NTA Abhyas JEE Main2020PhysicsElectrostaticsPractice
A non-conducting disc of radius R is uniformly charged with surface charge density σ . A disc of radius R 2 is cut from the disc, as shown in the figure. The electric potential at centre C of large disc will be
Options
- Aπ σ R 2 ε 0
- Bσ R 2 π ε 0
- Cσ R π - 1 2 π ε 0
- Dσ R π - 1 2 ε 0
Correct answer
C. σ R π - 1 2 π ε 0
Step-by-step solution
The electric potential at the centre of the complete disc, V 1 = σ R 2 ε 0 Let us assume that the potential at the circumference of the smaller disc is V 2 r = 2 R 2 cos θ 2 d r = - 2 R 2 sin θ 2 d θ 2 ∫ d V 2 = ∫ θ = 0 π σ × r θ d r 4 π ε 0 r = σ R 2 π ε 0 = V 2 ∴ Net potential at centre C is, V = V 1 – V 2 = σ R π - 1 2 π ε 0