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
The electric field is the negative gradient of the potential. Here \( V \) depends only on \( x \), so the field has only an \( x \)-component.
\( \displaystyle E_x = -\frac{dV}{dx} = -\frac{d}{dx}(4x^{2}) = -8x \)
At the point (1 m, 0, 2 m), \( x = 1\,\mathrm{m} \):
\( \displaystyle E_x = -8 \times 1 = -8\,\mathrm{V\,m^{-1}} \)
The magnitude is 8 V/m, and the negative sign means the field points along the negative X-axis (the field points towards decreasing potential).