A square loop of sides a=1 m is held normall | vedclass

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

Question

(A) The charge $q$ is placed at a distance $a/2$ from the center of the square loop of side $a$. By symmetry,we can enclose the charge $q$ in a cube o

Vedclass · Accepted Answer

$48$

Vedclass · Answer

(A) The charge $q$ is placed at a distance $a/2$ from the center of the square loop of side $a$. By symmetry,we can enclose the charge $q$ in a cube of side $a$ such that the square loop forms one of the faces of the cube. The total electric flux through the entire cube is $\Phi_{total} = \frac{q}{\varepsilon_0}$. Since the cube has $6$ identical faces,the flux through the square loop (which is one face) is $\Phi_{square} = \frac{1}{6} \frac{q}{\varepsilon_0}$. The square loop can be divided into $8$ identical triangular regions by drawing lines from the center to the corners and midpoints of the sides. Due to symmetry,the flux through each of these $8$ identical triangular regions is equal. Therefore,the flux through each triangular region is $\Phi_{triangle} = \frac{1}{8} \Phi_{square} = \frac{1}{8} 	imes \frac{1}{6} \frac{q}{\varepsilon_0} = \frac{1}{48} \frac{q}{\varepsilon_0}$. The shaded region in the figure consists of $5$ such identical triangular regions. Thus,the flux through the shaded region is $\Phi_{shaded} = 5 	imes \Phi_{triangle} = 5 	imes \frac{1}{48} \frac{q}{\varepsilon_0} = \frac{5}{48} \frac{q}{\varepsilon_0}$. Given $q = 1 \ C$,the flux is $\frac{5}{48} \frac{1}{\varepsilon_0}$. Comparing this with $\frac{5}{p} 	imes \frac{1}{\varepsilon_0}$,we get $p = 48$.

List-$I$	List-$II$
$(a)$ Net flux through $S_1$	$(i)$ $\frac{-2Q}{\epsilon_0}$
$(b)$ Net flux through $S_2$	(ii) $\frac{-Q}{\epsilon_0}$
$(c)$ Net flux through $S_3$	(iii) Zero

Similar Questions

$A$ charge '$Q$' $C$ is placed at the centre of a cube. The electric flux through two opposite faces of the cube is $(\epsilon_0 = \text{permittivity of free space})$

The electric field in a region of space is given by $\vec{E} = (5\hat{i} + 2\hat{j}) \text{ N/C}$. Calculate the electric flux through a surface of area $2 \text{ m}^2$ lying in the $YZ$-plane.

Consider a cylinder of length $L$ and radius $R$ placed in a uniform electric field $E$ such that its axis is parallel to the electric field. The total electric flux associated with the cylinder is .......

Two point charges $2q$ and $q$ are placed at vertex $A$ and the centre of face $CDEF$ of the cube, respectively, as shown in the figure. The electric flux passing through the cube is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module