The electric field due to electric potential \[V = 2{x^2} - 4x\]
(A) \[(4x + 4){\text{ }}i\]
(B) \[(4x - 4){\text{ }}i\]
(C) \[( - 4x + 4){\text{ }}i\]
(D) \[( - 4x - 4){\text{ }}i\]
VerifiedHint: The electric field at a point is defined as the negative of gradient of the potential at that point, where gradient is the dot product of Del operator with quantity V. Since only the x - component is here we get the final answer with unit vector $\hat i$.
Complete step-by-step answer:
As we know that,
\[\overrightarrow E = - \overrightarrow \nabla .V = {\text{ }} - \left( {\dfrac{{dV}}{{dx}}\widehat i + \dfrac{{dV}}{{dy}}\widehat j + \dfrac{{dV}}{{dz}}\widehat k} \right)\]
This means that change in potential of a point with respect to the distance in 3 axes is termed as electric field. The negative sign is present to show that the potential of a point decreases with distance which will create a negative potential gradient.
Solving the above equation we obtain:
\[\overrightarrow E = {\text{ }} - \dfrac{{d(2{x^2} - 4x)}}{{dx}}\widehat i\]
\[E = {\text{ }} - (4x{\text{ }} - {\text{ }}4)\hat i\]
\[E = ({\text{ }} - 4x{\text{ }} + {\text{ }}4)\widehat {{\text{ }}i}\]
As the expression of electric field is only dependent on x, i.e. the coordinate in the \[\widehat {\text{i}}\] direction, we will add the unit vector\[\widehat {\text{i}}\] to denote the direction of the Electric field.
Therefore, the correct answer is option C.
Note: The potential and potential difference of a point is a scalar quantity but electric field is a vector quantity. An electric field is a vector quantity equal to the negative of the potential gradient. $E = - \dfrac{{dV}}{{dr}}$
