Three point charges of… - JEE Main 2022 Physics PYQ

JEE Main 2022 — Physics Electromagnetism

2022integermedium

Three point charges of magnitude $5\mu C,0.16\mu C$ and $0.3\mu C$ are located at the vertices $A,B,C$ of a right angled triangle whose sides are $AB=3\mathrm{cm}$ , $BC=3\sqrt{2}\mathrm{cm}$ and $CA=3\mathrm{cm}$ and point $A$ is the right angle corner. Charge at point $A$ experiences _____ $N$ of electrostatic force due to the other two charges.

Official previous-year question

Held on 26 Jul 2022 · Verified 6 Jul 2026.

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Solution

JEE Main 2022 Physics, Electrostatics — solution figure

Using Coulomb's law the electrostatic force on $A$ due to $C$

${F}_{AC}=\frac{K{q}_{A}{q}_{C}}{{({r}_{AC})}^{2}}=\frac{k\times 5\times 0.3\times {10}^{-12}}{9\times {10}^{-4}}$ $=\frac{9\times {10}^{9}\times 5\times 0.3\times {10}^{-12}}{9\times {10}^{-4}}$

$=1.5\times 10=15N$

Similarly, the force on $A$ due to $B$

${F}_{AB}=\frac{K{q}_{A}{q}_{B}}{{({r}_{AB})}^{2}}=\frac{9\times {10}^{9}\times 5\times 0.16\times {10}^{-12}}{9\times {10}^{-4}}=8N$

We can see that the forces are perpendicular to each other. Therefore, the net force experienced by charge at $A$ is,

$F=\sqrt{{{F}_{AC}}^{2}+{{F}_{AB}}^{2}}$ $=\sqrt{{15}^{2}+{8}^{2}}=\sqrt{289}=17N$

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JEE Main 2022 — Physics Electromagnetism

2022integermedium

Official previous-year question

Held on 26 Jul 2022 · Verified 6 Jul 2026.

Did you get this right?

Solution

JEE Main 2022 Physics, Electrostatics — solution figure

Using Coulomb's law the electrostatic force on $A$ due to $C$

${F}_{AC}=\frac{K{q}_{A}{q}_{C}}{{({r}_{AC})}^{2}}=\frac{k\times 5\times 0.3\times {10}^{-12}}{9\times {10}^{-4}}$ $=\frac{9\times {10}^{9}\times 5\times 0.3\times {10}^{-12}}{9\times {10}^{-4}}$

$=1.5\times 10=15N$

Similarly, the force on $A$ due to $B$

${F}_{AB}=\frac{K{q}_{A}{q}_{B}}{{({r}_{AB})}^{2}}=\frac{9\times {10}^{9}\times 5\times 0.16\times {10}^{-12}}{9\times {10}^{-4}}=8N$

We can see that the forces are perpendicular to each other. Therefore, the net force experienced by charge at $A$ is,

$F=\sqrt{{{F}_{AC}}^{2}+{{F}_{AB}}^{2}}$ $=\sqrt{{15}^{2}+{8}^{2}}=\sqrt{289}=17N$

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Solution

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Solution

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JEE Main Electromagnetism in other years

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Other Physics topics for JEE Main

Mechanics

Modern Physics

Optics

Thermodynamics

Waves & Oscillations

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