A cubical region of side a has its centre at | vedclass

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

Question

(C) The total charge enclosed by the cube is $Q_{	ext{enclosed}} = -q + 3q - q = q$.According to Gauss's Law,the net electric flux through the entire

Vedclass · Accepted Answer

$(A, C, D)$

Vedclass · Answer

(C) The total charge enclosed by the cube is $Q_{	ext{enclosed}} = -q + 3q - q = q$.According to Gauss's Law,the net electric flux through the entire closed surface is $\phi = \frac{Q_{	ext{enclosed}}}{\varepsilon_0} = \frac{q}{\varepsilon_0}$. Thus,option $(C)$ is correct.The charges are located on the $y$-axis at $(0, -a/4, 0)$,$(0, 0, 0)$,and $(0, a/4, 0)$.Since the charge distribution is symmetric with respect to the $yz$-plane $(x=0)$,the flux through the plane $x = +a/2$ must be equal to the flux through the plane $x = -a/2$. Thus,option $(A)$ is correct.The charge distribution is not symmetric with respect to the $xz$-plane $(y=0)$. The charges are located at $y = -a/4, 0, a/4$. The plane $y = +a/2$ is closer to the charge at $y = +a/4$ than the plane $y = -a/2$ is to the charge at $y = -a/4$. However,by calculating the flux,one finds the flux through $y = +a/2$ and $y = -a/2$ are actually equal due to the specific arrangement of charges. Thus,$(B)$ is incorrect.By symmetry of the charge distribution with respect to the $xy$-plane $(z=0)$,the flux through $z = +a/2$ is equal to the flux through $z = -a/2$. Comparing this to the flux through $x = +a/2$,we find they are equal. Thus,$(D)$ is correct.

Similar Questions

$A$ point charge $+q$ is placed in front of a grounded metal plate. Points $P$ and $Q$ are between the charge $X$ and the plate $Y$ as shown in the figure. Let $E_P$ and $E_Q$ be the magnitudes of the electric fields at points $P$ and $Q$ respectively. Which of the following statements is true?

$A$ point charge $q$ is placed on the center of the flat circular base of a hemispherical surface as shown in the figure. The net flux of the electric field through the hemispherical surface is closest to:

If there is only one type of charge in the universe,then ($ \vec{E} $ = Electric field,$ \vec{d}s $ = Area vector):

The electric field in a region is given by $\vec E = \frac{3}{5}E_0\hat i + \frac{4}{5}E_0\hat j$ and $E_0 = 2 \times 10^3 \, N/C$. Then,the flux of this field through a rectangular surface of area $0.2 \, m^2$ parallel to the $y-z$ plane is ...... $\frac{N \cdot m^2}{C}$.

The $S.I.$ unit of electric flux is

Vedclass Test Series

Exam Paper Generator

Online Exam Module