Consider a sphere of radius R which carries a uniform charge density ρ. If a sphere of radius R2 is carved out of it, as shown, the ratio |→EA||→EB| of magnitude of electric field
|→EA| and |→EB|, respectively, at points A and B due to the remaining portion is
|EA|=0+(14πε0)ρ.43π(R2)3(R2)2
|EA|=(14πε0)ρ43π(R2)...(1)
E+B = Electric field due to sphere of radius R + Electric field due to cavity Again, assuming the charge to be concentrated at the centre
Electric field at B is given by
|EB|=E1−E2
|EB|=(14πε0)ρ.43πR3R2
−(14πε0)ρ.43π(R2)3(3R2)2
=(14πε0)ρ43πR−(14πε0)ρ.43π818
|EB|=(14πε0)ρ.43π(17R18)...(2)
[using (1) and (2)]
EAEB=917=1834
Hence, option (d) is correct.