A spherical shell with an inner radius a and | vedclass

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(B) $1$. According to the properties of a conductor in electrostatic equilibrium,the electric field inside the material of the conductor must be zero.

Vedclass · Accepted Answer

$-Q$ on inner surface,$(-q + Q)$ on outer surface

Vedclass · Answer

(B) $1$. According to the properties of a conductor in electrostatic equilibrium,the electric field inside the material of the conductor must be zero.$2$. Consider a Gaussian surface inside the conducting material of the shell. By Gauss's Law,the total charge enclosed by this surface must be zero.$3$. The total charge enclosed is the sum of the point charge $+Q$ at the center and the induced charge $q_{in}$ on the inner surface of the shell. Thus,$+Q + q_{in} = 0$,which implies $q_{in} = -Q$.$4$. The shell has a total charge of $-q$. Let $q_{out}$ be the charge on the outer surface. Since the total charge is the sum of the charges on the inner and outer surfaces,we have $q_{in} + q_{out} = -q$.$5$. Substituting $q_{in} = -Q$,we get $-Q + q_{out} = -q$,which gives $q_{out} = -q + Q$.$6$. Therefore,the charge on the inner surface is $-Q$ and the charge on the outer surface is $(-q + Q)$.

$A$ spherical shell with an inner radius $a$ and an outer radius $b$ is made of conducting material. $A$ point charge $+Q$ is placed at the centre of the spherical shell and a total charge $-q$ is placed on the shell. Find the final charge distribution on the surfaces.

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