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(C) $1$. Point $A$ is located inside the material of the conducting sphere. For an electrostatic condition, the electric field inside the material of

Vedclass · Accepted Answer

$E_A = 0$, $E_B > E_C$

Vedclass · Answer

(C) $1$. Point $A$ is located inside the material of the conducting sphere. For an electrostatic condition, the electric field inside the material of a conductor is always zero. Thus, $E_A = 0$.$2$. The charge $Q$ inside the cavity induces a charge $-Q$ on the inner surface of the cavity and a charge $+Q$ on the outer surface of the conducting sphere. $3$. Since the sphere is isolated and neutral, the total charge on the outer surface is $+Q$. This charge $+Q$ distributes itself on the outer surface of the sphere. Because the cavity is off-center, the charge distribution on the outer surface is non-uniform.$4$. The electric field at any point outside the sphere is due to the total charge $+Q$ on the outer surface. Since the outer surface is not spherically symmetric with respect to the center of the sphere, the electric field at points $B$ and $C$ (which are at the same distance from the center) will be different due to the non-uniform distribution of charge on the outer surface.$5$. Specifically, point $B$ is closer to the region of the outer surface where the induced charge density is higher (due to the proximity of the internal charge $Q$), resulting in a stronger electric field at $B$ compared to $C$. Therefore, $E_B > E_C$.

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