Two thin wire rings each having a radius R ar | vedclass

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Question

$\mathrm{V}_{\mathrm{A}}=\frac{\mathrm{kq}}{\mathrm{R}}-\frac{\mathrm{kq}}{\sqrt{\mathrm{R}^{2}+\mathrm{d}^{2}}}$

$\mathrm{V}_{\mathrm{B}}=\frac{-\

Vedclass · Accepted Answer

$\frac{q}{{2\pi {\varepsilon _0}}}\left[ {\frac{1}{R} - \frac{1}{{\sqrt {{R^2} + {d^2}} }}} \right]$

Vedclass · Answer

$\mathrm{V}_{\mathrm{A}}=\frac{\mathrm{kq}}{\mathrm{R}}-\frac{\mathrm{kq}}{\sqrt{\mathrm{R}^{2}+\mathrm{d}^{2}}}$

$\mathrm{V}_{\mathrm{B}}=\frac{-\mathrm{kq}}{\mathrm{R}}+\frac{\mathrm{kq}}{\sqrt{\mathrm{R}^{2}+\mathrm{d}^{2}}}$

$\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{B}}=2 \mathrm{kq}\left[\frac{1}{\mathrm{R}}-\frac{1}{\sqrt{\mathrm{R}^{2}+\mathrm{d}^{2}}}\right]$

$=\frac{\mathrm{q}}{2 \pi \varepsilon_{0}}\left[\frac{1}{\mathrm{R}}-\frac{1}{\sqrt{\mathrm{R}^{2}+\mathrm{d}^{2}}}\right]$

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

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