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

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

Question

(D) The electric field is $\overrightarrow{E} = \frac{3 E_{0}}{5} \hat{i} + \frac{4 E_{0}}{5} \hat{j} \, N/C$.For the first surface,the area vector is

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(D) The electric field is $\overrightarrow{E} = \frac{3 E_{0}}{5} \hat{i} + \frac{4 E_{0}}{5} \hat{j} \, N/C$.For the first surface,the area vector is $\overrightarrow{A}_{1} = 0.2 \hat{i} \, m^{2}$ (since it is parallel to the $y-z$ plane).The flux $\phi_{1} = \overrightarrow{E} \cdot \overrightarrow{A}_{1} = (\frac{3 E_{0}}{5} \hat{i} + \frac{4 E_{0}}{5} \hat{j}) \cdot (0.2 \hat{i}) = \frac{3 	imes 0.2}{5} E_{0} = 0.12 E_{0} \, V \cdot m$.For the second surface,the area vector is $\overrightarrow{A}_{2} = 0.3 \hat{j} \, m^{2}$ (since it is parallel to the $x-z$ plane).The flux $\phi_{2} = \overrightarrow{E} \cdot \overrightarrow{A}_{2} = (\frac{3 E_{0}}{5} \hat{i} + \frac{4 E_{0}}{5} \hat{j}) \cdot (0.3 \hat{j}) = \frac{4 	imes 0.3}{5} E_{0} = 0.24 E_{0} \, V \cdot m$.The ratio of the fluxes is $\frac{\phi_{1}}{\phi_{2}} = \frac{0.12 E_{0}}{0.24 E_{0}} = \frac{1}{2}$.Given the ratio is $a : b = 1 : 2$,therefore $a = 1$.

Similar Questions

Five charges $+q, +5q, -2q, +3q$ and $-4q$ are situated as shown in the figure. The electric flux due to this configuration through the surface $S$ is

If a charge $q$ is placed at the centre of the flat surface of a closed hemispherical non-conducting surface,what is the total electric flux passing through the flat surface?

How much electric flux will come out through a surface $S = 10\hat{j}$ kept in an electrostatic field $\vec{E} = 2\hat{i} + 4\hat{j} + 7\hat{k}$ units?

Three positive charges of equal value $q$ are placed at the vertices of an equilateral triangle. The resulting lines of force should be sketched as in:

What is called a Gaussian surface?

Vedclass Test Series

Exam Paper Generator

Online Exam Module