Equipotentials surfaces are shown in figure a and b. The field is uniform in 

In figure A as we move $x_0$ distance v increase by $V_0$ uniformly.

In figure B as we move $r_0$ distance initially v increased from $V_0$ to $\Delta V_0$ later from $\Delta V_{0}toq{V}_0$

$\therefore$  we know $E=-\frac{dv}{dx}$

$\frac{dv}{dx}$ is constant in figure a

$\therefore$ E is uniform in A