JEE Main 2016 — Physics Electromagnetism

2016mcqhard

The potential (in volts) of a charge distribution is given by

$V(z)=30-5{z}^{2}$ for $|z|\leq 1 m$

$V(z)=35-10 |z|$ for $|z|\geq 1 m$ .

$V(z)$ does not depend on x and y. If this potential is generated by a constant charge per unit volume ${\rho }_{0}$ (in units of ${\epsilon }_{0}$ ) which is spread over a certain region, then choose the correct statement.

Official previous-year question

Held on 9 Apr 2016 · Verified 6 Jul 2026.

Options

A
${\rho }_{0}=20 {\epsilon }_{0}$ in the entire region
B
${\rho }_{0}=10 {\epsilon }_{0}$ for $|z|\leq 1 m$ and ${\rho }_{0}=0$ else where
C
${\rho }_{0}=20 {\epsilon }_{0}$ for $|z|\leq 1 m$ and ${\rho }_{0}=0$ else where
D
${\rho }_{0}=40 {\epsilon }_{0}$ in the entire region

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Solution

Electric field due to given potentials is

${E}_{1}=\frac{-dV}{dz}=10 |z|$ for $\mid z\mid \leq 1m$ . Here ${E}_{1}$ is proportional to $z$ , so charge distribution is continuous for $|z|\leq 1m$

${E}_{2}=-\frac{dV}{dz}=10$ for $|z|\geq 1 m$ . Here, ${E}_{2}$ is constant.

So, using the Poisson’s equation ${\nabla }^{2}V=-\frac{\rho }{{\epsilon }_{0}}$ , we get $\rho =0$ .

$\therefore$ The source is an infinity large non-conducting thick plate.

Hence, consider a cylindrical Gaussian surface:

Applying Gauss's law for $\mid z\mid \leq 1m$

$E\times 2A=\frac{\rho \times 2z\times A}{{\epsilon }_{0}}$

$\Rightarrow (10z)(2A)=\frac{\rho (2\mathrm{Az})}{{\epsilon }_{0}}$

$\Rightarrow {\rho }_{0}=10{\epsilon }_{0}$ for $|z|\leq 1 m$ .

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Options

${\rho }_{0}=20 {\epsilon }_{0}$ in the entire region

${\rho }_{0}=10 {\epsilon }_{0}$ for $|z|\leq 1 m$ and ${\rho }_{0}=0$ else where

${\rho }_{0}=20 {\epsilon }_{0}$ for $|z|\leq 1 m$ and ${\rho }_{0}=0$ else where

${\rho }_{0}=40 {\epsilon }_{0}$ in the entire region

Solution

Electric field due to given potentials is

${E}_{1}=\frac{-dV}{dz}=10 |z|$ for $\mid z\mid \leq 1m$ . Here ${E}_{1}$ is proportional to $z$ , so charge distribution is continuous for $|z|\leq 1m$

${E}_{2}=-\frac{dV}{dz}=10$ for $|z|\geq 1 m$ . Here, ${E}_{2}$ is constant.

So, using the Poisson’s equation ${\nabla }^{2}V=-\frac{\rho }{{\epsilon }_{0}}$ , we get $\rho =0$ .

$\therefore$ The source is an infinity large non-conducting thick plate.

Hence, consider a cylindrical Gaussian surface:

Applying Gauss's law for $\mid z\mid \leq 1m$

$E\times 2A=\frac{\rho \times 2z\times A}{{\epsilon }_{0}}$

$\Rightarrow (10z)(2A)=\frac{\rho (2\mathrm{Az})}{{\epsilon }_{0}}$

$\Rightarrow {\rho }_{0}=10{\epsilon }_{0}$ for $|z|\leq 1 m$ .

JEE Main 2016 — Physics Electromagnetism

Options

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Solution

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JEE Main 2016 — 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