An infinitely long uniform line charge distri | vedclass

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

Question

(C) The electric flux through a surface is related to the solid angle subtended by the surface at the line charge. Alternatively,we can use the concep

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(C) The electric flux through a surface is related to the solid angle subtended by the surface at the line charge. Alternatively,we can use the concept of symmetry. The rectangular surface has a width $a$ and length $L$. The distance from the line charge to the center of the rectangle is $d = \frac{\sqrt{3}}{2} a$.The angle $	heta$ subtended by the width $a$ at the line charge is given by $	an(	heta/2) = \frac{a/2}{d} = \frac{a/2}{(\sqrt{3}/2)a} = \frac{1}{\sqrt{3}}$.Thus,$	heta/2 = 30^{\circ}$,which means $	heta = 60^{\circ}$.The total angle around the line charge is $360^{\circ}$. The number of such identical rectangular surfaces required to enclose the line charge completely is $n = \frac{360^{\circ}}{60^{\circ}} = 6$.According to Gauss's Law,the total flux through a closed surface enclosing a charge $q_{enclosed}$ is $\frac{q_{enclosed}}{\varepsilon_0}$. For a length $L$ of the line charge,the enclosed charge is $q = \lambda L$.Since the flux is distributed equally among the $6$ surfaces,the flux through one surface is $\phi = \frac{\lambda L}{6 \varepsilon_0}$.Comparing this with the given expression $\frac{\lambda L}{n \varepsilon_0}$,we get $n = 6$.

Similar Questions

In a region of space,the electric field is given by $\vec E = 8\hat i + 4\hat j + 3\hat k$. The electric flux through a surface of area $100 \text{ units}$ in the $x-y$ plane is...

The electric flux through surface $S_1$ is:

The figure shows electric field lines due to a charge configuration. From this,we conclude that:

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:

The figure shows the electric field lines. The spacing between the lines is parallel to the paper at every point. If the magnitude of the field at $A$ is $40 \ N/C$,then the approximate magnitude of the field at $B$ is ....... $N/C$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module