Equipotential surfaces are shown in fig, then the electric field strength will be

A. $100\,V{m^{ - 1}}$ along X-axis
B. $100\,V{m^{ - 1}}$ along Y-axis
C. $200\,V{m^{ - 1}}$ at an angle ${120^ \circ }$ with X-axis
D. $50\,V{m^{ - 1}}$ at an angle ${180^ \circ }$ with X-axis

Hint: The potential difference between two surfaces is given, the distance between two equipotential surfaces is also given. Potential difference is the dot product of electric field strength with the displacement vector. Substitute the values of potential difference and distance vector to find the electric field strength.

Complete step by step answer:
Equipotential surface is the locus of all points which are at the same potential level. Work done to move a charge from one point to another on the equipotential surface is zero. Any surface with the same electric potential at every point is termed as an equipotential surface.
Let $\vartriangle V$ be the potential difference between two equipotential surfaces.
$E$ be the electric field strength
$\vartriangle r$ be the distance between the two equipotential surfaces.
The potential difference is dot product of electric field and distance vector, given as:
$\vartriangle V = - E.\vartriangle r$
The negative sign indicates that work must be done against the field to move the charge apart.
$ \Rightarrow \vartriangle V = - E\vartriangle r\cos \theta $
$ \Rightarrow E = - \dfrac{{\vartriangle V}}{{\vartriangle r\cos \theta }}$
Let us consider the equipotential surface at $10\,cm$ and $20\,cm$ . These surfaces have potential of $20\,V$ and $10\,V$ respectively. Therefore, we can have
$\vartriangle V = 20V - 10V = 10V$ and $\vartriangle r = 20cm - 10cm = 0.1m$
The electric field is perpendicular to the equipotential surface, as shown below

It is clear that the electric field makes an angle of ${120^ \circ }$ with the positive X-axis, $\theta = {120^ \circ }$
Substituting the values, we get
$E = - \dfrac{{10}}{{0.1 \times \cos {{120}^ \circ }}}$
$\cos {120^ \circ } = - \dfrac{1}{2}$
$ \Rightarrow E = - \dfrac{{10}}{{0.1 \times \left( { - \dfrac{1}{2}} \right)}}$
$ \Rightarrow E = 200\,V{m^{ - 1}}$
Thus, the electric field strength is of $200\,V{m^{ - 1}}$ and it is at an angle of ${120^ \circ }$ with X-axis.

So, the correct answer is “Option C”.

Note:
Surfaces which are at same potential are known as equipotential surfaces.
The equipotential surface can be in any shape.
The potential difference is the dot product of the electric field strength and the distance.
The potential difference is taken as negative.