A spherical shell with an inner radius a and | vedclass

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(B) $1$. According to the property of conductors in electrostatic equilibrium,the electric field inside the material of the conductor must be zero.$2$

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$-Q$ on the inner surface,$-q + Q$ on the outer surface

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(B) $1$. According to the property of conductors in electrostatic equilibrium,the electric field inside the material of the conductor must be zero.$2$. By Gauss's Law,if we consider a Gaussian surface inside the conducting material,the net charge enclosed must be zero.$3$. $A$ point charge $+Q$ is placed at the center. To make the net charge inside the conducting material zero,an induced charge of $-Q$ must appear on the inner surface of the shell (at radius $a$).$4$. The total charge on the shell is given as $-q$. Let the charge on the outer surface be $q_{outer}$.$5$. Since the total charge on the shell is the sum of the charges on the inner and outer surfaces,we have: $-Q + q_{outer} = -q$.$6$. Solving for $q_{outer}$,we get: $q_{outer} = -q + Q$.$7$. Therefore,the charge distribution is $-Q$ on the inner surface and $-q + Q$ on the outer surface.

$A$ 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. Charge $-q$ is distributed on the surfaces as:

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