Three concentric metallic shells A, B and C o | vedclass

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(A) The charges on the shells are given by $q = \sigma 	imes 4\pi r^2$.Thus,$q_a = \sigma(4\pi a^2)$,$q_b = -\sigma(4\pi b^2)$,and $q_c = \sigma(4\pi

Vedclass · Accepted Answer

$\frac{\sigma}{\varepsilon_0}(a - b + c), \frac{\sigma}{\varepsilon_0}\left( \frac{a^2}{b} - b + c \right)$

Vedclass · Answer

(A) The charges on the shells are given by $q = \sigma 	imes 4\pi r^2$.Thus,$q_a = \sigma(4\pi a^2)$,$q_b = -\sigma(4\pi b^2)$,and $q_c = \sigma(4\pi c^2)$.The potential at any point is the sum of potentials due to all shells.For shell $A$ (at radius $a$): $V_A = \frac{1}{4\pi\varepsilon_0} \left[ \frac{q_a}{a} + \frac{q_b}{b} + \frac{q_c}{c} ight] = \frac{1}{4\pi\varepsilon_0} \left[ \frac{\sigma(4\pi a^2)}{a} + \frac{-\sigma(4\pi b^2)}{b} + \frac{\sigma(4\pi c^2)}{c} ight] = \frac{\sigma}{\varepsilon_0} (a - b + c)$.For shell $B$ (at radius $b$): $V_B = \frac{1}{4\pi\varepsilon_0} \left[ \frac{q_a}{b} + \frac{q_b}{b} + \frac{q_c}{c} ight] = \frac{1}{4\pi\varepsilon_0} \left[ \frac{\sigma(4\pi a^2)}{b} + \frac{-\sigma(4\pi b^2)}{b} + \frac{\sigma(4\pi c^2)}{c} ight] = \frac{\sigma}{\varepsilon_0} \left( \frac{a^2}{b} - b + c ight)$.

Three concentric metallic shells $A, B$ and $C$ of radii $a, b$ and $c$ $(a < b < c)$ have surface charge densities $\sigma, -\sigma$ and $\sigma$ respectively. Find the potentials ${V_A}$ and ${V_B}$.

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