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(A) Let the charge on the inner surface be $q_{in}$ and the charge on the outer surface be $q_{out}$.Due to the point charge $+Q$ at the center, the c

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$-5Q$

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(A) Let the charge on the inner surface be $q_{in}$ and the charge on the outer surface be $q_{out}$.Due to the point charge $+Q$ at the center, the charge induced on the inner surface is $q_{in} = -Q$.The charge density on the inner surface is $\sigma_{in} = \frac{-Q}{4 \pi a^2}$.Let the total charge of the shell be $q_{shell}$. Since $q_{shell} = q_{in} + q_{out}$, we have $q_{out} = q_{shell} - q_{in} = q_{shell} + Q$.The charge density on the outer surface is $\sigma_{out} = \frac{q_{shell} + Q}{4 \pi (2a)^2} = \frac{q_{shell} + Q}{16 \pi a^2}$.For the charge densities to be equal, $\sigma_{in} = \sigma_{out}$ (considering magnitude and sign, or just magnitude as implied by the context of such problems):$\frac{-Q}{4 \pi a^2} = \frac{q_{shell} + Q}{16 \pi a^2}$.$-Q = \frac{q_{shell} + Q}{4}$.$-4Q = q_{shell} + Q$.$q_{shell} = -5Q$.

$A$ solid spherical conducting shell has inner radius $a$ and outer radius $2a$. At the center of the shell, a point charge $+Q$ is located. What must the charge of the shell be in order for the charge density on the inner and outer surfaces of the shell to be exactly equal?

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