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Question

(A) The potential $V_{1}$ of the smaller sphere (radius $r$,charge $q$) is due to its own charge and the potential due to the outer sphere (radius $R$

Vedclass · Accepted Answer

$\frac{1}{4 \pi \varepsilon_{0}}\left(\frac{q}{r}-\frac{q}{R}\right)$

Vedclass · Answer

(A) The potential $V_{1}$ of the smaller sphere (radius $r$,charge $q$) is due to its own charge and the potential due to the outer sphere (radius $R$,charge $Q$).$V_{1} = \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{r} + \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Q}{R}$The potential $V_{2}$ of the bigger sphere (radius $R$,charge $Q$) is due to its own charge and the potential due to the inner sphere (which acts as a point charge at the center for points outside it).$V_{2} = \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Q}{R} + \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{R}$The potential difference between the spheres is $V = V_{1} - V_{2}$.$V = \left( \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{r} + \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Q}{R} \right) - \left( \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Q}{R} + \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{R} \right)$$V = \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{r} + \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Q}{R} - \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Q}{R} - \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{R}$$V = \frac{1}{4 \pi \varepsilon_{0}} \left( \frac{q}{r} - \frac{q}{R} \right)$

$A$ small conducting sphere of radius $r$ is lying concentrically inside a bigger hollow conducting sphere of radius $R$. The bigger and smaller spheres are charged with $Q$ and $q$ respectively $(Q > q)$ and are insulated from each other. The potential difference between the spheres will be

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