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(D) $1$. Due to the electrostatic induction,the inner surface of the spherical shell (at radius $b$) will develop an induced charge equal and opposite

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None of the above

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(D) $1$. Due to the electrostatic induction,the inner surface of the spherical shell (at radius $b$) will develop an induced charge equal and opposite to the charge on the inner sphere. Since the inner sphere has a charge of $+2Q$,the charge on the inner surface of the shell is $-2Q$.$2$. The surface charge density on the inner surface is $\sigma_{inner} = \frac{	ext{Charge}}{	ext{Area}} = \frac{-2Q}{4\pi b^2}$.$3$. The shell has a net charge of $-Q$. Let the charge on the outer surface (at radius $c$) be $q_{outer}$. Since the total charge on the shell is the sum of the charges on its inner and outer surfaces,we have: $-2Q + q_{outer} = -Q$,which gives $q_{outer} = +Q$.$4$. The surface charge density on the outer surface is $\sigma_{outer} = \frac{	ext{Charge}}{	ext{Area}} = \frac{Q}{4\pi c^2}$.$5$. Thus,the surface charge densities are $-\frac{2Q}{4\pi b^2}$ and $\frac{Q}{4\pi c^2}$. None of the given options match this result exactly,so the correct option is $D$.

$A$ solid conducting sphere of radius $a$ has a net positive charge $2Q$. $A$ conducting spherical shell of inner radius $b$ and outer radius $c$ is concentric with the solid sphere and has a net charge $-Q$. The surface charge density on the inner and outer surfaces of the spherical shell will be

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