An electric dipole with moment is held fixed at the origin O in the presence of an uniform electric field of magnitude . If the potential is constant on a circle of radius R centered at the origin as shown in the figure, then the correct statement(s) is/

Question

An electric dipole with 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 the figure, then the correct statement(s) is/are: ( $ε_{0}$ is permittivity of free space. $R ≫$ dipole size) [2019]

Smart Achievers · Accepted Answer

(2, 4)

$Given dipole moment of electric dipole$

$\vec{P} = \frac{P_{0}}{\sqrt{2}} (\hat{i} + \hat{j})$ and the circle is equipotential. Also, the electric field is normal to such a line. Thus, the direction of electric field is either radial or the magnitude of electric field should be zero at points on the circle. Now considering point A, the electric field due to dipole $\frac{2 K \vec{p}}{R^{3}}$ (directed from O to A) as point A lies on the axial line of electric dipole. The external electric field ${\vec{E}}_{0}$ should also be in the direction of O to A.

Now considering point B, which is a point on the equatorial line of the electric dipole. The electric field here due to dipole is $\frac{K \vec{p}}{R^{3}}$ in a direction opposite to the dipole. The external electric field should cancel out this field, i.e., ${\vec{E}}_{B} = 0$ .

$∴$ ${\vec{E}}_{0} = - \frac{K \vec{p}}{R^{3}}$ ...(i)

The electric field at A

${\vec{E}}_{A} = \frac{2 K \vec{p}}{R^{3}} + {\vec{E}}_{0} = - 2 {\vec{E}}_{0} + {\vec{E}}_{0} = - {\vec{E}}_{0}$

Again from Eq. (i) $E_{0} = \frac{1}{4 π ε_{0}} \frac{P_{0}}{\sqrt{2} (R^{3})} \times \sqrt{2}$

$∴$ $R^{3} = \frac{P_{0}}{4 π ε_{0} E_{0}}$

$∴$ $R = {[\frac{P_{0}}{4 π ε_{0} E_{0}}]}^{1 / 3}$

Solution

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

2 Total electric field at point $B$ is ${\vec{E}}_{B} = 0 .$

3 Total electric field at point $A$ is ${\vec{E}}_{A} = \sqrt{2} E_{0} (\hat{i} + \hat{j}) .$

4 $R = {(\frac{p_{0}}{4 π ε_{0} E_{0}})}^{1 / 3}$

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Solution

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

2 Total electric field at point B is E→B=0.

3 Total electric field at point A is E→A=2E0(i^+j^).

4 R=(p04πε0E0)1/3

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2 Total electric field at point $B$ is ${\vec{E}}_{B} = 0 .$

3 Total electric field at point $A$ is ${\vec{E}}_{A} = \sqrt{2} E_{0} (\hat{i} + \hat{j}) .$

4 $R = {(\frac{p_{0}}{4 π ε_{0} E_{0}})}^{1 / 3}$