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Question

(A) The surface charge densities are $\sigma_A = +\sigma$,$\sigma_B = -\sigma$,and $\sigma_C = +\sigma$. The charges on the shells are $Q_A = 4\pi a^2

Vedclass · Accepted Answer

$\frac{\sigma}{\epsilon_0}(a-b+c)$

Vedclass · Answer

(A) The surface charge densities are $\sigma_A = +\sigma$,$\sigma_B = -\sigma$,and $\sigma_C = +\sigma$. The charges on the shells are $Q_A = 4\pi a^2 \sigma$,$Q_B = 4\pi b^2 (-\sigma) = -4\pi b^2 \sigma$,and $Q_C = 4\pi c^2 \sigma$.The potential at the surface of shell $A$ is the sum of potentials due to all three shells: $V_A = V_{A,A} + V_{A,B} + V_{A,C}$.Since $A$ is inside $B$ and $C$,the potential due to $B$ and $C$ at the surface of $A$ is equal to the potential at their respective surfaces: $V_A = \frac{1}{4\pi\epsilon_0} \left( \frac{Q_A}{a} + \frac{Q_B}{b} + \frac{Q_C}{c} \right)$.Substituting the values: $V_A = \frac{1}{4\pi\epsilon_0} \left( \frac{4\pi a^2 \sigma}{a} + \frac{-4\pi b^2 \sigma}{b} + \frac{4\pi c^2 \sigma}{c} \right)$.$V_A = \frac{1}{4\pi\epsilon_0} (4\pi a \sigma - 4\pi b \sigma + 4\pi c \sigma)$.$V_A = \frac{\sigma}{\epsilon_0} (a - b + c)$.

Three concentric charged metallic spherical shells $A$,$B$,and $C$ have radii $a$,$b$,and $c$ (where $a < b < c$) and surface charge densities $+\sigma$,$-\sigma$,and $+\sigma$ respectively. The potential $V_A$ at the surface of shell $A$ is ($\epsilon_0$ = permittivity of free space).

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