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


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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