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(B) $1$. The electric field inside a conducting shell due to the charges on its surfaces is zero. The charge $q_A$ at the center is surrounded by the

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The force experienced by charge $q_A$ is zero.

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(B) $1$. The electric field inside a conducting shell due to the charges on its surfaces is zero. The charge $q_A$ at the center is surrounded by the shell. The electric field at the center due to the shell is zero,so the force on $q_A$ is zero.$2$. For the charge $q_B$ at distance $c (> b)$,the electric field is produced by the total charge enclosed by a Gaussian surface of radius $c$. The total charge on the shell is $Q_{shell} = (4\pi a^2)(-\sigma) + (4\pi b^2)(+\sigma) = 4\pi\sigma(b^2 - a^2)$.$3$. The total charge enclosed by the Gaussian surface is $Q_{enclosed} = q_A + Q_{shell} = q_A + 4\pi\sigma(b^2 - a^2)$.$4$. The electric field at distance $c$ is $E = \frac{1}{4\pi\varepsilon_0} \frac{Q_{enclosed}}{c^2} = \frac{q_A + 4\pi\sigma(b^2 - a^2)}{4\pi\varepsilon_0 c^2}$.$5$. The force on $q_B$ is $F = q_B E = \frac{q_B(q_A + 4\pi\sigma(b^2 - a^2))}{4\pi\varepsilon_0 c^2}$.$6$. Given the options,if we assume the shell is neutral ($Q_{shell} = 0$,implying $\sigma b^2 = \sigma a^2$),then $F = \frac{q_A q_B}{4\pi\varepsilon_0 c^2} = \frac{k q_A q_B}{c^2}$. Thus,option $B$ and $D$ are conceptually relevant depending on the shell's net charge. However,the force on $q_A$ is always zero due to the shell's symmetry.

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