An electric field overrightarrow{E} = 4x hat{ | vedclass

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

Question

(D) The electric field is given by $\overrightarrow{E} = 4x \hat{i} - (y^2 + 1) \hat{j}$.Surface $ABCD$ lies in the $xy$-plane at $z = 2$. The area ve

Vedclass · Accepted Answer

$-48$

Vedclass · Answer

(D) The electric field is given by $\overrightarrow{E} = 4x \hat{i} - (y^2 + 1) \hat{j}$.Surface $ABCD$ lies in the $xy$-plane at $z = 2$. The area vector for this surface is $\overrightarrow{S}_I = S \hat{k}$. Since $\overrightarrow{E} \cdot \hat{k} = 0$,the flux $\phi_I = 0$.Surface $BCGF$ lies in the $yz$-plane at $x = 3$. The area vector for this surface is $\overrightarrow{S}_{II} = 4 \hat{i}$ (area = $2 	imes 2 = 4$).At $x = 3$,the electric field is $\overrightarrow{E} = 4(3) \hat{i} - (y^2 + 1) \hat{j} = 12 \hat{i} - (y^2 + 1) \hat{j}$.The flux $\phi_{II} = \int \overrightarrow{E} \cdot d\overrightarrow{S} = \int_{0}^{2} \int_{0}^{2} (12 \hat{i} - (y^2 + 1) \hat{j}) \cdot (dy dz \hat{i}) = \int_{0}^{2} \int_{0}^{2} 12 \, dy dz = 12 	imes 4 = 48 	ext{ Nm}^2/C$.Thus,$\phi_I - \phi_{II} = 0 - 48 = -48 	ext{ Nm}^2/C$.

Similar Questions

Draw the electric field lines for a negative point charge.

The electric field in a region is given by $\overrightarrow{E} = \frac{2}{5} E_{0} \hat{i} + \frac{3}{5} E_{0} \hat{j}$ with $E_{0} = 4.0 \times 10^{3} \, N/C$. The flux of this field through a rectangular surface area $0.4 \, m^{2}$ parallel to the $Y-Z$ plane is ....... $N m^{2} C^{-1}$.

$A$ square Gaussian surface is placed in the $y-z$ plane. Its axis is along the $x-$axis and its center is at the origin. Two identical charges,each $Q$,are placed at points $(a, 0, 0)$ and $(-a, 0, 0)$. If each side length of the square is $2a$,then the electric flux passing through the square is:

In the figure,a $+Q$ charge is located at one of the corners of the cube. What is the electric flux through the cube due to the $+Q$ charge?

$A$ point charge $q$ is placed at the center $O$ of a cube of side length $L$. Another identical charge $q$ is placed at a distance $L$ from $O$ as shown in the figure. The electric flux through the face $ABCD$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module