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

Question

contestada

Respuesta :

okpalawalter8 okpalawalter8 · Answer 1 · 2021-02-26T01:35:53+01:00

Answer:

[tex]\varphi_1= 796.25 N m^2/C[/tex]

[tex]\varphi_2= 0 N m^2/C[/tex]

[tex]\varphi_3=686.1 N m^2/C[/tex]

Explanation:

From the question we are told that

Electric field of intensity [tex]E= 3.25 kN/C[/tex]

Rectangle parameter Width [tex]W=0.350 m[/tex] Length [tex]L=0.700 m[/tex]

Angle to the normal [tex]\angle=30.5 \textdegree[/tex]

Generally the equation for Electric flux at parallel to the yz plane [tex]\varphi_1[/tex] is mathematically given by

[tex]\varphi_1=EA cos theta[/tex]

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

[tex]\varphi_1= 796.25 N m^2/C[/tex]

Generally the equation for Electric flux at parallel to xy plane [tex]\varphi_2[/tex] is mathematically given by

[tex]\varphi_2=EA cos theta[/tex]

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

[tex]\varphi_2= 0 N m^2/C[/tex]

Generally the equation for Electric flux at angle 30 to x plane [tex]\varphi_3[/tex] is mathematically given by

[tex]\varphi_3=EA cos theta[/tex]

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

[tex]\varphi_3=686.072219 N m^2/C[/tex]

[tex]\varphi_3=686.1 N m^2/C[/tex]