Two thin coaxial rings,each of radius a and h | vedclass

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(D) Let $V_A$ be the potential at the centre of ring $A$ and $V_B$ be the potential at the centre of ring $B$.Potential at centre $A$ due to ring $A$

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Let $V_A$ be the potential at the centre of ring $A$ and $V_B$ be the potential at the centre of ring $B$.Potential at centre $A$ due to ring $A$ is $V_{A1} = \frac{KQ}{a}$.Potential at centre $A$ due to ring $B$ is $V_{A2} = \frac{K(-Q)}{\sqrt{a^2 + s^2}}$.So,$V_A = V_{A1} + V_{A2} = \frac{KQ}{a} - \frac{KQ}{\sqrt{a^2 + s^2}}$.Potential at centre $B$ due to ring $B$ is $V_{B1} = \frac{K(-Q)}{a}$.Potential at centre $B$ due to ring $A$ is $V_{B2} = \frac{KQ}{\sqrt{a^2 + s^2}}$.So,$V_B = V_{B1} + V_{B2} = -\frac{KQ}{a} + \frac{KQ}{\sqrt{a^2 + s^2}}$.The potential difference is $V_A - V_B = \left(\frac{KQ}{a} - \frac{KQ}{\sqrt{a^2 + s^2}}\right) - \left(-\frac{KQ}{a} + \frac{KQ}{\sqrt{a^2 + s^2}}\right)$.$V_A - V_B = \frac{2KQ}{a} - \frac{2KQ}{\sqrt{a^2 + s^2}} = 2KQ \left(\frac{1}{a} - \frac{1}{\sqrt{a^2 + s^2}}\right)$.Substituting $K = \frac{1}{4 \pi \varepsilon_0}$,we get $V_A - V_B = \frac{2}{4 \pi \varepsilon_0} Q \left(\frac{1}{a} - \frac{1}{\sqrt{a^2 + s^2}}\right) = \frac{Q}{2 \pi \varepsilon_0} \left(\frac{1}{a} - \frac{1}{\sqrt{a^2 + s^2}}\right)$.

Two thin coaxial rings,each of radius $a$ and having charges $+Q$ and $-Q$ respectively,are separated by a distance $s$. The potential difference between the centres of the two rings is:

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