An electric dipole with dipole moment frac p 0 √(2) (hat i +hat j ) is held fixed at the origin O in the presence of an uniform electric field of magnitude E 0 . If the potential is constant on a circle of radius R centered at the origin as shown in figure, then the correct statement(s) is/are : ( ε 0 is permittivity of free space. R gg dipole size)

Total electric field at point B is overrightarrow E B =0

Total electric field at point A is vec E A =√(2) E 0 (hat i +hat j )

The magnitude of total electric field on any two points of the circle will be same.

Physics · Electrostatics

JEE Advanced 2019 — Paper 2 — Question 6

An electric dipole with dipole moment $\frac{p_{0}}{\sqrt{2}}(\hat{i}+\hat{j})$ is held fixed at the origin O in the presence of an uniform electric field of magnitude $\mathrm{E}_{0}$ . If the potential is constant on a circle of radius R centered at the origin as shown in figure, then the correct statement(s) is/are : ( $\epsilon_{0}$ is permittivity of free space. $\mathrm{R} \gg$ dipole size)

Option A:
Total electric field at point $A$ is $\vec{E}_{A}=\sqrt{2} E_{0}(\hat{i}+\hat{j})$
Option B:
Total electric field at point $B$ is $\overrightarrow{\mathrm{E}}_{\mathrm{B}}=0$
Correct
Option C:
$R = \left( \frac{P_0}{4\pi \epsilon_0 E_0} \right)^{1/3}$
Correct
Option D:
The magnitude of total electric field on any two points of the circle will be same.

Answer: B, C

Step-by-step solution

At $B, \vec{E}$ due to dipole is tangential

$\Rightarrow$ The other field must atleast cancel this $\overrightarrow{\mathrm{E}}$ due to dipole.

Also, at B, if the other field also had a component in radial direction, then this component would contribute to

the net tangential component at A (since the other field is given to be uniform), which is not allowed since

the given sphere is equipotential.

$\Rightarrow \overrightarrow{\mathrm{E}}_{\text {other }}=\frac{1}{4 \pi \epsilon_{0}} \frac{\mathrm{P}_{0}}{\mathrm{R}^{3}}=\frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}}{\sqrt{2}}=$

$\therefore$ (B) and (C) only

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2019
Paper: Paper 2
Subject: Physics
Chapter: Electrostatics
Topic: Electric Dipole