The potential at a point $x$ (measured in $\mu m$) due to some charges situated on the $x$-axis is given by $V(x) = \frac{20}{x^2 - 4} \text{ volt}$. The electric field $E$ at $x = 4 \mu m$ is given by:

  • A
    $\frac{10}{9} \text{ V/}\mu m$ and in the $+ve \ x$ direction
  • B
    $\frac{5}{3} \text{ V/}\mu m$ and in the $-ve \ x$ direction
  • C
    $\frac{5}{3} \text{ V/}\mu m$ and in the $+ve \ x$ direction
  • D
    $\frac{10}{9} \text{ V/}\mu m$ and in the $-ve \ x$ direction

Explore More

Similar Questions

In a certain region,the electric potential is given by $V = 6x - 8xy^2 - 8y + 6yz - 4z^2 \, V$. The magnitude of the force on a charge of $2 \, C$ at the origin is ........ $N$.

Difficult
View Solution

The electric potential in some region is expressed by $V = 6x - 8xy^2 - 8y + 6yz - 4z^2 \text{ volt}$. The magnitude of the electric force acting on a charge of $2 \text{ C}$ situated at the origin will be $-$ (in $\text{ N}$)

Difficult
View Solution

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

Two large plane parallel conducting plates are kept $10 \ cm$ apart as shown in the figure. The potential difference between them is $V$. The potential difference between the points $A$ and $B$ (shown in the figure) is

The electric field and electric potential at a point due to a point charge are $500 \ NC^{-1}$ and $30 \ V$ respectively. The magnitude of the charge is:

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