a
For electrostatic field,
$\vec{E}_P=\vec{E}_1+\vec{E}_2 $
$=\frac{\rho}{3 \varepsilon_0} \overrightarrow{ C _1 P }+\frac{(-\rho)}{3 \varepsilon_0} \overrightarrow{ C _2 P } $
$=\frac{\rho}{3 \varepsilon_0}\left(\overrightarrow{ C _1 P }+\overrightarrow{ PC _2}\right)$
$\overrightarrow{ E }_{ p }=\frac{\rho}{3 \varepsilon_0} \overrightarrow{ C _1 C _2}$
For electrostatic potential, Since electric field is non zero so it is not equipotential.
