JEE Main 2019 — Physics Electromagnetism

2019mcqhard

A charge $Q$ is distributed over three concentric spherical shells of radii $a, b, c (a<b<c)$ such that their surface charge densities are equal to one another.

The total potential at a point at distance $r$ from their common centre, where $r<a,$ would be:

Official previous-year question

Held on 10 Jan 2019 · Verified 6 Jul 2026.

Options

A
$\frac{Q}{4\pi {\epsilon }_{0}(a+b+c)}$
B
$\frac{Q ({a}^{2}+{b}^{2}+{c}^{2})}{4\pi {\epsilon }_{0}({a}^{3}+{b}^{3}+{c}^{3})}$
C
$\frac{Q}{12\pi {\epsilon }_{0}}\frac{ab+bc+ca}{abc}$
D
$\frac{Q(a+b+c)}{4\pi {\epsilon }_{0}({a}^{2}+{b}^{2}+{c}^{2})}$

Did you get this right?

Solution

Surface charge densities of all sphere is given as $\frac{{q}_{1}}{{a}^{2}}=\frac{{q}_{2}}{{b}^{2}}=\frac{{q}_{3}}{{c}^{2}}=constant$ , where ${q}_{1}$ , ${q}_{2}$ and ${q}_{3}$ is charge on sphere of radius $a$ , $b$ and $c$ respectively.

${q}_{1}+{q}_{2}+{q}_{3}=Q$

${q}_{1}=\frac{{a}^{2}(Q)}{{a}^{2}+{b}^{2}+{c}^{2}}$

${q}_{2}=\frac{{b}^{2}(Q)}{{a}^{2}+{b}^{2}+{c}^{2}}$

${q}_{3}=\frac{{c}^{2}(Q)}{{a}^{2}+{b}^{2}+{c}^{2}}$

Potential at a point where $r<a$ is given as $V=\frac{K{q}_{1}}{a}+\frac{K{q}_{2}}{b}+\frac{K{q}_{3}}{c}$ , where $K$ is a constant.

$V=\frac{1}{4\pi {\epsilon }_{0}}[Q](\frac{a+b+c}{{a}^{2}+{b}^{2}+{c}^{2}})$

Your note

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Options

$\frac{Q}{4\pi {\epsilon }_{0}(a+b+c)}$

$\frac{Q ({a}^{2}+{b}^{2}+{c}^{2})}{4\pi {\epsilon }_{0}({a}^{3}+{b}^{3}+{c}^{3})}$

$\frac{Q}{12\pi {\epsilon }_{0}}\frac{ab+bc+ca}{abc}$

$\frac{Q(a+b+c)}{4\pi {\epsilon }_{0}({a}^{2}+{b}^{2}+{c}^{2})}$

Solution

${q}_{1}+{q}_{2}+{q}_{3}=Q$

${q}_{1}=\frac{{a}^{2}(Q)}{{a}^{2}+{b}^{2}+{c}^{2}}$

${q}_{2}=\frac{{b}^{2}(Q)}{{a}^{2}+{b}^{2}+{c}^{2}}$

${q}_{3}=\frac{{c}^{2}(Q)}{{a}^{2}+{b}^{2}+{c}^{2}}$

Potential at a point where $r<a$ is given as $V=\frac{K{q}_{1}}{a}+\frac{K{q}_{2}}{b}+\frac{K{q}_{3}}{c}$ , where $K$ is a constant.

$V=\frac{1}{4\pi {\epsilon }_{0}}[Q](\frac{a+b+c}{{a}^{2}+{b}^{2}+{c}^{2}})$

JEE Main 2019 — Physics Electromagnetism

Options

Did you get this right?

Solution

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JEE Main 2019 — 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