The electric potential at any point in space is given by $V= 4x^2$ volt. The electric field at the point (1 m, 0, 2 m) in SI unit is

8, along the negative X-axis

The correct answer is Option (3) → 8, along the negative X-axis

Electric field is related to potential by:

$\vec{E} = -\nabla V$

Given: $V = 4x^2$

$\frac{\partial V}{\partial x} = 8x$, $\frac{\partial V}{\partial y} = 0$, $\frac{\partial V}{\partial z} = 0$

So, $\vec{E} = -(8x)\hat{i}$

At $(1,0,2)$:

$\vec{E} = -8(1)\hat{i} = -8\hat{i}$