There are two thin wire rings,each of radius | vedclass

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(B) Let the centres of the rings be $A$ and $B$. The distance between them is $d = \sqrt{3} R$.Potential at centre $A$ due to ring $1$ (charge $q$) is

Vedclass · Accepted Answer

$\frac{q}{4 \pi \varepsilon_0 R}$

Vedclass · Answer

(B) Let the centres of the rings be $A$ and $B$. The distance between them is $d = \sqrt{3} R$.Potential at centre $A$ due to ring $1$ (charge $q$) is $V_{A1} = \frac{1}{4 \pi \varepsilon_0} \frac{q}{R}$.Potential at centre $A$ due to ring $2$ (charge $-q$) is $V_{A2} = \frac{1}{4 \pi \varepsilon_0} \frac{-q}{\sqrt{R^2 + d^2}} = \frac{1}{4 \pi \varepsilon_0} \frac{-q}{\sqrt{R^2 + 3R^2}} = \frac{1}{4 \pi \varepsilon_0} \frac{-q}{2R}$.Total potential at $A$ is $V_A = V_{A1} + V_{A2} = \frac{1}{4 \pi \varepsilon_0} (\frac{q}{R} - \frac{q}{2R}) = \frac{1}{4 \pi \varepsilon_0} \frac{q}{2R}$.Similarly,potential at centre $B$ due to ring $2$ (charge $-q$) is $V_{B2} = \frac{1}{4 \pi \varepsilon_0} \frac{-q}{R}$.Potential at centre $B$ due to ring $1$ (charge $q$) is $V_{B1} = \frac{1}{4 \pi \varepsilon_0} \frac{q}{\sqrt{R^2 + d^2}} = \frac{1}{4 \pi \varepsilon_0} \frac{q}{2R}$.Total potential at $B$ is $V_B = V_{B1} + V_{B2} = \frac{1}{4 \pi \varepsilon_0} (\frac{q}{2R} - \frac{q}{R}) = -\frac{1}{4 \pi \varepsilon_0} \frac{q}{2R}$.The magnitude of potential difference is $|V_A - V_B| = |\frac{1}{4 \pi \varepsilon_0} \frac{q}{2R} - (-\frac{1}{4 \pi \varepsilon_0} \frac{q}{2R})| = |\frac{1}{4 \pi \varepsilon_0} \frac{2q}{2R}| = \frac{q}{4 \pi \varepsilon_0 R}$.

There are two thin wire rings,each of radius $R$,whose axes coincide. The charges of the rings are $q$ and $-q$. The magnitude of potential difference between the centres of the rings separated by a distance $\sqrt{3} R$ is

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