Consider a sphere of radius R which carries a uniform charge density ρ. If a sphere of radius R/2 is carved out of it, as shown, the ratio (| E_A⃗ |)/(| E_B⃗ |) of magnitude of electric field E_A⃗ and E_B⃗ , respectively, at points A and B due to the remaining portion is :

Solution

Physics sir · 4 steps

Fill the hole with +ρ and cancel it with a -ρ sphere, then add the two simple fields.

1. Superposition. The remaining body is the same as a full sphere of density +ρ (radius R) plus a small sphere of density -ρ (radius R/2) sitting in the cavity. Each of these has a field you already know: (ρ r)/3ε₀ inside, and kQ/r² outside.
2. Locate the points. The cavity's centre C is R/2 above the big centre O, so A, the bottom of the cavity, is exactly at O. B is on the big surface, a distance R below O and 3R/2 below C. The cavity's charge is ρ × 4/3π(R/2)³ = (πρ R³)/6.

   Point	Full sphere (+ρ)	Cavity sphere (-ρ)	Net field
   A (at O)	0, since A is the centre	(ρ (R/2))/3ε₀ = (ρ R)/6ε₀, on its surface	(ρ R)/6ε₀ = (9ρ R)/54ε₀
   B (bottom)	(ρ R)/3ε₀, pointing down	1/4πε₀·(πρ R³/6)/(3R/2)² = (ρ R)/54ε₀, pointing up	(18ρ R)/54ε₀ - (ρ R)/54ε₀ = (17ρ R)/54ε₀

3. Directions at B. The full sphere pushes outward, so its field at B points down. The negative cavity pulls toward C, so its field at B points up. They oppose each other, which is why we subtract.
4. So. |E_A|/|E_B| = (9/54)/(17/54) = 9/17 = 18/34, which is option (c).

The idea to keep. A cavity in a uniform charged body is a full body plus a negative-density piece in the hole. Find each simple field separately and add them as vectors.

Answer from the NTA final answer key. Solution by PrepWiser.