Electrostatics

JEE Main level, numerical answer, medium

288 of 311

Question 288 of 311

The volume charge density of a sphere of radius 6m6m is 2μCcm−32\mu C{cm}^{-3}. The number of lines of force per unit surface area coming out from the surface of the sphere is _____ ×1010NC−1.\times {10}^{10}N{C}^{-1}.
[Given : Permittivity of vacuum ϵ0=8.85×10−12C2N−1−m−2{ϵ}_{0}=8.85\times {10}^{-12}{C}^{2}{N}^{-1}-{m}^{-2}]

Show answer and solution

Answer: 45

Given here, ρ=2μCcm−3\rho =2\mu C{cm}^{-3} and R=6mR=6m.
Number of lines of force per unit area == electric field at surface=KQR2=\dfrac{KQ}{{R}^{2}}
E=14πϵ0ρ43πR3R2E=\dfrac{1}{4\pi {\epsilon }_{0}}\dfrac{\rho \dfrac{4}{3}\pi {R}^{3}}{{R}^{2}}
E=ρR3ϵ0=2×63×8.85×10−12=0.45×1012NC−1E=\dfrac{\rho R}{3{ϵ}_{0}}=\dfrac{2\times 6}{3\times 8.85\times {10}^{-12}}=0.45\times {10}^{12}N{C}^{-1}
=45×1010NC−1=45\times {10}^{10}N{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.