Charge q is uniformly distributed over a thin | vedclass

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

Question

(A) Consider a small element of length $d\ell = R d	heta$ on the half ring.Charge on this element is $dq = \lambda d\ell = \lambda R d	heta$,where $

Vedclass · Accepted Answer

$\frac{q}{2{\pi ^2}{\varepsilon _0}{R^2}}$

Vedclass · Answer

(A) Consider a small element of length $d\ell = R d	heta$ on the half ring.Charge on this element is $dq = \lambda d\ell = \lambda R d	heta$,where $\lambda = \frac{q}{\pi R}$ is the linear charge density.The electric field at the centre due to this element is $dE = \frac{1}{4\pi \varepsilon_0} \frac{dq}{R^2} = \frac{1}{4\pi \varepsilon_0} \frac{\lambda R d	heta}{R^2} = \frac{\lambda}{4\pi \varepsilon_0 R} d	heta$.Due to symmetry,the horizontal components of the electric field cancel out,and only the vertical components add up.The vertical component is $dE_y = dE \cos 	heta$.Integrating from $-\pi/2$ to $\pi/2$ or $2 \int_{0}^{\pi/2} dE \cos 	heta$:$E = 2 \int_{0}^{\pi/2} \frac{\lambda}{4\pi \varepsilon_0 R} \cos 	heta d	heta = \frac{2\lambda}{4\pi \varepsilon_0 R} [\sin 	heta]_0^{\pi/2} = \frac{\lambda}{2\pi \varepsilon_0 R}$.Substituting $\lambda = \frac{q}{\pi R}$,we get $E = \frac{q/(\pi R)}{2\pi \varepsilon_0 R} = \frac{q}{2\pi^2 \varepsilon_0 R^2}$.

Charge $q$ is uniformly distributed over a thin half ring of radius $R$. The electric field at the centre of the ring is

Similar Questions

The unit $Volt/metre$ is the unit of:

Two charges $+Q$ and $-2 Q$ are located at points $A$ and $B$ on a horizontal line as shown below. The electric field is zero at a point which is located at a finite distance:

$A$ ring of radius $R$ is charged with a charge $Q$. The electric field at a point on its axis at a distance $r$ from the circumference of the ring is:

Consider two points $1$ and $2$ in a region outside a charged sphere. These two points are not very far away from the sphere. If $\overrightarrow{E}$ and $V$ represent the electric field vector and the electric potential respectively,which of the following is not possible?

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?

Vedclass Test Series

Exam Paper Generator

Online Exam Module