Two co-axial conducting cylinders of same len | vedclass

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

Question

(B) The electric field exists only in the region between the cylinders $(\sqrt{2}R  2R$,the

Vedclass · Accepted Answer

$\frac{ Q }{60 \epsilon_0}$

Vedclass · Answer

(B) The electric field exists only in the region between the cylinders $(\sqrt{2}R  2R$,the field is zero because the outer cylinder is grounded and encloses the inner charge.Using Gauss's Law for a cylinder of radius $r$ $(\sqrt{2}R The flux through an area element $dS = \ell dy$ on the plane is $d\phi = \vec{E} \cdot d\vec{S} = E \cos 	heta \ell dy$.From the geometry,$r = R \sec 	heta$ and $y = R 	an 	heta$,so $dy = R \sec^2 	heta d	heta$. Also $\cos 	heta = R/r$.Substituting these,$d\phi = \frac{\lambda}{2 \pi \kappa \epsilon_0 r} \cdot \frac{R}{r} \cdot \ell \cdot R \sec^2 	heta d	heta = \frac{\lambda \ell}{2 \pi \kappa \epsilon_0} d	heta$.The field is non-zero only for the segments $AB$ and $CD$ where $\sqrt{2}R For $r = \sqrt{2}R$,$\cos 	heta = R/(\sqrt{2}R) = 1/\sqrt{2} \Rightarrow 	heta = 45^\circ = \pi/4$.For $r = 2R$,$\cos 	heta = R/(2R) = 1/2 \Rightarrow 	heta = 60^\circ = \pi/3$.The flux through one segment is $\phi_{AB} = \int_{\pi/4}^{\pi/3} \frac{\lambda \ell}{2 \pi \kappa \epsilon_0} d	heta = \frac{Q}{2 \pi \kappa \epsilon_0} (\pi/3 - \pi/4) = \frac{Q}{2 \pi \kappa \epsilon_0} (\pi/12) = \frac{Q}{24 \kappa \epsilon_0}$.Given $\kappa = 5$,$\phi_{AB} = \frac{Q}{24 	imes 5 \epsilon_0} = \frac{Q}{120 \epsilon_0}$.Total flux through the plane is $\phi_{total} = \phi_{AB} + \phi_{CD} = 2 	imes \frac{Q}{120 \epsilon_0} = \frac{Q}{60 \epsilon_0}$.

Similar Questions

The electric flux from a cube of edge $l$ is $\phi$. If an edge of the cube is made $2l$ and the charge enclosed is halved,its value will be

An electric field,$\overrightarrow{E} = \frac{2 \hat{i} + 6 \hat{j} + 8 \hat{k}}{\sqrt{6}} \ V/m$,passes through a surface of $4 \ m^2$ area having a unit normal vector $\hat{n} = \left( \frac{2 \hat{i} + \hat{j} + \hat{k}}{\sqrt{6}} \right)$. The electric flux through that surface is:

How do electric field lines depend on the area or the solid angle subtended by the area?

$A$ charge $Q \; \mu C$ is placed at the centre of a cube. The flux coming out from any one of its faces will be:

Four charges $2 \mu C, -3 \mu C, 4 \mu C, -4 \mu C$ and $-1 \mu C$ are enclosed by a Gaussian surface of radius $2 \ m$. The net outward flux through the Gaussian surface is (in $\mu V-m$):

Vedclass Test Series

Exam Paper Generator

Online Exam Module