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Q.
Variation of electrostatic potential along the $x$-direction is shown in figure. The correct statement about electric field is
Electrostatic Potential and Capacitance
Solution:
We know, $E=-\frac{d V}{d x}$.
At $B, d V / d x=0$, hence $E_{x}=0$.
At $A, d V / d x$ is positive, hence $E_{x}$ is negative.
At $C, d V / d x$ is negative, hence $E_{x}$ is positive.