Obtain the equation of electric field at a point due to a system of $n$ point charges.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) According to the principle of superposition,the electric field at a point due to a system of charges is the vector sum of the electric fields produced by each individual charge at that point.
Let there be $n$ point charges $q_{1}, q_{2}, \ldots, q_{n}$ located at positions with position vectors $\vec{r}_{1}, \vec{r}_{2}, \ldots, \vec{r}_{n}$ relative to an origin $O$.
The electric field $\vec{E}_{1}$ at a point $P$ with position vector $\vec{r}$ due to charge $q_{1}$ is given by:
$\vec{E}_{1} = \frac{1}{4 \pi \epsilon_{0}} \cdot \frac{q_{1}}{r_{1P}^{2}} \hat{r}_{1P}$
where $r_{1P} = |\vec{r} - \vec{r}_{1}|$ and $\hat{r}_{1P}$ is the unit vector directed from $q_{1}$ to $P$.
Similarly,the electric field $\vec{E}_{i}$ at point $P$ due to any charge $q_{i}$ is:
$\vec{E}_{i} = \frac{1}{4 \pi \epsilon_{0}} \cdot \frac{q_{i}}{r_{iP}^{2}} \hat{r}_{iP}$
The total electric field $\vec{E}$ at point $P$ is the vector sum of individual fields:
$\vec{E} = \vec{E}_{1} + \vec{E}_{2} + \ldots + \vec{E}_{n} = \sum_{i=1}^{n} \vec{E}_{i}$
Substituting the expression for $\vec{E}_{i}$:
$\vec{E}(\vec{r}) = \frac{1}{4 \pi \epsilon_{0}} \sum_{i=1}^{n} \frac{q_{i}}{r_{iP}^{2}} \hat{r}_{iP}$
Here,$\vec{E}$ is a vector quantity that varies from point to point in space,determined by the positions and magnitudes of the source charges.

Explore More

Similar Questions

Three charges $2q, -q$ and $-q$ are located at the vertices of an equilateral triangle. At the center of the triangle,

Point charges $+8q$ and $-2q$ are placed at $x = 0$ and $x = L$ respectively. At which point on the $X$-axis will the resultant electric field intensity be zero?

$A$ uniformly charged semicircular arc of radius $r$ has a linear charge density $\lambda$. What is the electric field at its centre? ($\varepsilon_0$ is the permittivity of free space)

Two point charges $q_1 = \sqrt{10} \, \mu C$ and $q_2 = -25 \, \mu C$ are placed on the $x$-axis at $x = 1 \, m$ and $x = 4 \, m$ respectively. The electric field (in $V/m$) at a point $y = 3 \, m$ on the $y$-axis is,[ take $\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9 \, Nm^2C^{-2}$ ]

In the Millikan's experiment,the distance between two horizontal plates is $2.5 \, cm$ and the potential difference applied is $250 \, V$. The electric field between the plates will be ....... $V/m$.

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