = \frac{q}{4\pi \epsilon_0 \sqrt{x^2

Vedclass · Accepted Answer

$\frac{q}{2\pi \epsilon_0} \left[ \frac{1}{R} - \frac{1}{\sqrt{R^2 + d^2}} \right]$

Vedclass · Answer

(B) The electric potential at a distance $x$ from the center of a charged ring of radius $R$ along its axis is $V = \frac{q}{4\pi \epsilon_0 \sqrt{x^2 + R^2}}$.Let the centers of the rings be $C_1$ and $C_2$,separated by distance $d$. The potential at $C_1$ is the sum of the potential due to the ring at $C_1$ (charge $+q$) and the potential due to the ring at $C_2$ (charge $-q$):$V_1 = \frac{q}{4\pi \epsilon_0 R} + \frac{-q}{4\pi \epsilon_0 \sqrt{d^2 + R^2}}$.The potential at $C_2$ is the sum of the potential due to the ring at $C_2$ (charge $-q$) and the potential due to the ring at $C_1$ (charge $+q$):$V_2 = \frac{-q}{4\pi \epsilon_0 R} + \frac{q}{4\pi \epsilon_0 \sqrt{d^2 + R^2}}$.The potential difference between the centers is $\Delta V = V_1 - V_2$:$\Delta V = \left( \frac{q}{4\pi \epsilon_0 R} - \frac{q}{4\pi \epsilon_0 \sqrt{d^2 + R^2}} \right) - \left( -\frac{q}{4\pi \epsilon_0 R} + \frac{q}{4\pi \epsilon_0 \sqrt{d^2 + R^2}} \right)$$\Delta V = \frac{q}{4\pi \epsilon_0} \left( \frac{1}{R} - \frac{1}{\sqrt{d^2 + R^2}} + \frac{1}{R} - \frac{1}{\sqrt{d^2 + R^2}} \right)$$\Delta V = \frac{q}{4\pi \epsilon_0} \cdot 2 \left( \frac{1}{R} - \frac{1}{\sqrt{d^2 + R^2}} \right) = \frac{q}{2\pi \epsilon_0} \left( \frac{1}{R} - \frac{1}{\sqrt{d^2 + R^2}} \right)$.

Two thin circular rings of radius $R$ are placed with their axes coinciding. The charges on the two rings are $+q$ and $-q$. The potential difference between the centers of the two rings is:

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