Electrostatics

JEE Main level, single correct, medium

309 of 311

Question 309 of 311

The flat base of a hemisphere of radius a with no charge inside it lies in a horizontal plane. A uniform electric field E⃗\vec{E} is applied at an angle π4\frac{\pi}{4} with the vertical direction. The electric flux through the curved surface of the hemisphere is

Show answer and solution

Answer: (B) πa2E2\frac{\pi a^2 E}{\sqrt{2}}

We know that,
ϕ=∮E⋅dS=E∮dScos⁡45∘\phi=\oint E \cdot d S=E \oint d S \cos 45^{\circ}
In case of hemisphere
ϕcurved =ϕcircular \phi_{\text {curved }}=\phi_{\text {circular }}
Therefore, ϕcurved =Eπa2⋅12=Eπa22\phi_{\text {curved }}=E \pi a^2 \cdot \frac{1}{\sqrt{2}}=\frac{E \pi a^2}{\sqrt{2}}

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.
See it move: Electric field lines simulationDrag the sliders and watch the numbers in this question change.Revise Electrostatics firstFormula sheet, solved examples and the traps students fall into.

More Physics practice

JEE Main Physics practice question on Electrostatics (Electric Flux and Gauss Law), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.