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Electrostatics

JEE Main level, single correct, medium

190 of 311

Question 190 of 311

The two thin coaxial rings, each of radius $a$ and having charges $+Q$ and $-Q$ respectively are separated by a distance of $s$ . The potential difference between the centres of the two rings is :

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Answer: (A) $\dfrac{Q}{2\pi {\epsilon }_{0}}\left[\dfrac{1}{a}-\dfrac{1}{\sqrt{{s}^{2}+{a}^{2}}}\right]$

{V}_{1}=\dfrac{KQ}{a}+\dfrac{-KQ}{\sqrt{{s}^{2}+{a}^{2}}}

{V}_{2}=\dfrac{KQ}{\sqrt{{s}^{2}+{a}^{2}}}+\dfrac{-KQ}{a}

{V}_{1}-{V}_{2}=\dfrac{Q}{2\pi {\epsilon }_{0}}\left[\dfrac{1}{a}-\dfrac{1}{\sqrt{{s}^{2}+{a}^{2}}}\right]

JEE Main question paper PDF

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JEE Main 2025 Apr 2 Shift 1 question paper

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

The two thin coaxial rings, each of radius aaa and having charges +Q+Q+Q and −Q-Q−Q respectively are separated by a distance of sss. The potential difference between the centres of the two rings is :

JEE Main question paper PDF

More Physics practice

The two thin coaxial rings, each of radius $a$ and having charges $+Q$ and $-Q$ respectively are separated by a distance of $s$ . The potential difference between the centres of the two rings is :