JEE Main · 2025 · 04 Apr 2025 · Shift 1 · Electrostatics

Each infinite sheet sets up a uniform field $\sigma /2{\epsilon }_0$ directed away from it, the same at every distance, while the charged sphere gives a field that falls off with distance from its centre. Points A and B lie between the two sheets, where the two sheet fields are equal and opposite and cancel, leaving only the sphere's field; since A is nearer the sphere than B, $E_A>E_B$. Points C and D lie outside both sheets, where the two sheet fields add to $\sigma /{\epsilon }_0$, and to this is added the sphere's field, whose value differs at C and D because they stand at different distances from the sphere, so $E_C\ne E_D$. Thus $E_C\ne E_D$ while $E_A>E_B$.