Write an equation for the electric potential due to a linear charge distribution.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) The electric potential $V$ at a point $P$ due to a continuous linear charge distribution with linear charge density $\lambda$ along a line element $dl$ is given by the integral:
$V = \frac{1}{4\pi\epsilon_0} \int \frac{\lambda dl}{r}$
where:
$1. \epsilon_0$ is the permittivity of free space.
$2. \lambda$ is the linear charge density (charge per unit length).
$3. dl$ is the infinitesimal length element of the distribution.
$4. r$ is the distance from the charge element $dl$ to the point $P$ where the potential is being calculated.

Explore More

Similar Questions

$A$ uniformly charged conducting sphere of $2.4 \, m$ diameter has a surface charge density of $80.0 \, \mu C m^{-2}$. The charge on the sphere is nearly

Obtain the expression for the electric field at any point due to a continuous distribution of charge on a $(i)$ line,$(ii)$ surface,and $(iii)$ volume.

Difficult
View Solution

Two spherical hollow spheres of radii $R_1$ and $R_2$ are charged with the same charge $Q$. If $\sigma_1$ and $\sigma_2$ are their respective surface charge densities, then the ratio $\sigma_1 : \sigma_2$ is:

Give definitions of linear,surface,and volume charge densities and write their $SI$ units.

Each of two large conducting parallel plates has one-sided surface area $A$. If one of the plates is given a charge $Q$ whereas the other is neutral,then the electric field at a point in between the plates is given by

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