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30 - Excercises And Problems

Expert-verified
Found in: Page 725

### Physics For Scientists & Engineers

Book edition 9th Edition
Author(s) Raymond A. Serway, John W. Jewett
Pages 1624 pages
ISBN 9781133947271

# 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?

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$=0$

See the step by step solution

## part(a) step 1: given informaiton

cube with side of 3.0 cm

${\Phi }_{E}=AE\mathrm{cos}\theta$

$\Phi =ElectricFlux$

Again, the flux is${\Phi }_{E}=AE\mathrm{cos}\theta .$

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 ${\theta }_{4}=120°$

$\text{Thus,}{\Phi }_{1}=EA\mathrm{cos}{\theta }_{1}$

$=\left(500\right)\left(.03{\right)}^{2}\mathrm{cos}150°$

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

${\Phi }_{2}=EA\mathrm{cos}{\theta }_{2}$

$=\left(500\right)\left(.03\right)2\mathrm{cos}60°$

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

${\Phi }_{3}=EA\mathrm{cos}\theta 3$

$=\left(500\right)\left(.03\right)2\mathrm{cos}30°$

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

${\Phi }_{4}=EA\mathrm{cos}{\theta }_{4}$

$=\left(500\right)\left(.03\right)2\mathrm{cos}120°$

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

## part(b) step 1: given informaiton

cube with side 3.0

Formula Used:

${\Phi }_{E}=AE\mathrm{cos}\theta$

$\Phi =ElectricFlux$

A= Cross-sectional area of the loop

Again, the flux is ${\Phi }_{E}=AE\mathrm{cos}\theta$

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.

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