Verified Solution • 2025–26 Session moderate

QJEEM23Ph3db1: JEE Main 2023 Physics - Class 12 JEEM23Ph3db1 Solution: A point charge is placed at origin. Another point charge is…

A point charge q1=4q0q_1=4q_0 is placed at origin. Another point charge q2=−q0q_2=-q_0 is placed at x=12x=12 cm. Charge of proton is q0q_0 . The proton is placed on xx axis so that the electrostatic force on the proton is zero. In this situation, the position of the proton from the origin is ___________ cm.

Step-by-Step Verified Solution

JEE Main 2023 (Online) 29th January Morning Shift Physics - Electrostatics Question 72 English Explanation
Let a proton having charge q0q_0 on the xx axis at distance xx from q2q_2 and (12+x)(12+x) distance from q1q_1 .

Now, balance the force between them F⃗1+F⃗2=0\vec{F}_1+\vec{F}_2=0

K4q0(q0)(12+x)2+K(−q0)(q0)x2=0\frac{K 4 q_0\left(q_0\right)}{(12+x)^2}+\frac{K\left(-q_0\right)\left(q_0\right)}{x^2}=0

⇒\Rightarrow 4Kq02(12+x)2=K(q0)2x2\frac{4 K q_0^2}{(12+x)^2}=\frac{K\left(q_0\right)^2}{x^2}

⇒\Rightarrow 4(12+x)2=1x2\frac{4}{(12+x)^2}=\frac{1}{x^2}

⇒\Rightarrow 212+x=1x\frac{2}{12+x}=\frac{1}{x}

⇒\Rightarrow 2x=12+x2 x=12+x

⇒\Rightarrow x=12 cmx=12 \mathrm{~cm}

The charge q0q_0 is at distance of 24 cm24 \mathrm{~cm} from charge q1q_1 on xx -axis.

∴\therefore Distance from origin is 12+12=24 cm12+12=24 \mathrm{~cm}

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring