NTA Abhyas JEE Main2020PhysicsElectrostaticsPractice
A solid ball of radius R has a charge density ρ given by ρ = ρ o 1 - r R for 0 ≤ r ≤ R . The electric field outside the ball is
Options
- Aρ 0 R 3 ε 0 r 2
- B4 ρ 0 R 3 3 ε 0 r 2
- C3 ρ 0 R 3 4 ε 0 r 2
- Dρ 0 R 3 12 ε 0 r 2
Correct answer
D. ρ 0 R 3 12 ε 0 r 2
Step-by-step solution
ρ = ρ 0 1 - r R d q = ρ d v q i n = ∫ d q = ρ d v ⇒ q in = ∫ 0 R ρ 0 1 - r R 4 π r 2 d r = 4 π ρ 0 ∫ 0 R 1 - r R r 2 d r = 4 π ρ 0 ∫ 0 R r 2 d r - r 3 R d r = 4 π ρ 0 r 3 3 0 R - r 4 4 R 0 R = 4 π ρ 0   R 3 3 - R 4 4 R = 4 π ρ 0 R 3 12 ⇒ q in = π ρ 0 R 3 3 E × 4 π r 2 = π ρ 0 R 3 3 ε 0 ⇒ E = ρ 0 R 3 12 ε 0 r 2