An electric dipole with dipole moment $\frac{p_0}{\sqrt{2}}\left(\hat{i}+\hat{j}\right)$ 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 : (${\epsilon }_0$​ is permittivity of free space. $R\gg$dipole size)

$E_0$ is the external field in the direction of $\overrightarrow{p}=\frac{p_0}{\sqrt{2}}\left(\hat{i}+\hat{j}\right)$

For an equipotential surface of radius $r$, the point $B$ is the point at the equatorial line of the dipole. Since, on an equipotential surface, no tangential field exists, so we have

$E_0=E_{eq}$ (in magnitude)

but both in opposite direction (as shown)

$$\begin{gathered} \Rightarrow \frac{p_0}{4\pi {\epsilon }_0{R}^3}=E_0 \\ \Rightarrow R={\left(\frac{p_0}{4\pi {\epsilon }_0{E}_0}\right)}^{\frac{1}{3}} \end{gathered}$$

Also, $A$ lies at axial line of dipole, so

${\overrightarrow{E}}_A=\frac{2\overrightarrow{p}}{4\pi {\epsilon }_0{R}^3}+{\overrightarrow{E}}_0=2{\overrightarrow{E}}_0+{\overrightarrow{E}}_0=3{\overrightarrow{E}}_0$

Since $E_{eq}=E_0$

$\Rightarrow \left|{\overrightarrow{E}}_B\right|=E_{eq}-E_0=0$