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30 - Excercises And Problems

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Physics For Scientists & Engineers
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Illustration

Short Answer

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

Φ1=-0.39 N m2

Φ2=0.225 Nm2/C

Φ3=0.39 Nm2/C

Φ4=-0.225 N m2

net flux through the four sides of cube is=0

See the step by step solution

Step by Step Solution

part(a) step 1: given informaiton

cube with side of 3.0 cm

ΦE=AEcosθ

Φ=ElectricFlux

Again, the flux isΦE=AEcosθ.

On side 1 , the angle betweenE~ and the normal is

θ1=150°

For side 2 , it's θ2=60°

For side 3 it's θ3=30°

and for side 4 it's θ4=120°

Thus, Φ1=EAcosθ1

=(500)(.03)2cos150°

=-0.39 N/Cm2

Φ2=EAcosθ2

=(500)(.03)2cos60°

=0.225 N/Cm2

Φ3=EAcosθ3

=(500)(.03)2cos30°

=0.39 N/Cm2

Φ4=EAcosθ4

=(500)(.03)2cos120°

=-0.225 N/Cm2

part(b) step 1: given informaiton

cube with side 3.0

Formula Used:

ΦE=AEcosθ

Φ=ElectricFlux

A= Cross-sectional area of the loop

Again, the flux is ΦE=AEcosθ

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

θ1=150°.

Forside2,it'sθ2=60°

For side 3 it'sθ3=30°

and for side 4 it's θ4=120°

Taking the values from part A,

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

Φ=Φ1+Φ2+Φ3+Φ4=0,

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

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