A spherical conducting shell of inner radius | vedclass

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(A) $1$. According to the principle of electrostatic induction,when a charge $-q$ is placed at the center of a conducting shell,an equal and opposite

Vedclass · Accepted Answer

$\frac{q}{4 \pi r_1^2}$ and $\frac{Q-q}{4 \pi r_2^2}$

Vedclass · Answer

(A) $1$. According to the principle of electrostatic induction,when a charge $-q$ is placed at the center of a conducting shell,an equal and opposite charge $+q$ is induced on the inner surface of the shell to ensure the electric field inside the conductor is zero.$2$. The inner surface of the shell now has a charge of $+q$. The surface charge density on the inner surface is given by $\sigma_{inner} = \frac{q}{4 \pi r_1^2}$.$3$. Since the shell has a total charge $Q$ and the inner surface has a charge $+q$,the charge on the outer surface must be $Q_{outer} = Q - q$ to conserve the total charge.$4$. The surface charge density on the outer surface is given by $\sigma_{outer} = \frac{Q-q}{4 \pi r_2^2}$.$5$. Therefore,the surface charge densities are $\frac{q}{4 \pi r_1^2}$ and $\frac{Q-q}{4 \pi r_2^2}$.

$A$ spherical conducting shell of inner radius $r_1$ and outer radius $r_2$ has a net charge $Q$. $A$ point charge $-q$ is placed at the center of the shell. Determine the surface charge density on the inner and outer surfaces of the shell.

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