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

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(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

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

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(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:

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