Question

An electric field of intensity 3.25 kN/C is applied along the x-axis. Calculate the electric flux through a rectangular plane 0.350 m wide and 0.700 m long if the following conditions are true. (a) The plane is parallel to the yz-plane. N · m2/C (b) The plane is parallel to the xy-plane. N · m2/C (c) The plane contains the y-axis, and its normal makes an angle of 30.5° with the x-axis. N · m2/C

Answer 1

Answer:

$\varphi_1= 796.25 N m^2/C$

$\varphi_2= 0 N m^2/C$

$\varphi_3=686.1 N m^2/C$

Explanation:

From the question we are told that

Electric field of intensity $E= 3.25 kN/C$

Rectangle parameter Width $W=0.350 m$  Length $L=0.700 m$

Angle to the normal $\angle=30.5 \textdegree$

Generally the equation for Electric flux at parallel to the yz plane $\varphi_1$ is mathematically given by

$\varphi_1=EA cos theta$

$\varphi_1=3.25* 10^3 N/C * ( 0.350)(0.700) cos 0$

$\varphi_1= 796.25 N m^2/C$

Generally the equation for Electric flux at parallel to xy  plane $\varphi_2$ is mathematically given by

$\varphi_2=EA cos theta$

$\varphi_2=3.25* 10^3 N/C * ( 0.350)(0.700) cos 90$

$\varphi_2= 0 N m^2/C$

Generally the equation for Electric flux at angle 30 to x plane $\varphi_3$ is mathematically given by

$\varphi_3=EA cos theta$

$\varphi_3=3.25* 10^3 N/C * ( 0.350)(0.700) cos 30.5$

$\varphi_3=686.072219 N m^2/C$

$\varphi_3=686.1 N m^2/C$