Two thin coaxial rings, each of radius 'a' and having charges +{Q} and -{Q} respectively are

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2. Electric Potential and Capacitance

hard

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 :

A

$\frac{{Q}}{2 \pi \varepsilon_{0}}\left[\frac{1}{{a}}+\frac{1}{\sqrt{{s}^{2}+{a}^{2}}}\right]$

B

$\frac{{Q}}{4 \pi \varepsilon_{0}}\left[\frac{1}{{a}}+\frac{1}{\sqrt{{s}^{2}+{a}^{2}}}\right]$

C

$\frac{{Q}}{4 \pi \varepsilon_{0}}\left[\frac{1}{{a}}-\frac{1}{\sqrt{{s}^{2}+{a}^{2}}}\right]$

D

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

(JEE MAIN-2021)

Solution

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

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

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

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

Standard 12

Physics

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