The equipotential surfaces due to a constant electric field in the x-axis have the shape of:

CONCEPT:

• Electric potential is equal to the amount of work done per unit charge by an external force to move the charge q from infinity to a specific point in an electric field.

\(⇒ V=\frac{W}{q}\)

• Potential due to a single charged particle Q at a distance r from it is given by:

\(⇒ V=\frac{Q}{4\piϵ_{0}r}\)

Where, 

ϵ0 is the permittivity of free space and has a value of 8.85 × 10-12 F/m in SI units

• Potential at a point P due to a system of charged particles Q1, Q2, Q3, ... Qn having distances r1, r2, r3, ... rn respectively from point P is given by:

\(⇒ V=\frac{Q_1}{4\piϵ_{0}r} + \frac{Q_2}{4\piϵ_{0}r} + \frac{Q_3}{4\piϵ_{0}r} + ...+\frac{Q_n}{4\piϵ_{0}r}\)

\(⇒ V=\sum_{i=1}^{n}\frac{Q_i}{4\piϵ_{0}r_i}\)

• An Equipotential surface is a surface with a constant value of the potential at all points on the surface. 
• The following figures show equipotential surfaces due to a point charge and an electric dipole.

• The equipotential surfaces due to a constant electric field in the x-axis have the shape of parallel planes in the y-z plane. Therefore option 3 is correct