JEE Advanced 2019 · Paper 2 · Physics Q4

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PhysicsMultiple correct+4 / -1Question 4 of 54

An electric dipole with dipole moment $\dfrac{p_0}{\sqrt2}(\hat i + \hat j)$ 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:

( $\epsilon_0$ is permittivity of free space. $R \gg$ dipole size)

A circle of radius R centred at the origin O, where a dipole points along the 45° direction (î + ĵ). Point A is on the circle along the dipole's direction (45° from the y-axis on the right), and point B is at 45° from the y-axis on the left, perpendicular to the dipole.

R = \left(\dfrac{p_0}{4\pi\epsilon_0E_0}\right)^{1/3}

Total electric field at point A is

\vec E_A = \sqrt2E_0(\hat i + \hat j)

Total electric field at point B is

\vec E_B = 0

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

Answer and worked solution

Official answer: A, C

Condition for an equipotential sphere. With the uniform field $\vec E_0$ parallel to $\vec p$ , the potential is
$V=\frac{kp\cos\theta}{r^2}-E_0r\cos\theta=\cos\theta\left(\frac{kp}{r^2}-E_0r\right).$
It is zero on the whole sphere $r^3=kp/E_0$ . (An antiparallel field would give no such sphere.)

(A)
$R=\left(\frac{p_0}{4\pi\epsilon_0E_0}\right)^{1/3}.$
True.

(B) A lies on the dipole's axis. There the dipole field is $\dfrac{2kp}{R^3}=2E_0$ along $\hat p$ , so the total is $3E_0\hat p=\dfrac{3E_0}{\sqrt2}(\hat i+\hat j)$ . The option's $\sqrt2E_0(\hat i+\hat j)$ is $2E_0\hat p$ . False.

(C) B lies on the dipole's equatorial line. There the dipole field is $-\dfrac{kp}{R^3}\hat p=-E_0\hat p$ , which cancels the applied field. True.

(D) The field strength is $3E_0$ at A and 0 at B, so it is not the same everywhere on the circle. False.

(A), (C)

Topics:Electric dipole Equipotential surfaces Electric field

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JEE Advanced 2019 · Paper 2, Physics question 4