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(C) The potential at any point on a conducting spherical shell is the sum of the potentials due to all the charges present on the shells.For a point a

Vedclass · Accepted Answer

$\frac{1}{4 \pi \epsilon_0}\left(\frac{q_1}{a}+\frac{q_2}{b}+\frac{q_3}{c}\right), \frac{1}{4 \pi \epsilon_0}\left(\frac{q_1+q_2}{b}+\frac{q_3}{c}\right), \frac{1}{4 \pi \epsilon_0}\left(\frac{q_1+q_2+q_3}{c}\right)$

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(C) The potential at any point on a conducting spherical shell is the sum of the potentials due to all the charges present on the shells.For a point at distance $r$ from the center,the potential due to a shell of radius $R$ and charge $q$ is $\frac{kq}{R}$ if $r \le R$ and $\frac{kq}{r}$ if $r > R$.For sphere $A$ (radius $a$): It is inside $B$ and $C$,so the potential is $V_A = \frac{kq_1}{a} + \frac{kq_2}{b} + \frac{kq_3}{c} = \frac{1}{4 \pi \epsilon_0} \left( \frac{q_1}{a} + \frac{q_2}{b} + \frac{q_3}{c} \right)$.For sphere $B$ (radius $b$): It is on the surface of $B$,inside $C$,and outside $A$. So,$V_B = \frac{kq_1}{b} + \frac{kq_2}{b} + \frac{kq_3}{c} = \frac{1}{4 \pi \epsilon_0} \left( \frac{q_1+q_2}{b} + \frac{q_3}{c} \right)$.For sphere $C$ (radius $c$): It is on the surface of $C$,and outside $A$ and $B$. So,$V_C = \frac{kq_1}{c} + \frac{kq_2}{c} + \frac{kq_3}{c} = \frac{1}{4 \pi \epsilon_0} \left( \frac{q_1+q_2+q_3}{c} \right)$.

There are three concentric conducting spherical shells $A$,$B$,and $C$ of radii $a$,$b$,and $c$ respectively. The potentials of the spheres $A$,$B$,and $C$ respectively are:

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