Electrostatics

JEE Main level, numerical answer, medium

298 of 311

Question 298 of 311

A point charge of +12μC+12\mu C is at a distance 6cm6cm vertically above the centre of a square of side 12cm12cm as shown in figure. The magnitude of the electric flux through the square will be _______ ×103Nm2C−1.\times {10}^{3}N{m}^{2}{C}^{-1}.

Show answer and solution

Answer: 226

From symmetry ϕ=16(qϵ0)\phi =\dfrac{1}{6}\left(\dfrac{q}{{\epsilon }_{0}}\right)
=12×10−66×8.85×10−12=\dfrac{12\times {10}^{-6}}{6\times 8.85\times {10}^{-12}}
=225.98×103Nm2C−1=225.98\times {10}^{3}N{m}^{2}{C}^{-1}≃226×103Nm2C−1≃226\times {10}^{3}N{m}^{2}{C}^{-1}

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.