The distance between the plates of a capacitor is $1 \ mm$. An electric field of $5 \times 10^5 \ V/m$ is produced between the two plates. What will be the change in the potential energy $(PE)$ of an electron if it is moved from one plate to the other?

  • A
    $5 \times 10^{-19} \ J$
  • B
    $8 \times 10^{-17} \ J$
  • C
    $5 \times 10^{-17} \ J$
  • D
    $5 \times 10^2 \ J$

Explore More

Similar Questions

Two charges $7 \ \mu C$ and $-2 \ \mu C$ are placed at $(-9, 0, 0) \ cm$ and $(9, 0, 0) \ cm$ respectively in an external field $E = \frac{A}{r^2} \hat{r}$,where $A = 9 \times 10^5 \ N/C \cdot m^2$. Considering the potential at infinity is $0$,the electrostatic energy of the configuration is . . . . . . $J$.

Three particles of each charge $q$ are placed at the vertices of an equilateral triangle of side $L$. The work to be done to decrease the side of the triangle to $\frac{L}{2}$ is

Given $q_1 = +2 \times 10^{-8} \ C$ and $q_2 = -0.4 \times 10^{-8} \ C$. When a charge $q_3 = 0.2 \times 10^{-8} \ C$ is moved from $C$ to $D$,what is the change in the potential energy of $q_3$?

Difficult
View Solution

The displacement of a charge $Q$ in the electric field $\vec{E} = e_1 \hat{i} + e_2 \hat{j} + e_3 \hat{k}$ is $\vec{r} = a \hat{i} + b \hat{j}$. The work done by the electric field is:

Two point charges $A = +3 \text{ nC}$ and $B = +1 \text{ nC}$ are placed $5 \text{ cm}$ apart in air. The work done to move charge $B$ towards $A$ by $1 \text{ cm}$ 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