Question

figure shows four sides of a 3.0 cm * 3.0 cm * 3.0 cm cube. a. What are the electric fluxes Φ1 to Φ4 through sides 1 to 4? b. What is the net flux through these four sides?

Short answer

Electric flux on the all four side of the cube

${\Phi }_1=-0.39\mathrm{N}\mathrm{m}2$

${\Phi }_2=0.225{\mathrm{Nm}}^2/\mathrm{C}$

${\Phi }_{3=}0.39{\mathrm{Nm}}^2/\mathrm{C}$

${\Phi }_4=-0.225\mathrm{N}\mathrm{m}2$

net flux through the four sides of cube is$=0$

Step-by-step solution

part(a) step 1: given informaiton

cube with side of 3.0 cm

${\Phi }_E=AE\cos \theta$

$\Phi =ElectricFlux$

Again, the flux is${\Phi }_E=AE\cos \theta .$

On side 1 , the angle between$E~$and the normal is

${\theta }_1=150°$

For side 2 , it's ${\theta }_2=60°$

For side 3 it's ${\theta }_3=30°$

and for side 4 it's ${\theta }_4=120°$

$\text{Thus,}{\Phi }_1=EA\cos {\theta }_1$

$=\left(500\right)(.03{)}^2\cos 150°$

$=-0.39\mathrm{N}/{\mathrm{Cm}}^2$

${\Phi }_2=EA\cos {\theta }_2$

$=\left(500\right)(.03)2\cos 60°$

$=0.225\mathrm{N}/{\mathrm{Cm}}^2$

${\Phi }_3=EA\cos \theta 3$

$=\left(500\right)(.03)2\cos 30°$

$=0.39\mathrm{N}/{\mathrm{Cm}}^2$

${\Phi }_4=EA\cos {\theta }_4$

$=\left(500\right)(.03)2\cos 120°$

$=-0.225\mathrm{N}/{\mathrm{Cm}}^2$

part(b) step 1: given informaiton

cube with side 3.0

Formula Used:

${\Phi }_E=AE\cos \theta$

$\Phi =ElectricFlux$

A= Cross-sectional area of the loop

Again, the flux is ${\Phi }_E=AE\cos \theta$

On side 1 , the angle between $E~$and the normal is

${\theta }_1=150°\text{.}$

$Forside2,it\text{'}s{\theta }_2=60°$

For side 3 it's${\theta }_3=30°$

and for side 4 it's ${\theta }_4=120°$

Taking the values from part A,

The net flux through the four sides is just the sum of the individual fluxes,

$\Phi ={\Phi }_1+{\Phi }_2+{\Phi }_3+{\Phi }_4=0,$

which we know has to be the case, since there is no enclosed net charge.