Variation of electrostatic potential along the x-direction is shown in the graph. The correct statement about the electric field is:

(A) $x$ component at the point $B$ is maximum
(B) The $x$ component of point $A$ is towards the positive x-axis
(C) The $x$ component of point $C$ is along the negative x-axis
(D) The $x$ component of point $C$ is along the positive x-axis
Answer
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Hint The first half of the curve has a positive increase with respect to the x-axis and the second half of the curve has a negative increase or a decrease with respect to the x-axis. The midpoint of the curve has no increase or decrease with respect to the x-axis. Use this along with the relation between the electric field and the potential to arrive at the answer.
Complete Step by step answer
From the given graph, we can see that the point $A$ is situated in the left half of the curve that has a positive increase with the x-axis. This means that the tangent at this point will be positive. Similarly, point $C$ is situated in the right half of the curve that has a negative increase or positive decrease with the x-axis, which means that in the second half, the tangent is negative. The midpoint has a tangent equal to zero.
The tangent at any point is given by $\dfrac{{dV}}{{dx}}$, where $V$ is the potential in the y-axis in this case and $x$ is the distance of separation of the two charges in the x-axis.
As stated above, the tangent at point $A$, $\dfrac{{dV}}{{dx}} = + ve$
The tangent at point $B$, $\dfrac{{dV}}{{dx}} = 0$
And the tangent at point $C$, $\dfrac{{dV}}{{dx}} = - ve$
But, the change in potential with the change in distance of separation between two charges is defined as the electric field. Mathematically,
$E = - \dfrac{{dV}}{{dx}}$
Therefore, the electric field at point $A$ will be $ - ve$ , the electric field at point $B$ will be $0$ and the electric field at point $C$ will be $ + ve$ .
The $x$-component of the electric field at point is towards $A$ negative $x$-direction, the $x$ component of point $B$ is zero and the $x$ component of point $C$ is towards the positive $x$-direction.
Hence, option (D) is the correct option.
Note
The electric potential is defined as the work done to move a unit charge from a reference point in space to the desired point in an electric field. Or the electric field is defined as the negative gradient of electric potential. This means that the electric field points the direction in which electric potential is decreasing.
Complete Step by step answer
From the given graph, we can see that the point $A$ is situated in the left half of the curve that has a positive increase with the x-axis. This means that the tangent at this point will be positive. Similarly, point $C$ is situated in the right half of the curve that has a negative increase or positive decrease with the x-axis, which means that in the second half, the tangent is negative. The midpoint has a tangent equal to zero.
The tangent at any point is given by $\dfrac{{dV}}{{dx}}$, where $V$ is the potential in the y-axis in this case and $x$ is the distance of separation of the two charges in the x-axis.
As stated above, the tangent at point $A$, $\dfrac{{dV}}{{dx}} = + ve$
The tangent at point $B$, $\dfrac{{dV}}{{dx}} = 0$
And the tangent at point $C$, $\dfrac{{dV}}{{dx}} = - ve$
But, the change in potential with the change in distance of separation between two charges is defined as the electric field. Mathematically,
$E = - \dfrac{{dV}}{{dx}}$
Therefore, the electric field at point $A$ will be $ - ve$ , the electric field at point $B$ will be $0$ and the electric field at point $C$ will be $ + ve$ .
The $x$-component of the electric field at point is towards $A$ negative $x$-direction, the $x$ component of point $B$ is zero and the $x$ component of point $C$ is towards the positive $x$-direction.
Hence, option (D) is the correct option.
Note
The electric potential is defined as the work done to move a unit charge from a reference point in space to the desired point in an electric field. Or the electric field is defined as the negative gradient of electric potential. This means that the electric field points the direction in which electric potential is decreasing.
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