Dipole in a uniform electric field

JEE Advanced 2019 Paper 2, Question 4

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 $E_{0}$ . 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. $R \gg$ dipole size)

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$R=\left(\frac{p_{0}}{4 \pi \epsilon_{0} E_{0}}\right)^{1 / 3}$
Total electric field at point $A$ is ${\bf E}^A=\sqrt{2} E_{0}(\hat{i}+\hat{j})$
Total electric field at point $B$ is ${\bf E}^B=0$
The magnitude of total electric field on any two points of the circle will be same.

Solution

The potential due to the dipole kept at the origin is

(1) $\begin{equation*} \phi_{\rm dip} ({\bf r}) = \frac{1}{4 \pi \varepsilon_0} \frac{{\bf p} \cdot {\bf r}}{r^3} . \end{equation*}$

We’re also told that there is a uniform electric field of magnitude $E_0$ present, but we don’t know its direction. The combination of this electric field with that due to the dipole creates an equipotential surface, which is the circle of radius $R$ . Let us take ${\bf E}$ to point along the unit vector $a\hat{i}+b\hat{j}$ where $\sqrt{a^2+b^2}=1$ . That is,

(2) $\begin{equation*} {\bf E}_{\rm ext} = E_0 (a\hat{i}+b\hat{j}) . \end{equation*}$

If we denote the potential associated with this field as $\phi_E({\bf r})$ , then ${\bf E}_{\rm ext} = -\nabla \phi_E$ , which means

(3) $\begin{equation*} \phi_E({\bf r}) = -E_0 (ax+by) + \phi_0 \end{equation*}$

where $\phi_0$ is a constant. We’re unconcerned with what happens in the $z$ direction, so we ignore the potential along $z$ . A point on the circle of radius $R$ is given

(4) $\begin{equation*} {\bf r}_c = R \cos \theta \hat{i} + R \sin \theta \hat{j} \equiv (R \cos \theta, R \sin \theta), \end{equation*}$

where $\theta$ is the angle with respect to the positive $x$ -axis. By superposition, the total potential at this point is

(5) $\begin{equation*} \phi_{\rm dip} ({\bf r}_c) + \phi_{E} ({\bf r}_c) = \frac{1}{4 \pi \varepsilon_0} \frac{p_0}{\sqrt{2}} \frac{\cos \theta + \sin \theta}{R^2} - E_0 (a R \cos \theta + b R \sin \theta) + \phi_0 \end{equation*}$

where we have used $r = |\{\bf r}_c| = R$ . This potential will be a constant if we choose

(6) $\begin{equation*} E_0 a R = \frac{1}{4 \pi \varepsilon_0 R^2} \frac{p_0}{\sqrt{2}} = E_0 b R . \end{equation*}$

That means $a=b$ , and $\sqrt{a^2+b^2}=1 \implies a = 1/\sqrt{2}$ . Therefore

(7) $\begin{equation*} \boxed{E_0 = \frac{p_0}{4 \pi \varepsilon_0 R^3}} \implies \text{ option (A) is correct. } \end{equation*}$

Substituting $a=b=1/\sqrt{2}$ in (2) we also find that

(8) $\begin{equation*} {\bf E}_{\rm ext} = \frac{E_0}{\sqrt{2}} (\hat{i}+\hat{j}) . \end{equation*}$

To find the electric fields at points $A$ and $B$ we need to add the field due to the dipole to ${\bf E}_{\rm ext}$ . This is

(9) $\begin{equation*} {\bf E}_{\rm dip} = \frac{1}{4 \pi \varepsilon_0} \left( \frac{3 ({\bf p} \cdot {\bf r}) {\bf r}}{r^5} - \frac{{\bf p}}{r^3} \right) = \frac{1}{4 \pi \varepsilon_0} \frac{p_0}{\sqrt{2}} \left( \frac{3 (x+y)(x\hat{i}+y\hat{j})}{r^5} - \frac{(\hat{i}+\hat{j})}{r^3} \right) \end{equation*}$

The point $A$ has the co-ordinates ${\bf r}_A \equiv (\frac{R}{\sqrt{2}},\frac{R}{\sqrt{2}})$ . Therefore, the field at this point is

(10) $\begin{equation*} {\bf E}_{\rm dip}^A = \sqrt{2} \frac{p_0}{4 \pi \varepsilon_0 R^3} (\hat{i}+\hat{j}) = \sqrt{2} E_0 (\hat{i}+\hat{j}) \end{equation*}$

Similarly, the dipole field at ${\bf r}_B \equiv (-\frac{R}{\sqrt{2}},\frac{R}{\sqrt{2}})$ is

(11) $\begin{equation*} {\bf E}_{\rm dip}^B = \frac{1}{\sqrt{2}} \frac{p_0}{4 \pi \varepsilon_0 R^3} (\hat{i}+\hat{j}) = \frac{E_0}{\sqrt{2}} (\hat{i}+\hat{j}) \end{equation*}$

Adding the uniform field from (8), we find that the total electric field at these points are

(12) $\begin{align*} {\bf E}^A = \left( \sqrt{2} + \frac{1}{\sqrt{2}} \right) E_0 (\hat{i}+\hat{j}) &\implies \text{ option (B) is incorrect. } \\[1em] {\bf E}^B = 0 &\implies \text{ option (C) is correct. } \end{align*}$

Just from examining the points $A$ and $B$ we also see that the magnitude of the electric field cannot be the same at all point of the circle. That means, option (D) is incorrect as well.

Commentary: This problem seems particularly lengthy for the JEE exam. If there is a simpler way to do this problem, please let me know. It is certainly possible to notice some symmetries of the setup to calculate ${\bf E}_{\rm dip}$ without using the formula in (9), but I doubt if it will save much time.

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