Let a total charge 2Q be… - JEE Main 2019 Physics PYQ

JEE Main 2019 — Physics Electromagnetism

2019mcqhard

Let a total charge $2Q$ be distributed in a sphere of radius $R,$ with the charge density given by $\rho (r)=kr,$ where $r$ is the distance from the centre. Two charges $A$ and $B$ , of $-Q$ each, are placed on diametrically opposite points, at equal distance, a, from the centre. If $A$ and $B$ do not experience any force, then:

Official previous-year question

Held on 12 Apr 2019 · Verified 6 Jul 2026.

Options

A
$a=\frac{3R}{{2}^{\frac{1}{4}}}$
B
$a=\frac{R}{\sqrt{3}}$
C
$a={2}^{\frac{-1}{4}}R$
D
$a={8}^{\frac{-1}{4}}R$

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Solution

As total charge on sphere is $=2Q$

$2Q=\int _{0}^{R}(kr)(4\pi {r}^{2}dr)\Rightarrow k{\pi r}^{4}=2Q$

Charged enclosed in the sphere of radius $a$

$q=\int _{0}^{a}kr\times 4\pi {r}^{2}dr=k\pi {a}^{4}$

As force on $Q$ is zero

$\frac{1}{4\pi {\epsilon }_{0}}\times \frac{{Q}^{2}}{{(2a)}^{2}}=\frac{k\pi {a}^{4}}{4\pi {\epsilon }_{0}\times {a}^{2}}\times Q$

$\Rightarrow \frac{{Q}^{2}}{4}=k\pi {a}^{4}Q$

$={a}^{4}(\frac{2Q}{{R}^{4}})Q$

${R}^{4}=8{a}^{4}\Rightarrow R={8}^{\frac{1}{4}} a$ or $a={8}^{-\frac{1}{4}}R$

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Solution

As total charge on sphere is $=2Q$

$2Q=\int _{0}^{R}(kr)(4\pi {r}^{2}dr)\Rightarrow k{\pi r}^{4}=2Q$

Charged enclosed in the sphere of radius $a$

$q=\int _{0}^{a}kr\times 4\pi {r}^{2}dr=k\pi {a}^{4}$

As force on $Q$ is zero

$\frac{1}{4\pi {\epsilon }_{0}}\times \frac{{Q}^{2}}{{(2a)}^{2}}=\frac{k\pi {a}^{4}}{4\pi {\epsilon }_{0}\times {a}^{2}}\times Q$

$\Rightarrow \frac{{Q}^{2}}{4}=k\pi {a}^{4}Q$

$={a}^{4}(\frac{2Q}{{R}^{4}})Q$

${R}^{4}=8{a}^{4}\Rightarrow R={8}^{\frac{1}{4}} a$ or $a={8}^{-\frac{1}{4}}R$

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Options

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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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