# Electric field / Gauss' Law --

bluesteels
Homework Statement:
The electric field at 2 cm from the center of long copper rod of radius 1 cm has a magnitude 3 N/C and directed outward from the axis of the rod.

(b) What would be the electric flux through a cube of side 5 cm situated such that the rod passes through opposite sides of the cube perpendicularly?
Relevant Equations:
E = 3 N/C
λ = 3.3*10^-12 C/m
r = 2cm
Area of cube = 6a^2
electric flux= q/e0
Electric Flux = E*A = 5*6(0.05)^2.

when i look up at other sources they use Electric flux = q/ (8.854*10^-12 [this is e]) equation

but I am confused on why the E*A equation don't work. The answer is 0.02Nm^2/C

Last edited:

Mentor
How is the electric field oriented around a (long) charged rod? Is there a symmetry involved?

Would you expect the field to be of even magnitude over the surface of each side of the cube? How about the sides that the rod passes through?

What does Gauss' Law have to say about the total electric flux passing through a closed surface? Hint: your formula "electric flux= q/e0" will come in handy.

• vcsharp2003 and bluesteels
bluesteels
How is the electric field oriented around a (long) charged rod? Is there a symmetry involved?

Would you expect the field to be of even magnitude over the surface of each side of the cube? How about the sides that the rod passes through?

What does Gauss' Law have to say about the total electric flux passing through a closed surface? Hint: your formula "electric flux= q/e0" will come in handy.
wait so your saying that the enclosed area (the cube) is equal to the rod that out of the cube?

Staff Emeritus
Homework Helper
wait so your saying that the enclosed area (the cube) is equal to the rod that out of the cube?
Your sentence doesn't make any sense. Did you accidentally leave some words out?

• Delta2
bluesteels
Your sentence doesn't make any sense. Did you accidentally leave some words out?
it kinda hard to explain so basically i watch "organic chemistry tutor" and he say like if you know the electric field and the area you can use the EA formula.

Like i know that if you use
Gauss law you can find the total charge enclosed by that surface.

Staff Emeritus
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Gold Member
it kinda hard to explain so basically i watch "organic chemistry tutor" and he say like if you know the electric field and the area you can use the EA formula.
That is only the case if the field is constant in magnitude on the area and perpendicular to the surface.

• vcsharp2003
PhDeezNutz
I'm getting the same exact answer as @vcsharp2003 assuming a solid copper rod with uniform volume density.

But part of me thinks this question is a bad one. It specifically mentions copper which is a conductor so all the charge would be on the outside in which case I don't get the stated answer.

Homework Helper
Sorry @PhDeezNutz , am I missing the link with the current thread ?

##\ ##

Gold Member
Homework Statement:: The electric field at 2 cm from the center of long copper rod of radius 1 cm has a magnitude 3 N/C and directed outward from the axis of the rod.

(b) What would be the electric flux through a cube of side 5 cm situated such that the rod passes through opposite sides of the cube perpendicularly?
Relevant Equations:: E = 3 N/C
λ = 3.3*10^-12 C/m
r = 2cm
Area of cube = 6a^2
electric flux= q/e0

Electric Flux = E*A = 5*6(0.05)^2.

when i look up at other sources they use Electric flux = q/ (8.854*10^-12 [this is e]) equation

but I am confused on why the E*A equation don't work. The answer is 0.02Nm^2/C
What value are you putting for E? Is it right?

Mentor
Sorry @PhDeezNutz , am I missing the link with the current thread ?

##\ ##
A full solution has been removed.

• BvU
vcsharp2003
I'm getting the same exact answer as @vcsharp2003 assuming a solid copper rod with uniform volume density.

But part of me thinks this question is a bad one. It specifically mentions copper which is a conductor so all the charge would be on the outside in which case I don't get the stated answer.
The 4 faces (besides the 2 faces through which the copper rod goes through) of the cube would surround this copper rod, so you shouldn't worry whether the charge resides on surface of rod as long as it's within the cube.

• PhDeezNutz
PhDeezNutz
The 4 faces (besides the 2 faces through which the copper rod goes through) of the cube would surround this copper rod, so you shouldn't worry whether the charge resides on surface of rod as long as it's within the cube.
You're right I dun goof'd.

If I did my math right the 3 cases (uniform line charge, uniform volume charge, uniform surface charge) all lead to

##Q_{enc} = 2 \pi E_0 \epsilon_0 r_2 \ell_c##

Derp. Guess I got lost in the numbers.

@bluesteels you already did the hard part (computing the line charge density)

Remember that Gauss's Law says

##\Phi = \frac{Q_{enc}}{\epsilon_0}##

Instead of manually computing the flux of ##E## find the total charge enclosed (again you already have ##\lambda##)

How much charge is in a 5 cm segment of the line charge? What is that number divided by ##\epsilon_0##? Doesn't this line charge segment contain the same amount of charge as the larger cube? (since the only part in the cube that contains charge IS the line charge)

• vcsharp2003
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2022 Award
If we can visualise the lines of flux the problem is simple.

The field has cylindrical symmetry. Replace the 5cm-sided cube by a cylinder of radius 2cm and length 5cm, coaxial with the rod.

Find the flux through the cylinder’s curved surface using its area and the given value of E at 2cm. This flux is the same as the flux through the cube.

• vcsharp2003 and PhDeezNutz
vcsharp2003
If we can visualise the lines of flux the problem is simple.

The field has cylindrical symmetry. Replace the 5cm-sided cube by a cylinder of radius 2cm and length 5cm, coaxial with the rod.

Find the flux through the cylinder’s curved surface using its area and the given value of E at 2cm. This flux is the same as the flux through the cube.
Is this because flux through an area can be thought as proportional to number of lines of force going through this area?

All lines of force through the curved surface of concentric cylinder will eventually come out of the faces of the cube and thus flux through concentric cylinder is same as flux through faces of the cube.

• Delta2, Steve4Physics and PhDeezNutz
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Is this because flux through an area can be thought as proportional to number of lines of force going through this area?

All lines of force through the curved surface of concentric cylinder will eventually come out of the faces of the cube and thus flux through concentric cylinder is same as flux through faces of the cube.
Yes. And that's a very nice explanation.

• Delta2, vcsharp2003 and PhDeezNutz