Two concentric hollow spherical shells have r | vedclass

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(A) Let the surface charge density be $\sigma$. Since the surface charge densities are equal for both shells,we have $\sigma_1 = \sigma_2 = \sigma$.Th

Vedclass · Accepted Answer

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

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(A) Let the surface charge density be $\sigma$. Since the surface charge densities are equal for both shells,we have $\sigma_1 = \sigma_2 = \sigma$.The total charge $Q$ is the sum of charges on the two shells: $Q = q_1 + q_2$.$Q = \sigma(4 \pi r^2) + \sigma(4 \pi R^2) = 4 \pi \sigma (r^2 + R^2)$.Thus,$\sigma = \frac{Q}{4 \pi (r^2 + R^2)}$.The electric potential at the centre of a spherical shell of radius $a$ with charge $q$ is $V = \frac{1}{4 \pi \varepsilon_0} \frac{q}{a}$.For two concentric shells,the potential at the centre is the sum of potentials due to each shell:$V = V_r + V_R = \frac{1}{4 \pi \varepsilon_0} \left( \frac{q_1}{r} + \frac{q_2}{R} \right)$.Substituting $q_1 = \sigma(4 \pi r^2)$ and $q_2 = \sigma(4 \pi R^2)$:$V = \frac{1}{4 \pi \varepsilon_0} \left( \frac{\sigma 4 \pi r^2}{r} + \frac{\sigma 4 \pi R^2}{R} \right) = \frac{\sigma}{\varepsilon_0} (r + R)$.Substituting the value of $\sigma$:$V = \frac{Q}{4 \pi (r^2 + R^2) \varepsilon_0} (r + R) = \frac{Q(R + r)}{4 \pi \varepsilon_0 (R^2 + r^2)}$.

Two concentric hollow spherical shells have radii $r$ and $R$ $(R \gg r)$. $A$ charge $Q$ is distributed on them such that the surface charge densities are equal. The electric potential at the centre is

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