If the electric potential at any point $(x, y, z) m$ in space is given by $\mathrm{V}=3 \mathrm{x}^{2}$ volt. The electric field at the point $(1,0,3) \mathrm{m}$ will be :

(D) $6 \mathrm{Vm}^{-1}$, directed along negative x -axis.

(A) $3 \mathrm{Vm}^{-1}$, directed along positive x -axis.

(B) $3 \mathrm{Vm}^{-1}$, directed along negative $x$-axis.

(C) $6 \mathrm{Vm}^{-1}$, directed along positive $x$-axis.

JEE Main · 2022 · 29 Jun 2022 · Shift 2 · Electrostatics

If the electric potential at any point

(x, y, z) m

in space is given by

V = 3 x^{2}

volt. The electric field at the point

(1, 0, 3) m

will be :

Answer: (D)

The electric field is related to potential by

E = - \nabla V

E_{x} = - \frac{\partial V}{\partial x} = - \frac{\partial ( 3 x ^{2} )}{\partial x} = - 6 x

At point

(1, 0, 3)

E_{x} = - 6 (1) = - 6

V/m

The magnitude is 6 V/m, directed along the negative x-axis (indicated by the negative sign).

JEE Main · 2022 · 29 Jun 2022 · Shift 2 · Single correct option