Two identical thin rings, each of radius R m | vedclass

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Question

work done $W=q\left(V_{1}-V_{2}\right)$

potential at $A$ is

$V_{1}=\frac{1}{4 \pi \varepsilon_{0}}\left(\frac{Q_{1}}{R}+\frac{Q_{2}}{\sqrt{2} R}

Vedclass · Accepted Answer

$q$$\,\left( {{Q_1}\, - \,{Q_2}} \right)\left( {\sqrt 2 \, - \,1} \right)/\left( {\sqrt {2\,} \,.\,4\pi {\varepsilon _0}R} \right)$

Vedclass · Answer

work done $W=q\left(V_{1}-V_{2}\right)$

potential at $A$ is

$V_{1}=\frac{1}{4 \pi \varepsilon_{0}}\left(\frac{Q_{1}}{R}+\frac{Q_{2}}{\sqrt{2} R}\right)$

potential at $\mathrm{B}$ is

$V_{2}=\frac{1}{4 \pi \varepsilon_{0}}\left(\frac{Q_{2}}{R}+\frac{Q_{1}}{\sqrt{2} R}\right)$

$W=q\left(V_{1}-V_{2}\right)$

$W=\frac{q}{4 \pi \varepsilon_{0}}\left[\left(\frac{Q_{1}}{R}+\frac{Q_{2}}{\sqrt{2} R}\right)-\left(\frac{Q_{2}}{R}+\frac{Q_{1}}{\sqrt{2} R}\right)\right]$

$W=\frac{q\left(Q_{1}-Q_{2}\right)(\sqrt{2}-1)}{4 \pi \varepsilon_{0} \sqrt{2} R}$

Two identical thin rings, each of radius $R $ meter are coaxially placed at distance $R$ meter apart. If $Q_1$ and $Q_2$ coulomb are respectively the charges uniformly spread on the two rings, the work done in moving a charge $q$ from the centre of one ring to that of the other is

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