An electric dipole with dipole moment $\frac{\mathrm{p}_{0}}{\sqrt{2}}(\hat{\mathrm{i}}+\hat{\mathrm{j}})$ 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: ( $\varepsilon_{0}$ is permittivity of free space, $\mathrm{R} \gg$ dipole size)

(A) $\mathrm{R}=\left(\frac{\mathrm{p}_{0}}{4 \pi \varepsilon_{0} \mathrm{E}_{0}}\right)^{1 / 3}$; (D) Total electric field at point $B$ is $\overrightarrow{\mathrm{E}}_{\mathrm{B}}=0$

(B) The magnitude of total electric field on any two points of the circle will be same

(C) Total electric field at point A is $\overrightarrow{\mathrm{E}}_{\mathrm{A}}=\sqrt{2} \mathrm{E}_{0}(\hat{\mathrm{i}}+\hat{\mathrm{j}})$

JEE Advanced · 2019 · 27 May 2019 · Paper 2 · Electrostatics

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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 ≫

dipole size)

(A) $R = (\frac{p _{0}}{4 π ε _{0} E _{0}})^{1/3}$ ✓ Correct
(B)The magnitude of total electric field on any two points of the circle will be same
(C)Total electric field at point A is $E_{A} = 2 E_{0} (\hat{i} + \hat{j})$
(D)Total electric field at point $B$ is $E_{B} = 0$ ✓ Correct

Answer: (A), (D)

Solution

For the circle to be an equipotential, the tangential component of the electric field at point B must be zero. The dipole field at B along the tangent equals

E_{0}

, so:

E_{0} = \frac{K ∣ P ∣}{R ^{3}} = \frac{K p _{0}}{R ^{3}}

Solving:

R^{3} = \frac{p _{0}}{4 π ϵ _{0} E _{0}}

, giving

R = (\frac{p _{0}}{4 π ϵ _{0} E _{0}})^{1/3}

. Option (A) is correct.

At point B, the uniform field and dipole field cancel tangentially, and the radial components also cancel, so

E_{B} = 0

. Option (D) is correct.

Since

E_{0}

is uniform but the dipole field varies at different points, the total field magnitude varies around the circle. Option (B) is incorrect.

At point A, the field is

E_{A} = \frac{2 K P}{R ^{3}} + \frac{K P}{R ^{3}} = 3 E_{0}

along the dipole direction, not

2 E_{0} (\hat{i} + \hat{j})

. Option (C) is incorrect.

JEE Advanced · 2019 · 27 May 2019 · Paper 2 · One or more correct options

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JEE Advanced · 2019 · 27 May 2019 · Paper 2 · Electrostatics

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