A thin semi-circular ring of radius r has a p | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Consider a small differential element $dl$ on the semi-circular ring. The charge $dq$ on this element is given by $dq = \lambda dl$,where $\lambda

Vedclass · Accepted Answer

$-\frac{q}{2\pi^2\varepsilon_0 r^2}\hat{j}$

Vedclass · Answer

(D) Consider a small differential element $dl$ on the semi-circular ring. The charge $dq$ on this element is given by $dq = \lambda dl$,where $\lambda = \frac{q}{\pi r}$ is the linear charge density.Since $dl = r d	heta$,we have $dq = \left(\frac{q}{\pi r}ight) (r d	heta) = \frac{q}{\pi} d	heta$.The magnitude of the electric field $dE$ at the centre $O$ due to this element is $dE = \frac{1}{4\pi\varepsilon_0} \frac{dq}{r^2} = \frac{1}{4\pi\varepsilon_0} \frac{q}{\pi r^2} d	heta = \frac{q}{4\pi^2\varepsilon_0 r^2} d	heta$.Due to symmetry,the horizontal components $(dE \cos	heta)$ of the electric field from elements on opposite sides of the vertical axis cancel each other out.The net electric field is the integral of the vertical components $(dE \sin	heta)$ from $	heta = 0$ to $\pi$.$E = \int_0^{\pi} dE \sin	heta = \int_0^{\pi} \frac{q}{4\pi^2\varepsilon_0 r^2} \sin	heta d	heta$.$E = \frac{q}{4\pi^2\varepsilon_0 r^2} [-\cos	heta]_0^{\pi} = \frac{q}{4\pi^2\varepsilon_0 r^2} (-(-1) - (-1)) = \frac{q}{4\pi^2\varepsilon_0 r^2} (2) = \frac{q}{2\pi^2\varepsilon_0 r^2}$.Since the charge is positive and distributed on the upper semi-circle,the field at the centre points downwards,i.e.,in the $-\hat{j}$ direction.Therefore,$\vec{E} = -\frac{q}{2\pi^2\varepsilon_0 r^2} \hat{j}$.

$A$ thin semi-circular ring of radius $r$ has a positive charge $q$ distributed uniformly over it. The net electric field $\vec{E}$ at the centre $O$ is

Similar Questions

Two particles $A$ and $B$ having charges $20\, \mu C$ and $-5\, \mu C$ respectively are held fixed with a separation of $5\, cm$. At what position should a third charged particle be placed so that it does not experience a net electric force?

The magnitude of a point charge due to which the electric field $30 \ cm$ away has a magnitude of $2 \ N \ C^{-1}$ is:

Two point charges $+8q$ and $-2q$ are placed at $x = 0$ and $x = L$ respectively. The net electric field due to these two point charges will be zero at a point on the $x$-axis at:

Vedclass Test Series

Exam Paper Generator

Online Exam Module