An electric field is uniform,and in the positive $x$ direction for positive $x,$ and uniform with the same magnitude but in the negative $x$ direction for negative $x$. It is given that $E = 200 \hat{i} \; N/C$ for $x > 0$ and $E = -200 \hat{i} \; N/C$ for $x < 0$. $A$ right circular cylinder of length $20 \; cm$ and radius $5 \; cm$ has its centre at the origin and its axis along the $x$-axis so that one face is at $x = +10 \; cm$ and the other is at $x = -10 \; cm$.
$(a)$ What is the net outward flux through each flat face?
$(b)$ What is the flux through the side of the cylinder?
$(c)$ What is the net outward flux through the cylinder?
$(d)$ What is the net charge inside the cylinder?

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) On the left face $(x = -10 \; cm)$,the electric field is $E = -200 \hat{i} \; N/C$ and the area vector is $\Delta S = -\Delta S \hat{i}$. The outward flux is:
$\phi_{L} = E \cdot \Delta S = (-200 \hat{i}) \cdot (-\Delta S \hat{i}) = 200 \Delta S$
$\phi_{L} = 200 \times \pi (0.05)^2 = 200 \times 3.1416 \times 0.0025 = 1.57 \; N \cdot m^2/C$
On the right face $(x = +10 \; cm)$,the electric field is $E = 200 \hat{i} \; N/C$ and the area vector is $\Delta S = \Delta S \hat{i}$. The outward flux is:
$\phi_{R} = E \cdot \Delta S = (200 \hat{i}) \cdot (\Delta S \hat{i}) = 200 \Delta S = 1.57 \; N \cdot m^2/C$
$(b)$ For any point on the curved side of the cylinder,the electric field $E$ is parallel to the surface and perpendicular to the area vector $\Delta S$. Thus,$E \cdot \Delta S = 0$. The flux through the side is zero.
$(c)$ The net outward flux $\phi$ is the sum of the fluxes through all surfaces:
$\phi = \phi_{L} + \phi_{R} + \phi_{side} = 1.57 + 1.57 + 0 = 3.14 \; N \cdot m^2/C$
$(d)$ Using Gauss's law,the net charge $q$ inside the cylinder is:
$q = \varepsilon_{0} \phi = (8.854 \times 10^{-12} \; C^2/N \cdot m^2) \times (3.14 \; N \cdot m^2/C) \approx 2.78 \times 10^{-11} \; C$

Explore More

Similar Questions

Electric field in a region is given by $\overrightarrow{E} = a \hat{i} + b \hat{j}$,where $a$ and $b$ are constants. The net flux passing through a square area of side $l$ parallel to the $y-z$ plane is

$A$ square surface of side $L$ metres is in the plane of the paper. $A$ uniform electric field $\vec{E} \text{ (V/m)}$,also in the plane of the paper,is limited only to the lower half of the square surface,(see figure). The electric flux in $SI$ units associated with the surface is

If a charge $Q$ is placed at the corner of a cube,what is the electric flux passing through one of its faces?

$A$ point charge $q$ is placed at a distance $a/2$ directly above the center of a square of side $a$. The electric flux through the square is:

$A$ uniformly charged conducting sphere of diameter $14 \ cm$ has a surface charge density of $40 \ \mu C/m^2$. The total electric flux leaving the surface of the sphere is nearly (Permittivity of free space $\varepsilon_0 = 8.85 \times 10^{-12} \ C^2/N \cdot m^2$)

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo