A spherical metal shell A of radius R_A and a | vedclass

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(B) When two conductors are connected by a wire,charge flows until their potentials become equal. Let the final charges be $Q_A$ and $Q_B$.Since $V_A

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$(A, B, C, D)$

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(B) When two conductors are connected by a wire,charge flows until their potentials become equal. Let the final charges be $Q_A$ and $Q_B$.Since $V_A = V_B$,we have $\frac{kQ_A}{R_A} = \frac{kQ_B}{R_B}$,which implies $\frac{Q_A}{Q_B} = \frac{R_A}{R_B}$.Since $R_A > R_B$,it follows that $Q_A > Q_B$. Thus,$(B)$ is correct.For a spherical shell,the electric field inside is always zero due to Gauss's Law. Thus,$(A)$ is correct.The surface charge density is $\sigma = \frac{Q}{4\pi R^2}$. Therefore,$\frac{\sigma_A}{\sigma_B} = \frac{Q_A}{4\pi R_A^2} \cdot \frac{4\pi R_B^2}{Q_B} = \frac{Q_A}{Q_B} \cdot \frac{R_B^2}{R_A^2} = \frac{R_A}{R_B} \cdot \frac{R_B^2}{R_A^2} = \frac{R_B}{R_A}$. Thus,$(C)$ is correct.The electric field on the surface is $E = \frac{\sigma}{\epsilon_0}$. Since $\frac{\sigma_A}{\sigma_B} = \frac{R_B}{R_A} Therefore,all options $(A, B, C, D)$ are correct.

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