Logo
Jeeify
MCQEasyJEE 2023Gauss's Law Applications

JEE Physics 2023 Question with Solution

A capacitor of capacitance CC is charge to a potential VV. The flux of the electric field through a closed surface enclosing the positive plate of the capacitor is :

  • A

    Zero

  • B

    CVε0\frac{CV}{\varepsilon_0}

  • C

    2CVε0\frac{2CV}{\varepsilon_0}

  • D

    CV2ε0\frac{CV}{2\varepsilon_0}

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: A capacitor has capacitance CC and potential difference VV. A closed surface encloses the positive plate.

Find: The electric flux through the closed surface.

Using Gauss's law, the electric flux through any closed surface is

Φ=qε0\Phi = \frac{q}{\varepsilon_0}

where qq is the enclosed charge.

For a capacitor, the charge on each plate is

q=CVq = CV

Substituting into Gauss's law,

Φ=qε0=CVε0\Phi = \frac{q}{\varepsilon_0} = \frac{CV}{\varepsilon_0}

Therefore, the flux of the electric field through the closed surface is CVε0\frac{CV}{\varepsilon_0}. The correct option is B. The solution labels option A, but its worked result matches the second listed option.

Direct Relation

Given: The closed surface encloses the positive plate only.

Find: The flux through that surface.

The shortcut is to combine the two standard relations immediately:

Φ=qε0,q=CV\Phi = \frac{q}{\varepsilon_0}, \qquad q = CV

Hence,

Φ=CVε0\Phi = \frac{CV}{\varepsilon_0}

This works because electric flux through a closed surface depends only on the net enclosed charge, not on the detailed shape of the surface.

Common mistakes

  • Using the potential difference VV directly in Gauss's law is incorrect because flux depends on enclosed charge, not on potential. First use q=CVq = CV, then apply Φ=qε0\Phi = \frac{q}{\varepsilon_0}.

  • Taking the enclosed charge as CV2\frac{CV}{2} is incorrect. Each plate of a capacitor carries charge of magnitude CVCV, not half of it. The closed surface encloses the full positive plate charge.

  • Answering zero is incorrect because the closed surface encloses a non-zero net charge. Flux is zero only when the net enclosed charge is zero.

Practice more Gauss's Law Applications questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Start free trialMore on Gauss's Law Applications

Related questions