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
Answer & explanation
Correct answer: option 3
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 $(1,0,2)$:
$\vec{E} = -8(1)\hat{i} = -8\hat{i}$