Two identical thin rings each of radius R tex | vedclass

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(B) Let the charges on the rings be $Q_1 = 10 	ext{ C}$ and $Q_2 = 5 	ext{ C}$. The distance between the centres $O$ and $O'$ is $R$.The potential a

Vedclass · Accepted Answer

$\frac{5 q}{4 \pi \varepsilon_0 R}\left[1-\frac{1}{\sqrt{2}}ight] 	ext{ J}$

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(B) Let the charges on the rings be $Q_1 = 10 	ext{ C}$ and $Q_2 = 5 	ext{ C}$. The distance between the centres $O$ and $O'$ is $R$.The potential at the centre $O$ of the first ring is due to its own charge $Q_1$ and the charge $Q_2$ on the second ring:$V_O = \frac{k Q_1}{R} + \frac{k Q_2}{\sqrt{R^2 + R^2}} = \frac{k Q_1}{R} + \frac{k Q_2}{\sqrt{2}R} = \frac{k}{R} \left( Q_1 + \frac{Q_2}{\sqrt{2}} ight)$The potential at the centre $O'$ of the second ring is due to its own charge $Q_2$ and the charge $Q_1$ on the first ring:$V_{O'} = \frac{k Q_2}{R} + \frac{k Q_1}{\sqrt{R^2 + R^2}} = \frac{k Q_2}{R} + \frac{k Q_1}{\sqrt{2}R} = \frac{k}{R} \left( Q_2 + \frac{Q_1}{\sqrt{2}} ight)$The potential difference is $\Delta V = V_O - V_{O'}$:$\Delta V = \frac{k}{R} \left( Q_1 + \frac{Q_2}{\sqrt{2}} - Q_2 - \frac{Q_1}{\sqrt{2}} ight) = \frac{k}{R} \left( Q_1(1 - \frac{1}{\sqrt{2}}) - Q_2(1 - \frac{1}{\sqrt{2}}) ight)$$\Delta V = \frac{k}{R} (Q_1 - Q_2) \left( 1 - \frac{1}{\sqrt{2}} ight)$Substituting $Q_1 = 10 	ext{ C}$,$Q_2 = 5 	ext{ C}$,and $k = \frac{1}{4 \pi \varepsilon_0}$:$W = q \Delta V = \frac{q}{4 \pi \varepsilon_0 R} (10 - 5) \left( 1 - \frac{1}{\sqrt{2}} ight) = \frac{5 q}{4 \pi \varepsilon_0 R} \left( 1 - \frac{1}{\sqrt{2}} ight) 	ext{ J}$.

Two identical thin rings each of radius $R \text{ m}$ are kept on the same axis at a distance of $R \text{ m}$ apart. If the charges on them are $10 \text{ C}$ and $5 \text{ C}$ respectively,calculate the work done in moving a charge '$q$' coulomb from the centre of one ring to that of another.

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