The two thin coaxial rings, each of radius \('a'\) and having charges \(+{Q}\) and \(-{Q}\) respectively are separated by a distance of \('s'.\) The potential difference between the centres of the two rings is :

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Answer & Solution

Correct Answer

(D) \(\frac{{Q}}{2 \pi \varepsilon_{0}}\left[\frac{1}{{a}}-\frac{1}{\sqrt{{s}^{2}+{a}^{2}}}\right]\)

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Detailed explanation

\({V}_{{A}}=\frac{{KQ}}{{a}}-\frac{{KQ}}{\sqrt{{a}^{2}+{s}^{2}}}\)

\({V}_{{B}}=\frac{-{KQ}}{{a}}+\frac{{KQ}}{\sqrt{{a}^{2}+{s}^{2}}}\)

\({V}_{{A}}-{V}_{{B}}=\frac{2 {KQ}}{{a}}-\frac{2 {KQ}}{\sqrt{{a}^{2}+{s}^{2}}}\)

\(=\frac{{Q}}{2 \pi \varepsilon_{0}}\left(\frac{1}{{a}}-\frac{1}{{s}^{2}+{a}^{2}}\right)\)

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