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The two thin coaxial rings, each of radius '

Question:

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 :

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

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

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

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

Correct Option: , 4

Solution:

$V_{A}=\frac{K Q}{a}-\frac{K Q}{\sqrt{a^{2}+s^{2}}}$

$V_{B}=\frac{-K Q}{a}+\frac{K Q}{\sqrt{a^{2}+s^{2}}}$

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

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

Ans 4

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