JEE Main 2022 — Physics Electromagnetism

2022mcqmedium

Two point charges $Q$ each are placed at a distance $d$ apart. A third point charge $q$ is placed at a distance $x$ from mid-point on the perpendicular bisector. The value of $x$ at which charge $q$ will experience the maximum Coulomb's force is:

Official previous-year question

Held on 29 Jun 2022 · Verified 6 Jul 2026.

Options

A
$d$
B
$\frac{d}{2}$
C
$\frac{d}{\sqrt{2}}$
D
$\frac{d}{2\sqrt{2}}$

Did you get this right?

Solution

JEE Main 2022 Physics, Electrostatics — solution figure

The distance between $q$ and $Q$ is $r=\sqrt{{x}^{2}+{(\frac{d}{2})}^{2}}=\sqrt{{x}^{2}+\frac{{d}^{2}}{4}}$ .

The force between $q$ and $Q$ will be $F=\frac{kqQ}{{r}^{2}}$ .

From the given figure we can see that only vertical components of the forces will add, and the horizontal components will get cancelled out. Therefore,

${F}_{net}=\frac{2kqQ}{{r}^{2}}\mathrm{cos}\theta =\frac{2kqQ}{{r}^{2}}\times \frac{x}{r}=\frac{kqQx}{{({x}^{2}+\frac{{d}^{2}}{4})}^{\frac{3}{2}}}$

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

$\Rightarrow kqQ[\frac{{({x}^{2}+\frac{{d}^{2}}{4})}^{\frac{3}{2}}-x\times \frac{3}{2}{({x}^{2}+\frac{{d}^{2}}{4})}^{\frac{1}{2}}\times 2x}{{({({x}^{2}+\frac{{d}^{2}}{4})}^{\frac{3}{2}})}^{2}}]=0$

$\Rightarrow ({x}^{2}+\frac{{d}^{2}}{4})=3{x}^{2}\Rightarrow 2{x}^{2}=\frac{{d}^{2}}{4}\Rightarrow x=\frac{d}{2\sqrt{2}}$

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JEE Main 2022 — Physics Electromagnetism

2022mcqmedium

Official previous-year question

Held on 29 Jun 2022 · Verified 6 Jul 2026.

Options

A
$d$
B
$\frac{d}{2}$
C
$\frac{d}{\sqrt{2}}$
D
$\frac{d}{2\sqrt{2}}$

Did you get this right?

Solution

JEE Main 2022 Physics, Electrostatics — solution figure

The distance between $q$ and $Q$ is $r=\sqrt{{x}^{2}+{(\frac{d}{2})}^{2}}=\sqrt{{x}^{2}+\frac{{d}^{2}}{4}}$ .

The force between $q$ and $Q$ will be $F=\frac{kqQ}{{r}^{2}}$ .

From the given figure we can see that only vertical components of the forces will add, and the horizontal components will get cancelled out. Therefore,

${F}_{net}=\frac{2kqQ}{{r}^{2}}\mathrm{cos}\theta =\frac{2kqQ}{{r}^{2}}\times \frac{x}{r}=\frac{kqQx}{{({x}^{2}+\frac{{d}^{2}}{4})}^{\frac{3}{2}}}$

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

$\Rightarrow kqQ[\frac{{({x}^{2}+\frac{{d}^{2}}{4})}^{\frac{3}{2}}-x\times \frac{3}{2}{({x}^{2}+\frac{{d}^{2}}{4})}^{\frac{1}{2}}\times 2x}{{({({x}^{2}+\frac{{d}^{2}}{4})}^{\frac{3}{2}})}^{2}}]=0$

$\Rightarrow ({x}^{2}+\frac{{d}^{2}}{4})=3{x}^{2}\Rightarrow 2{x}^{2}=\frac{{d}^{2}}{4}\Rightarrow x=\frac{d}{2\sqrt{2}}$

JEE Main 2022 — Physics Electromagnetism

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JEE Main 2022 — Physics Electromagnetism

Options

Did you get this right?

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