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(D) Let the charges on the shells be $Q_1, Q_2, Q_3$ and surface charge density be $\sigma$. Since $\sigma$ is equal for all,$\sigma = \frac{Q_1}{4\pi

Vedclass · Accepted Answer

$\frac{Q(a+b+c)}{4\pi \epsilon_0(a^2+b^2+c^2)}$

Vedclass · Answer

(D) Let the charges on the shells be $Q_1, Q_2, Q_3$ and surface charge density be $\sigma$. Since $\sigma$ is equal for all,$\sigma = \frac{Q_1}{4\pi a^2} = \frac{Q_2}{4\pi b^2} = \frac{Q_3}{4\pi c^2}$.Thus,$Q_1 = 4\pi a^2 \sigma$,$Q_2 = 4\pi b^2 \sigma$,$Q_3 = 4\pi c^2 \sigma$.The total charge $Q = Q_1 + Q_2 + Q_3 = 4\pi \sigma (a^2 + b^2 + c^2)$,so $\sigma = \frac{Q}{4\pi (a^2 + b^2 + c^2)}$.The potential at $r Substituting the values of $Q_1, Q_2, Q_3$: $V = \frac{1}{4\pi \epsilon_0} (\frac{4\pi a^2 \sigma}{a} + \frac{4\pi b^2 \sigma}{b} + \frac{4\pi c^2 \sigma}{c}) = \frac{\sigma}{\epsilon_0} (a + b + c)$.Substituting $\sigma$: $V = \frac{Q}{4\pi \epsilon_0 (a^2 + b^2 + c^2)} (a + b + c)$.

$A$ charge $Q$ is distributed over three concentric spherical shells of radii $a, b, c$ $(a < b < c)$ such that their surface charge densities are equal to one another. The total potential at a point at distance $r$ from their common centre,where $r < a$,would be

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