Consider three concentric metal shells A,B,an | vedclass

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(C) The electric potential at the surface of shell $A$ is the sum of the potentials due to all three shells $A$,$B$,and $C$.$V_A = V_a + V_b + V_c$Sin

Vedclass · Accepted Answer

$\frac{\sigma}{\varepsilon_0}(a-b+c)$

Vedclass · Answer

(C) The electric potential at the surface of shell $A$ is the sum of the potentials due to all three shells $A$,$B$,and $C$.$V_A = V_a + V_b + V_c$Since the point on the surface of shell $A$ is inside shells $B$ and $C$,the potential due to shells $B$ and $C$ at this point is equal to the potential at their respective surfaces.$V_A = \frac{1}{4 \pi \varepsilon_0} \left[ \frac{q_a}{a} + \frac{q_b}{b} + \frac{q_c}{c} \right]$Given surface charge densities: $\sigma_a = \sigma$,$\sigma_b = -\sigma$,$\sigma_c = \sigma$.The charges are $q_a = 4 \pi a^2 \sigma$,$q_b = 4 \pi b^2(-\sigma)$,and $q_c = 4 \pi c^2 \sigma$.Substituting these into the potential formula:$V_A = \frac{1}{4 \pi \varepsilon_0} \left[ \frac{4 \pi a^2 \sigma}{a} + \frac{4 \pi b^2(-\sigma)}{b} + \frac{4 \pi c^2 \sigma}{c} \right]$$V_A = \frac{1}{4 \pi \varepsilon_0} [4 \pi a \sigma - 4 \pi b \sigma + 4 \pi c \sigma]$$V_A = \frac{\sigma}{\varepsilon_0} (a - b + c)$

Consider three concentric metal shells $A$,$B$,and $C$ with radii $a$,$b$,and $c$ respectively,as shown in the figure $(a < b < c)$. Their surface charge densities are $\sigma$,$-\sigma$,and $\sigma$ respectively. Calculate the electric potential on the surface of shell $A$.

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