Electrostatics

JEE Main level, single correct, medium

117 of 311

Question 117 of 311

A positive point charge is released from rest at a distance r0{r}_{0} from a positive line charge with uniform charge density. The speed (v)(v) of the point charge, as a function of instantaneous distance rr from line charge, is proportional to

Show answer and solution

Answer: (B) v∝ln⁡⁡(rr0)v\propto \sqrt{\ln ⁡\left(\dfrac{r}{{r}_{0}}\right)}

12mv2=q∆V\dfrac{1}{2}{mv}^{2}=q∆V
12mv2=q×λ2πϵ0(ln⁡⁡(rr0))\dfrac{1}{2}{mv}^{2}=q\times \dfrac{\lambda }{2{\pi \epsilon }_{0}}\left(\ln ⁡\left(\dfrac{r}{{r}_{0}}\right)\right)
v2=qλmπϵ0(ln⁡⁡(rr0)){v}^{2}=\dfrac{q\lambda }{{m\pi \epsilon }_{0}}\left(\ln ⁡\left(\dfrac{r}{{r}_{0}}\right)\right)
v=qλmπϵ0(ln⁡⁡(rr0))v=\sqrt{\dfrac{q\lambda }{{m\pi \epsilon }_{0}}\left(\ln ⁡\left(\dfrac{r}{{r}_{0}}\right)\right)}
v∝ln⁡⁡(rr0)v\propto \sqrt{\ln ⁡\left(\dfrac{r}{{r}_{0}}\right)}

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 Field and Electric Field Lines), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.