The electric field in a region is given by ve | vedclass

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

Question

(D) The electric field is given by $\vec{E} = (2 \hat{i} + 4 \hat{j} + 6 \hat{k}) 	imes 10^3 \ N/C$. Since the surface is parallel to the $x-z$ plane

Vedclass · Accepted Answer

$15$

Vedclass · Answer

(D) The electric field is given by $\vec{E} = (2 \hat{i} + 4 \hat{j} + 6 \hat{k}) 	imes 10^3 \ N/C$. Since the surface is parallel to the $x-z$ plane,its area vector $\vec{A}$ must be perpendicular to the $x-z$ plane,which means it is directed along the $y$-axis. Thus,$\vec{A} = A \hat{j}$. The electric flux $\phi$ is given by the dot product $\phi = \vec{E} \cdot \vec{A}$. Substituting the values: $\phi = (2 \hat{i} + 4 \hat{j} + 6 \hat{k}) 	imes 10^3 \cdot (A \hat{j}) = 4 	imes 10^3 	imes A$. Given $\phi = 6.0 \ N m^2 C^{-1}$,we have $6.0 = 4 	imes 10^3 	imes A$. Solving for $A$: $A = \frac{6.0}{4 	imes 10^3} = 1.5 	imes 10^{-3} \ m^2$. Converting to $cm^2$: $A = 1.5 	imes 10^{-3} 	imes (10^2 \ cm)^2 = 1.5 	imes 10^{-3} 	imes 10^4 \ cm^2 = 15 \ cm^2$.

The electric field in a region is given by $\vec{E}=(2 \hat{i}+4 \hat{j}+6 \hat{k}) \times 10^3 \ N/C$. The flux of the field through a rectangular surface parallel to the $x-z$ plane is $6.0 \ N m^2 C^{-1}$. The area of the surface is . . . . . . $cm^2$.

Similar Questions

The charge distribution is shown in the figure below. The electric flux through the surface $S$ due to these charges is .........

The distribution of some charges on two Gaussian surfaces $A$ and $B$ are as shown in the figure. If $\phi_A$ and $\phi_B$ are electric fluxes linked with the surfaces $A$ and $B$ respectively,then $\frac{\phi_A}{\phi_B}=$

$A$ charged particle $q$ is placed at the centre $O$ of a cube of side length $L$ $(A, B, C, D, E, F, G, H)$. Another identical charge $q$ is placed at a distance $L$ from $O$, outside the cube, as shown in the figure. The electric flux through the face $BGFC$ is

$A$ hollow cylinder has a charge '$q$' $C$ within it. If '$\phi$' is the electric flux associated with the curved surface $B$,the flux linked with the plane surface $A$ will be

$q_1, q_2, q_3$ and $q_4$ are point charges located 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?

Vedclass Test Series

Exam Paper Generator

Online Exam Module