Two concentric hollow spheres of radii r and | vedclass

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Question

(B) Let $\sigma$ be the surface charge density. The charges on the spheres are $q_1 = 4\pi r^2 \sigma$ and $q_2 = 4\pi R^2 \sigma$.The total charge $Q

Vedclass · Accepted Answer

$\frac{1}{4\pi \varepsilon_0} \frac{(R + r)Q}{(R^2 + r^2)}$

Vedclass · Answer

(B) Let $\sigma$ be the surface charge density. The charges on the spheres are $q_1 = 4\pi r^2 \sigma$ and $q_2 = 4\pi R^2 \sigma$.The total charge $Q = q_1 + q_2 = 4\pi \sigma (r^2 + R^2)$,so $\sigma = \frac{Q}{4\pi (r^2 + R^2)}$.The potential at the center due to the inner sphere is $V_1 = \frac{1}{4\pi \varepsilon_0} \frac{q_1}{r} = \frac{1}{4\pi \varepsilon_0} \frac{4\pi r^2 \sigma}{r} = \frac{r \sigma}{\varepsilon_0}$.The potential at the center due to the outer sphere is $V_2 = \frac{1}{4\pi \varepsilon_0} \frac{q_2}{R} = \frac{1}{4\pi \varepsilon_0} \frac{4\pi R^2 \sigma}{R} = \frac{R \sigma}{\varepsilon_0}$.The total potential at the center is $V = V_1 + V_2 = \frac{\sigma}{\varepsilon_0} (r + R)$.Substituting $\sigma = \frac{Q}{4\pi (r^2 + R^2)}$,we get $V = \frac{1}{4\pi \varepsilon_0} \frac{(R + r)Q}{(R^2 + r^2)}$.

Two concentric hollow spheres of radii $r$ and $R$ $(R > r)$ have the same surface charge density $\sigma$. The total charge $Q$ is distributed between them. The electric potential at the common center is:

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