An infinite line charge is at the axis of a c | vedclass

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

Question

(A) The electric flux $\phi$ through the curved surface of a cylinder of radius $r$ and length $L$ is given by $\phi = E 	imes A$,where $A$ is the cu

Vedclass · Accepted Answer

$1.1 	imes 10^2$

Vedclass · Answer

(A) The electric flux $\phi$ through the curved surface of a cylinder of radius $r$ and length $L$ is given by $\phi = E 	imes A$,where $A$ is the curved surface area.The curved surface area $A = 2 \pi r L$.Given: $E = 250 \,NC^{-1}$,$r = 7 \,cm = 0.07 \,m$,and $L = 1 \,m$.Substituting the values:$\phi = 250 	imes (2 	imes \pi 	imes 0.07 	imes 1)$$\phi = 250 	imes (0.14 \pi)$$\phi = 35 \pi$Using $\pi \approx 3.14159$:$\phi \approx 35 	imes 3.14159 \approx 109.95 \,Nm^2C^{-1}$Rounding to two significant figures,we get $\phi \approx 1.1 	imes 10^2 \,Nm^2C^{-1}$.

An infinite line charge is at the axis of a cylinder of length $1 \,m$ and radius $7 \,cm$. If the electric field at any point on the curved surface of the cylinder is $250 \,NC^{-1}$,then the net electric flux through the cylinder is ............ $Nm^2C^{-1}$.

Similar Questions

The electric field in a region is given by $\vec E = a\hat i + b\hat j$. Here $a$ and $b$ are constants. Find the net flux passing through a square area of side $l$ parallel to the $y-z$ plane.

The distribution of some charges on two Gaussian surfaces $A$ and $B$ are as shown in the figure. If $\phi_A$ and $\phi_B$ are electric fluxes linked with the surfaces $A$ and $B$ respectively,then $\frac{\phi_A}{\phi_B}=$

The charge distribution is shown in the figure below. The electric flux through the surface $S$ due to these charges is .........

$A$ charge '$Q$ $\mu C$' is placed at the centre of a cube. The flux through one face and two opposite faces of the cube is respectively:

Consider the charge configuration and spherical Gaussian surface as shown in the figure. When calculating the flux of the electric field over the spherical surface,the electric field will be due to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module