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Question

(A) Let $\sigma$ be the surface charge density. Since $\sigma_1 = \sigma_2$,we have $\frac{Q_1}{4 \pi R^2} = \frac{Q_2}{4 \pi (4R)^2}$,which implies $

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$\frac{3 Q_1}{16 \pi \varepsilon_0 R}$

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(A) Let $\sigma$ be the surface charge density. Since $\sigma_1 = \sigma_2$,we have $\frac{Q_1}{4 \pi R^2} = \frac{Q_2}{4 \pi (4R)^2}$,which implies $Q_2 = 16 Q_1$.The potential at the surface of the inner sphere $(r=R)$ is $V(R) = \frac{k Q_1}{R} + \frac{k Q_2}{4R}$.The potential at the surface of the outer sphere $(r=4R)$ is $V(4R) = \frac{k Q_1}{4R} + \frac{k Q_2}{4R}$.The potential difference is $V(R) - V(4R) = (\frac{k Q_1}{R} + \frac{k Q_2}{4R}) - (\frac{k Q_1}{4R} + \frac{k Q_2}{4R}) = \frac{k Q_1}{R} - \frac{k Q_1}{4R} = \frac{3 k Q_1}{4R}$.Substituting $k = \frac{1}{4 \pi \varepsilon_0}$,we get $V(R) - V(4R) = \frac{3 Q_1}{16 \pi \varepsilon_0 R}$.

Concentric metallic hollow spheres of radii $R$ and $4R$ hold charges $Q_1$ and $Q_2$ respectively. Given that surface charge densities of the concentric spheres are equal,the potential difference $V(R) - V(4R)$ is

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