NTA Abhyas JEE Main2020PhysicsElectrostaticsPractice
Work done by an external agent to move slowly a charge Q from rim of a uniformly charged horizontal disc of radius a and charge per unit area R , to center of this disc is -
Options
- Aσ R Q ε 0 2 π - 1 2
- Bσ R Q ε 0 1 2 - 1 π
- Cσ a Q ε 0 1 π - 1 2
- Dσ a Q ε 0 1 2 - 2 π
Correct answer
B. σ R Q ε 0 1 2 - 1 π
Step-by-step solution
2 R cos θ = r d r = − 2 R sin θ d θ when r = 0 θ = 90 o = π 2 r = 2 R θ = 0 d V = K d θ r d V = K ( σ r 2 θ ) d r r d V = 2 K σ θ d r ∫ d V = ∫ 2 K σ θ ( − 2 R sin θ d θ ) V = − 4 K σ R ∫ π 2 0 θ sin θ d θ V = 4 K σ R ∫ 0 π 2 θ sin θ d θ V = 4 K σ R [ − θ cos θ − ∫ 1 ( − cos θ ) d θ ] V = 4 K σ R [ − θ cos θ + sin θ ] 0 π 2 V = 4 K σ R = σ π ε 0 R For Rim point V R = V = σ R π ε 0 W = Q ( V f − V i ) W = Q ( V C = V R ) W = Q ( σ R 2 ε 0 − σ R π ε 0 ) W = Q σ R ε 0 ( 1 2 − 1 π )