A semi-infinite non-conducting rod lies along | vedclass

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

Question

(C) Consider an infinitesimal section of the rod of length $dx$,at a distance $x$ from the origin,as shown in the figure.It contains charge $dq = \lam

Vedclass · Accepted Answer

$\frac{\lambda}{2 \sqrt{2} \pi \varepsilon_0 L}$

Vedclass · Answer

(C) Consider an infinitesimal section of the rod of length $dx$,at a distance $x$ from the origin,as shown in the figure.It contains charge $dq = \lambda dx$ and is at a distance $r = \sqrt{x^2 + L^2}$ from the point $(0, L)$.The magnitude of the electric field produced by $dx$ at the point $(0, L)$ is $dE = \frac{k dq}{r^2} = \frac{k \lambda dx}{x^2 + L^2}$.The $x$ and $y$ components of the electric field are $dE_x = dE \sin 	heta$ and $dE_y = dE \cos 	heta$,where $	an 	heta = \frac{x}{L}$.Thus,$x = L 	an 	heta$ and $dx = L \sec^2 	heta d	heta$. Also,$r^2 = L^2 \sec^2 	heta$.Substituting these into the components:$dE_x = \frac{k \lambda (L \sec^2 	heta d	heta)}{L^2 \sec^2 	heta} \sin 	heta = \frac{k \lambda}{L} \sin 	heta d	heta$.$dE_y = \frac{k \lambda (L \sec^2 	heta d	heta)}{L^2 \sec^2 	heta} \cos 	heta = \frac{k \lambda}{L} \cos 	heta d	heta$.Integrating from $x=0$ to $x=\infty$ corresponds to $	heta$ from $0$ to $\frac{\pi}{2}$:$E_x = \int_0^{\pi/2} \frac{k \lambda}{L} \sin 	heta d	heta = \frac{k \lambda}{L} [-\cos 	heta]_0^{\pi/2} = \frac{k \lambda}{L}$.$E_y = \int_0^{\pi/2} \frac{k \lambda}{L} \cos 	heta d	heta = \frac{k \lambda}{L} [\sin 	heta]_0^{\pi/2} = \frac{k \lambda}{L}$.The magnitude of the resultant electric field is $E = \sqrt{E_x^2 + E_y^2} = \sqrt{(\frac{k \lambda}{L})^2 + (\frac{k \lambda}{L})^2} = \sqrt{2} \frac{k \lambda}{L}$.Substituting $k = \frac{1}{4 \pi \varepsilon_0}$,we get $E = \frac{\sqrt{2} \lambda}{4 \pi \varepsilon_0 L} = \frac{\lambda}{2 \sqrt{2} \pi \varepsilon_0 L}$.

$A$ semi-infinite non-conducting rod lies along the $+x$-axis with its left end at the origin. The rod has a uniform linear charge density $\lambda$. The magnitude of the electric field $|\vec{E}|$ at a point on the $y$-axis at a distance $L$ from the origin will be:

Similar Questions

Four positive point charges $+q$ are kept at the four corners of a square of side $l$. The net electric field at the midpoint of any one side of the square is (take $\frac{1}{4 \pi \epsilon_0}=k$ ).

$4 \times 10^{10}$ electrons are removed from a neutral metal sphere of diameter $20 \text{ cm}$ placed in air. The magnitude of the electric field (in $\text{N C}^{-1}$) at a distance of $20 \text{ cm}$ from its centre is:

In the given figure,the electric field at the center $O$ due to section $AB$ of a uniformly charged ring is $\overrightarrow{E}$. What will be the electric field at $O$ due to the remaining section $ACB$?

Two point charges $-5 \mu C$ and $+5 \mu C$ are placed at $A$ and $B$ respectively on a straight line separated by $5 \ cm$. At a point $C$,if the resultant electric field is parallel to the line joining the points $A$ and $B$,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module