Question

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 the figure, then the correct statement (s) is/are : (${\in }_0$ is permittivity of free space. R >> dipole size)

• A. The magnitude of total electric field on any two points of the circle will be same.
• B. Total electric field at point A is ${\overrightarrow{E}}_A=\sqrt{2}{E}_0(\hat{i}+\hat{j})$
• C. $R={\left(\frac{P_0}{4\pi {\epsilon }_0{E}_0}\right)}^{1/3}$
• D. Total electric field at point B is ${\overrightarrow{E}}_B=0$

Answer 1

Answer: (C), (D)

Total electric field at point B is ${\overrightarrow{E}}_B=0$

B and B' are on equitorial line of dipole. As circle is equipotential, $E_B=0$
$\Rightarrow E_{eq}-E_o=0$
$E_o=\frac{K{P}_o}{R^3}$
So, $R^3=\frac{K{P}_0}{E_0}=\left(\frac{P_0}{4\pi {ϵ}_0{E}_0}\right)$

So $R={\left(\frac{P_0}{4\pi {ϵ}_0{E}_0}\right)}^{1/3}$
As $E_0$ is uniform & electric field due to dipole is different at different points, magnitude of total electric field will also be variable.
$E_A=E_{axial}+E_o$
$E_A=\frac{2K\overrightarrow{P}}{R^3}+\frac{K\overrightarrow{P}}{R^3}=3\frac{K}{R^3}\frac{P_0}{\sqrt{2}}(\hat{i}+\hat{j})$ $(\because \overrightarrow{P}=\frac{P_0}{\sqrt{2}}(\hat{i}+\hat{j}\left)\right)$
So, option C and D are correct.