JEE Main 2023 — Physics Electromagnetism

2023mcqeasy

Two isolated metallic solid spheres of radii $R$ and $2R$ are charged such that both have same charge density $\sigma$ . The spheres are then connected by a thin conducting wire. If the new charge density of the bigger sphere is ${\sigma }^{'}$ . The ratio $\frac{{\sigma }^{'}}{\sigma }$ is :

Official previous-year question

Held on 30 Jan 2023 · Verified 6 Jul 2026.

Options

A
$\frac{9}{4}$
B
$\frac{4}{3}$
C
$\frac{5}{3}$
D
$\frac{5}{6}$

Did you get this right?

Solution

Charge and charge density on both isolated metallic solid spheres are show below.

JEE Main 2023 Physics, Electrostatics — solution figure 1

Let the charges and the charge densities on both the spheres after connected by thin conducting wire be ${Q}_{1}^{'},{Q}_{2}^{'},\sigma \text{and}{\sigma }^{'}$ as shown below.

JEE Main 2023 Physics, Electrostatics — solution figure 2

Once connected, charge on both the spheres will redistribute such that potentials at the surface of both spheres is the same.

$\therefore \frac{k{Q}_{1}^{'}}{R}=\frac{k{Q}_{2}^{'}}{2R}$

$\Rightarrow \frac{{Q}_{1}^{'}}{4{\pi \epsilon }_{0}R}=\frac{{Q}_{2}^{'}}{4\pi {\epsilon }_{0}(2R)}$

$\therefore {Q}_{2}^{'}=2{Q}_{1}^{'}$

Using charge conservation,

${Q}_{1}^{'}+{Q}_{2}^{'}={Q}_{1}+{Q}_{2}$

$\therefore \frac{{Q}_{2}^{'}}{2}+{Q}_{2}^{'}=20{\pi R}^{2}\sigma$

$\frac{3}{2}{Q}_{2}^{'}=20{\pi R}^{2}\sigma$

$\therefore \frac{{Q}_{2}^{'}}{4\pi {(2R)}^{2}}=\frac{2}{3}\times \frac{20{\pi R}^{2}\sigma }{16{\pi R}^{2}}$

$\therefore \frac{\sigma '}{\sigma }=\frac{5}{6}$

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Solution

Charge and charge density on both isolated metallic solid spheres are show below.

JEE Main 2023 Physics, Electrostatics — solution figure 1

Let the charges and the charge densities on both the spheres after connected by thin conducting wire be ${Q}_{1}^{'},{Q}_{2}^{'},\sigma \text{and}{\sigma }^{'}$ as shown below.

JEE Main 2023 Physics, Electrostatics — solution figure 2

Once connected, charge on both the spheres will redistribute such that potentials at the surface of both spheres is the same.

$\therefore \frac{k{Q}_{1}^{'}}{R}=\frac{k{Q}_{2}^{'}}{2R}$

$\Rightarrow \frac{{Q}_{1}^{'}}{4{\pi \epsilon }_{0}R}=\frac{{Q}_{2}^{'}}{4\pi {\epsilon }_{0}(2R)}$

$\therefore {Q}_{2}^{'}=2{Q}_{1}^{'}$

Using charge conservation,

${Q}_{1}^{'}+{Q}_{2}^{'}={Q}_{1}+{Q}_{2}$

$\therefore \frac{{Q}_{2}^{'}}{2}+{Q}_{2}^{'}=20{\pi R}^{2}\sigma$

$\frac{3}{2}{Q}_{2}^{'}=20{\pi R}^{2}\sigma$

$\therefore \frac{{Q}_{2}^{'}}{4\pi {(2R)}^{2}}=\frac{2}{3}\times \frac{20{\pi R}^{2}\sigma }{16{\pi R}^{2}}$

$\therefore \frac{\sigma '}{\sigma }=\frac{5}{6}$

JEE Main 2023 — Physics Electromagnetism

Options

Did you get this right?

Solution

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JEE Main 2023 — 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