Obtain the equation for the electric potential energy of a system of two electric charges in an external electric field.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) Consider two electric charges $q_{1}$ and $q_{2}$ brought from infinity to positions defined by position vectors $\overrightarrow{r_{1}}$ and $\overrightarrow{r_{2}}$ respectively in an external electric field $\overrightarrow{E}$.
The work done in bringing the charge $q_{1}$ from infinity to position $\overrightarrow{r_{1}}$ is:
$W_{1} = q_{1} V(\overrightarrow{r_{1}}) \quad \dots (1)$
Next,the work done in bringing the charge $q_{2}$ from infinity to position $\overrightarrow{r_{2}}$ is done against two fields: the external electric field $\overrightarrow{E}$ and the electric field produced by charge $q_{1}$.
The work done against the external field is:
$W_{2} = q_{2} V(\overrightarrow{r_{2}}) \quad \dots (2)$
The work done against the field produced by $q_{1}$ is:
$W_{3} = \frac{k q_{1} q_{2}}{r_{12}} \quad \dots (3)$
where $r_{12}$ is the distance between $q_{1}$ and $q_{2}$.
The total potential energy $U$ of the system is the sum of the total work done in assembling the configuration:
$U = W_{1} + W_{2} + W_{3}$
$U = q_{1} V(\overrightarrow{r_{1}}) + q_{2} V(\overrightarrow{r_{2}}) + \frac{k q_{1} q_{2}}{r_{12}}$

Explore More

Similar Questions

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 is

Charges $5 \mu C$ and $10 \mu C$ are placed $1 \ m$ apart. Work done to bring these charges at a distance $0.5 \ m$ from each other is . . . . . . .
$(k = 9 \times 10^9 \ SI)$

$A$ point charge of $10^{-8} \text{ C}$ is placed at the origin. The work done in moving a point charge of $2 \mu\text{C}$ from point $A(4, 4, 2) \text{ m}$ to $B(2, 2, 1) \text{ m}$ is . . . . . . $\text{J}$.

$A$ mass of $1 \ kg$ carrying a charge of $2 \ C$ is accelerated through a potential difference of $1 \ V$. The velocity acquired by it is:

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:

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