Question

The electric field due to electric potential $$V=2x^2-4x$$ 

(A)  $$(4x+4)i$$

(B) $$(4x-4)i$$

(C) $$(-4x+4)i$$

(D) $$(-4x-4)i$$

Answer 1

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 $\hat{i}$.

Complete step-by-step answer:

As we know that,

$$\overrightarrow{E}=-\overrightarrow{\mathrm{\nabla }}.V=-\left({\displaystyle \frac{dV}{dx}}\hat{i}+{\displaystyle \frac{dV}{dy}}\hat{j}+{\displaystyle \frac{dV}{dz}}\hat{k}\right)$$

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:

$$\overrightarrow{E}=-{\displaystyle \frac{d(2x^2-4x)}{dx}}\hat{i}$$

$$E=-(4x-4)\hat{i}$$

$$E=(-4x+4)\hat{i}$$

As the expression of electric field is only dependent on x, i.e. the coordinate in the $$\hat{\text{i}}$$ direction, we will add the unit vector$$\hat{\text{i}}$$ 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. $E=-{\displaystyle \frac{dV}{dr}}$