A system of three charges are placed as shown in the figure: ![JEE Main 2019 Physics, Electrostatics — question figure](https://prepforbharat.s3.ap-south-1.amazonaws.com/exam/615f0e999476412f48314daf/Physics/images/Electrostatics/648b5a66417cc3fb48d6647b/question_1__q_648b5a66417cc3fb48d6647b__cdn-question-pool.getmarks.app__5f6e26f6-4757-4e4d-ab08-145cb986046c-image5__4f69c0febc_final_ppt_sync.png) If $D >> d,$ the potential energy of the system is best given by:

A system of three charges are… - JEE Main 2019 Physics PYQ

JEE Main 2019 — Physics Electromagnetism

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A system of three charges are placed as shown in the figure:

JEE Main 2019 Physics, Electrostatics — question figure

If $D >> d,$ the potential energy of the system is best given by:

Official previous-year question

Held on 9 Apr 2019 · Verified 6 Jul 2026.

Options

A
$\frac{1}{4\pi {\epsilon }_{0}}[-\frac{{q}^{2}}{d}-\frac{q Q d}{{D}^{2}}]$
B
$\frac{1}{4\pi {\epsilon }_{0}}[+\frac{{q}^{2}}{d}+\frac{q Q d}{{D}^{2}}]$
C
$\frac{1}{4\pi {\epsilon }_{0}}[-\frac{{q}^{2}}{d}-\frac{q Q d}{2{D}^{2}}]$
D
$\frac{1}{4\pi {\epsilon }_{0}}[-\frac{{q}^{2}}{d}+\frac{2q Q d}{{D}^{2}}]$

Did you get this right?

Solution

$U=\frac{k(-q)(q)}{d}+\frac{k(-q)(Q)}{(\frac{d}{2}-D)}+\frac{k(q)(Q)}{(D+\frac{d}{2})}$

$=\frac{-k{q}^{2}}{d}-\frac{kQq}{(D-\frac{d}{2})}+\frac{KQq}{(D+\frac{d}{2})}$

$=\frac{-k{q}^{2}}{d}+kQq(\frac{-D-\frac{d}{2}+ D-\frac{d}{2}}{{D}^{2}-\frac{{d}^{2}}{4}})$

$=\frac{-k{q}^{2}}{d}-\frac{kQqd}{{D}^{2}-\frac{{d}^{2}}{4}}$

Since, D>>d

$U=\frac{-k{q}^{2}}{d}-\frac{kQqd}{{D}^{2}}=\frac{1}{4\pi {\epsilon }_{0}}[-\frac{{q}^{2}}{d}-\frac{q Q d}{{D}^{2}}]$

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Options

$\frac{1}{4\pi {\epsilon }_{0}}[-\frac{{q}^{2}}{d}-\frac{q Q d}{{D}^{2}}]$

$\frac{1}{4\pi {\epsilon }_{0}}[+\frac{{q}^{2}}{d}+\frac{q Q d}{{D}^{2}}]$

$\frac{1}{4\pi {\epsilon }_{0}}[-\frac{{q}^{2}}{d}-\frac{q Q d}{2{D}^{2}}]$

$\frac{1}{4\pi {\epsilon }_{0}}[-\frac{{q}^{2}}{d}+\frac{2q Q d}{{D}^{2}}]$

Solution

$U=\frac{k(-q)(q)}{d}+\frac{k(-q)(Q)}{(\frac{d}{2}-D)}+\frac{k(q)(Q)}{(D+\frac{d}{2})}$

$=\frac{-k{q}^{2}}{d}-\frac{kQq}{(D-\frac{d}{2})}+\frac{KQq}{(D+\frac{d}{2})}$

$=\frac{-k{q}^{2}}{d}+kQq(\frac{-D-\frac{d}{2}+ D-\frac{d}{2}}{{D}^{2}-\frac{{d}^{2}}{4}})$

$=\frac{-k{q}^{2}}{d}-\frac{kQqd}{{D}^{2}-\frac{{d}^{2}}{4}}$

Since, D>>d

$U=\frac{-k{q}^{2}}{d}-\frac{kQqd}{{D}^{2}}=\frac{1}{4\pi {\epsilon }_{0}}[-\frac{{q}^{2}}{d}-\frac{q Q d}{{D}^{2}}]$

Options

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Options

Did you get this right?

Solution

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JEE Main Electromagnetism in other years

More JEE Main Electromagnetism questions

Other Physics topics for JEE Main

Mechanics

Modern Physics

Optics

Thermodynamics

Waves & Oscillations

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