A line charge of length frac{a}{2} is kept at | vedclass

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

Question

(A) The total length of the line charge is $L = \frac{a}{2}$.The total charge $q$ of the line charge is $q = \lambda L = \lambda \left( \frac{a}{2} \r

Vedclass · Accepted Answer

$\frac{\lambda a}{8 \epsilon_0}$

Vedclass · Answer

(A) The total length of the line charge is $L = \frac{a}{2}$.The total charge $q$ of the line charge is $q = \lambda L = \lambda \left( \frac{a}{2} \right) = \frac{\lambda a}{2}$.This line charge is placed at the center of an edge of the cube. An edge is shared by $4$ identical cubes in a symmetric arrangement.Therefore, the fraction of the charge enclosed by the given cube is $\frac{1}{4}$ of the total charge.Thus, the charge enclosed by the cube is $q_{in} = \frac{q}{4} = \frac{\lambda a / 2}{4} = \frac{\lambda a}{8}$.According to Gauss's Law, the total electric flux $\phi$ through the cube is $\phi = \frac{q_{in}}{\varepsilon_0}$.Substituting the value of $q_{in}$, we get $\phi = \frac{\lambda a}{8 \varepsilon_0}$.

Similar Questions

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:

$A$ point charge of $2.0 \; \mu C$ is at the centre of a cubic Gaussian surface $9.0 \; cm$ on edge. What is the net electric flux through the surface?

$q_1, q_2, q_3$ and $q_4$ are point charges located at points as shown in the figure and $S$ is a spherical Gaussian surface of radius $R$. Which of the following is true according to Gauss's law?

$A$ charge $q$ is placed at the centre of the open end of a cylindrical vessel. The flux of the electric field through the surface of the vessel is:

The flux of the electric field $\overrightarrow{E} = 24 \hat{i} + 30 \hat{j} + 28 \hat{k} \text{ NC}^{-1}$ through an area of $20 \text{ m}^2$ on the $yz$ plane is

Vedclass Test Series

Exam Paper Generator

Online Exam Module