The electric potential is given by $V = 4x^2 \ V$. What is the electric field at the point $(1 \ m, 0, 2 \ m)$?

  • A
    $8 \ V/m$ along $-X$-axis
  • B
    $8 \ V/m$ along $+X$-axis
  • C
    $16 \ V/m$ along $-X$-axis
  • D
    $16 \ V/m$ along $+Z$-axis

Explore More

Similar Questions

When a charge of $3 \, C$ is placed in a uniform electric field, it experiences a force of $3000 \, N$. Within this field, the potential difference between two points separated by a distance of $1 \, cm$ is: (in $V$)

Electric potential at any point is $V = -5x + 3y + \sqrt{15} z$,then the magnitude of the electric field is

In a certain region of space,the potential is given by: $V = k[2x^2 - y^2 + z^2]$. The electric field at the point $(1, 1, 1)$ has magnitude =

The electric potential $V$ at any point $(x, y, z)$ in space is given by the equation $V = 4x^2 \, V$,where $x, y$,and $z$ are all in meters. The electric field at the point $(1 \, m, 0, 2 \, m)$ in $V/m$ is:

Difficult
View Solution

The electric potential is given by $V = 6x - 8xy^2 - 8y + 6yz - 4z^2$. Find the electric force in $N$ acting on a point charge of $2 \ C$ placed at the origin.

Difficult
View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo