JEE Main 2023 · 11 April · evening · Single correct · +4 / −1
A capacitor of capacitance C is charged to a potential V. The flux of the electric field through a closed surface enclosing the positive plate of the capacitor is :
- (a) Zero
- (b) (C V)/ε₀
- (c) (C V)/(2 ε₀)
- (d) (2 C V)/ε₀
Answer and solution
Answer (b) (C V)/ε₀
Solution
Physics sir · 3 steps
Gauss only cares about the charge inside your surface, and here that is the plate's charge Q = CV.
- Charge enclosed. A capacitor charged to potential V holds +Q on the positive plate and -Q on the negative plate, with Q = CV. A surface around the positive plate alone encloses only +Q.
- Apply Gauss's law. The total flux through any closed surface depends only on the net charge inside it, not on its shape or on charges outside it.
φ = q_(enclosed)/ε₀ = Q/ε₀ = CV/ε₀
- So. The flux through the surface is CV/ε₀, which is option (b).
The idea to keep. For flux through a closed surface, first ask what charge it encloses. The field from outside charges may pass through, but it adds nothing to the net flux.
Answer from the NTA final answer key. Solution by PrepWiser.