Two identical thin rings each of radius R met | vedclass

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(B) The work done $W$ in moving a charge $q$ from the centre of the first ring $(O_1)$ to the centre of the second ring $(O_2)$ is given by $W = q(V_{

Vedclass · Accepted Answer

$\frac{q(Q_2 - Q_1)(\sqrt{2} - 1)}{\sqrt{2} \cdot 4\pi \varepsilon_0 R}$

Vedclass · Answer

(B) The work done $W$ in moving a charge $q$ from the centre of the first ring $(O_1)$ to the centre of the second ring $(O_2)$ is given by $W = q(V_{O_2} - V_{O_1})$.The potential at the centre of the first ring $(O_1)$ due to both rings is:$V_{O_1} = \frac{Q_1}{4\pi \varepsilon_0 R} + \frac{Q_2}{4\pi \varepsilon_0 \sqrt{R^2 + R^2}} = \frac{1}{4\pi \varepsilon_0 R} \left( Q_1 + \frac{Q_2}{\sqrt{2}} \right)$.The potential at the centre of the second ring $(O_2)$ due to both rings is:$V_{O_2} = \frac{Q_2}{4\pi \varepsilon_0 R} + \frac{Q_1}{4\pi \varepsilon_0 \sqrt{R^2 + R^2}} = \frac{1}{4\pi \varepsilon_0 R} \left( Q_2 + \frac{Q_1}{\sqrt{2}} \right)$.The potential difference is:$V_{O_2} - V_{O_1} = \frac{1}{4\pi \varepsilon_0 R} \left( Q_2 + \frac{Q_1}{\sqrt{2}} - Q_1 - \frac{Q_2}{\sqrt{2}} \right) = \frac{1}{4\pi \varepsilon_0 R} \left( (Q_2 - Q_1) - \frac{1}{\sqrt{2}}(Q_2 - Q_1) \right) = \frac{Q_2 - Q_1}{4\pi \varepsilon_0 R} \left( 1 - \frac{1}{\sqrt{2}} \right) = \frac{(Q_2 - Q_1)(\sqrt{2} - 1)}{\sqrt{2} \cdot 4\pi \varepsilon_0 R}$.Therefore,the work done is $W = \frac{q(Q_2 - Q_1)(\sqrt{2} - 1)}{\sqrt{2} \cdot 4\pi \varepsilon_0 R}$.

Two identical thin rings each of radius $R$ meters are coaxially placed at a distance $R$ meters apart. If $Q_1$ coulomb 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 other is

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