Two equal positive point… - JEE Main 2023 Physics PYQ

JEE Main 2023 — Physics Electromagnetism

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Two equal positive point charges are separated by a distance $2a$ . The distance of a point from the centre of the line joining two charges on the equatorial line (perpendicular bisector) at which force experienced by a test charge ${q}_{0}$ becomes maximum is $\frac{a}{\sqrt{x}}$ . The value of $x$ is ______.

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Held on 1 Feb 2023 · Verified 6 Jul 2026.

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Solution

JEE Main 2023 Physics, Electrostatics — solution figure

Let us assume at distance $y$ , force is maximum.

Now, force at a distance $y$ is given by, ${F}_{net}=2F\mathrm{cos}\theta$

So, ${F}_{net}=\frac{2Kq{q}_{0}y}{{({y}^{2}+{a}^{2})}^{\frac{3}{2}}}$

For ${F}_{net}$ to be maximum, $\frac{d{F}_{net}}{dy}=0$

Or $Kq{q}_{0}\frac{{({y}^{2}+{a}^{2})}^{\frac{3}{2}}-y\frac{3}{2}\times 2y{({y}^{2}+{a}^{2})}^{\frac{1}{2}}}{{({y}^{2}+{a}^{2})}^{3}}=0$

Simplifying, we get $y=\frac{a}{\sqrt{2}}$ .

Hence, the value of $x=2$ .

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Solution

JEE Main 2023 Physics, Electrostatics — solution figure

Let us assume at distance $y$ , force is maximum.

Now, force at a distance $y$ is given by, ${F}_{net}=2F\mathrm{cos}\theta$

So, ${F}_{net}=\frac{2Kq{q}_{0}y}{{({y}^{2}+{a}^{2})}^{\frac{3}{2}}}$

For ${F}_{net}$ to be maximum, $\frac{d{F}_{net}}{dy}=0$

Or $Kq{q}_{0}\frac{{({y}^{2}+{a}^{2})}^{\frac{3}{2}}-y\frac{3}{2}\times 2y{({y}^{2}+{a}^{2})}^{\frac{1}{2}}}{{({y}^{2}+{a}^{2})}^{3}}=0$

Simplifying, we get $y=\frac{a}{\sqrt{2}}$ .

Hence, the value of $x=2$ .

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Solution

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Solution

Your note

JEE Main Electromagnetism in other years

More JEE Main Electromagnetism questions

Other Physics topics for JEE Main

Mechanics

Modern Physics

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Electromagnetism