Obtain the equation for the electric potential energy of a single charge in an external electric field.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) The external electric field $\overrightarrow{E}$ and the corresponding external potential $V$ may vary from point to point.
According to the definition of electric potential,$V$ at a point $P$ is the work done in bringing a unit positive charge from infinity to the point $P$. (We assume the potential at infinity to be zero.)
Thus,the work done in bringing a charge $q$ from infinity to the point $P$ in the external field is $W = qV$.
This work is stored in the form of potential energy $U$ of the charge $q$.
$\therefore U = qV$.
If the point $P$ has a position vector $\vec{r}$ relative to the origin,then the potential energy at point $P$ is $U(\vec{r}) = qV(\vec{r})$.
This means the potential energy in an external field is equal to the product of the electric charge and the electric potential at that point in the external field.

Explore More

Similar Questions

Two point charges of $5 \ \mu C$ and $10 \ \mu C$ are separated by a distance of $1 \ m$. The work done to bring them to a distance of $0.5 \ m$ is:

Explain electric potential energy. Show that the sum of kinetic energy and electric potential energy remains constant.

$A$ particle $A$ has charge $+q$ and a particle $B$ has charge $+4q$,with each of them having the same mass $m$. When allowed to fall from rest through the same electric potential difference,the ratio of their speeds $\frac{v_A}{v_B}$ will be:

Two point charges $100\,\mu C$ and $5\,\mu C$ are placed at points $A$ and $B$ respectively,with $AB = 40\,cm$. Calculate the work done by an external force in displacing the charge $5\,\mu C$ from $B$ to $C$,where $BC = 30\,cm$ and $\angle ABC = \frac{\pi}{2}$. Given $\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9\,N m^2/C^2$.

Difficult
View Solution

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