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There are two thin wire rings, each of radius $R$, whose axes coincide. The charges of the rings are $q$ and $-\mathrm{q}$. The magnitude of potential difference between the centres of the rings separated by a distance $\sqrt{3} \mathrm{R}$ is
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$\frac{\mathrm{q}}{4 \pi \varepsilon_0 \mathrm{R}}$
$\begin{aligned} & \mathrm{V}_{\mathrm{A}}=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}}{\mathrm{R}}-\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}}{\sqrt{\mathrm{R}^2+3 \mathrm{R}^2}} \\ & \mathrm{~V}_{\mathrm{B}}=\frac{-1}{4 \pi \varepsilon_0} \frac{\mathrm{q}}{\mathrm{R}}+\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}}{\sqrt{\mathrm{R}^2+3 \mathrm{R}^2}} \\ & \text { So, } \mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{B}}=\frac{1}{4 \pi \varepsilon_0} \frac{2 \mathrm{q}}{\mathrm{R}}-\frac{1}{4 \pi \varepsilon_0} \frac{2 \mathrm{q}}{2 \mathrm{R}} \\ & =\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}}{\mathrm{R}}\end{aligned}$
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