JEE Main 2020 — Physics Electromagnetism

2020mcqmedium

Concentric metallic hollow spheres of radii R and 4R hold charges ${Q}_{1}$ and ${Q}_{2}$ respectively. Given that surface charge densities of the concentric spheres are equal, the potential difference $V(R)-V(4R)$ is:

Official previous-year question

Held on 3 Sept 2020 · Verified 6 Jul 2026.

Options

A
$\frac{{Q}_{2}}{4\pi {\epsilon }_{0}R}$
B
$\frac{3{Q}_{2}}{4\pi {\epsilon }_{0}R}$
C
$\frac{3{Q}_{1}}{16\pi {\epsilon }_{0}R}$
D
$\frac{3{Q}_{1}}{4\pi {\epsilon }_{0}R}$

Did you get this right?

Solution

JEE Main 2020 Physics, Electrostatics — solution figure

As given the surface charge density for both the spheres is same, hence

${\sigma }_{1}={\sigma }_{2} \therefore \frac{{Q}_{1}}{{A}_{1}}=\frac{{Q}_{2}}{{A}_{2}}\Rightarrow \frac{{Q}_{1}}{4\pi {R}^{2}}=\frac{{Q}_{2}}{4\pi {(4R)}^{2}} \therefore {Q}_{2}=16{Q}_{1}$

From above diagram, the electric potential for both the spheres is

${V}_{\mathrm{Inner}}=\frac{{\mathrm{KQ}}_{1}}{R}+\frac{{\mathrm{KQ}}_{2}}{4R}=\frac{K}{R}({Q}_{1}+\frac{16{Q}_{1}}{4})=\frac{5{\mathrm{KQ}}_{1}}{R}.$

In the same way for outer sphere,

${V}_{\mathrm{Outer}}=\frac{{\mathrm{KQ}}_{1}}{4R}+\frac{{\mathrm{KQ}}_{2}}{4R}=\frac{K}{4R}(16{Q}_{1}+{Q}_{1})=\frac{17{\mathrm{KQ}}_{1}}{4R}$ .

Now the potential difference is,

$\Delta V={V}_{\mathrm{inner}}-{V}_{\mathrm{outer}}$

$=\frac{5{\mathrm{KQ}}_{1}}{R}-\frac{17{\mathrm{KQ}}_{1}}{4R}=\frac{3{Q}_{1}}{16{\pi \epsilon }_{0}R}$ .

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JEE Main 2020 — Physics Electromagnetism

2020mcqmedium

Official previous-year question

Held on 3 Sept 2020 · Verified 6 Jul 2026.

Options

A
$\frac{{Q}_{2}}{4\pi {\epsilon }_{0}R}$
B
$\frac{3{Q}_{2}}{4\pi {\epsilon }_{0}R}$
C
$\frac{3{Q}_{1}}{16\pi {\epsilon }_{0}R}$
D
$\frac{3{Q}_{1}}{4\pi {\epsilon }_{0}R}$

Did you get this right?

Solution

JEE Main 2020 Physics, Electrostatics — solution figure

As given the surface charge density for both the spheres is same, hence

${\sigma }_{1}={\sigma }_{2} \therefore \frac{{Q}_{1}}{{A}_{1}}=\frac{{Q}_{2}}{{A}_{2}}\Rightarrow \frac{{Q}_{1}}{4\pi {R}^{2}}=\frac{{Q}_{2}}{4\pi {(4R)}^{2}} \therefore {Q}_{2}=16{Q}_{1}$

From above diagram, the electric potential for both the spheres is

${V}_{\mathrm{Inner}}=\frac{{\mathrm{KQ}}_{1}}{R}+\frac{{\mathrm{KQ}}_{2}}{4R}=\frac{K}{R}({Q}_{1}+\frac{16{Q}_{1}}{4})=\frac{5{\mathrm{KQ}}_{1}}{R}.$

In the same way for outer sphere,

${V}_{\mathrm{Outer}}=\frac{{\mathrm{KQ}}_{1}}{4R}+\frac{{\mathrm{KQ}}_{2}}{4R}=\frac{K}{4R}(16{Q}_{1}+{Q}_{1})=\frac{17{\mathrm{KQ}}_{1}}{4R}$ .

Now the potential difference is,

$\Delta V={V}_{\mathrm{inner}}-{V}_{\mathrm{outer}}$

$=\frac{5{\mathrm{KQ}}_{1}}{R}-\frac{17{\mathrm{KQ}}_{1}}{4R}=\frac{3{Q}_{1}}{16{\pi \epsilon }_{0}R}$ .

JEE Main 2020 — Physics Electromagnetism

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Solution

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JEE Main 2020 — 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

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