JEE Advanced2015PhysicsElectrostaticsActual
Consider a uniform spherical charge distribution of radius R 1 centred at the origin O. In this distribution, a spherical cavity of radius R 2 , centred at P with distance O P = a = R 1 - R 2 (see figure) is made. If the electric field inside the cavity at position r → is E → ( r → ) , then the correct statement(s) is (are)
Options
- AE → is uniform, its magnitude is independent of R 2 but its direction depends on r →
- BE → is uniform, its magnitude is depends on R 2 and its direction depends on r →
- CE → is uniform, its magnitude is dependent of a but its direction depends on a →
- DE → is uniform and both its magnitude and direction depend on a →
Correct answer
D. E → is uniform and both its magnitude and direction depend on a →
Step-by-step solution
Considering any point A inside the cavity electric field can be calculated at A by using super position principle. E → A = E → A D u e t o s p h e r e w i t h o u t c a v i t y - E → A D u e t o s p h e r e o f r a d i u s R 2 & c e n t r e a t P E → A = ρ 3 ϵ 0 O A → = ρ 3 ϵ 0 P A → But O P → + P A → = O P → ∴ O A → - P A → = O P → ∴ E → A = ρ 3 ϵ 0 O P → = ρ 3 ϵ 0 a → So E.F. is independent of location of A inside the cavity.