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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) $( - 4x - 4){\text{ }}i$

Last updated date: 21st Jun 2024
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Hint: 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 $\hat i$.

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. $E = - \dfrac{{dV}}{{dr}}$