# Electric field / Gauss' Law --

• bluesteels
In summary: The answer is 0.2 Nm^2/C. If the line charge is 5C then the total charge in the 5 cm segment is 20.2 Nm^2.
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:
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
gneill said:
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?

bluesteels said:
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
vela said:
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.

bluesteels said:
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
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.

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

##\ ##

bluesteels said:
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?

BvU said:
Sorry @PhDeezNutz , am I missing the link with the current thread ?

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

BvU
PhDeezNutz said:
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
vcsharp2003 said:
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
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
Steve4Physics said:
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
vcsharp2003 said:
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

## What is an electric field?

An electric field is a physical field that surrounds electrically charged particles and exerts a force on other charged particles within its range. It is a fundamental concept in electromagnetism and is responsible for the flow of electricity.

## How is an electric field measured?

The strength of an electric field is measured in units of volts per meter (V/m). This value represents the amount of force exerted on a unit of charge at a specific point in the field.

## What is Gauss' Law?

Gauss' Law is a fundamental law in electromagnetism that describes the relationship between electric charges and electric fields. It states that the electric flux through a closed surface is proportional to the total charge enclosed by the surface.

## How is Gauss' Law used in calculations?

Gauss' Law can be used to calculate the electric field at a point due to a distribution of charges. By using the symmetry of the charge distribution, the electric field can be simplified and calculated using the enclosed charge and the geometry of the surface.

## What are some real-life applications of electric fields and Gauss' Law?

Electric fields and Gauss' Law have numerous applications in everyday life, such as in the design of electrical circuits, the functioning of electronic devices, and the behavior of lightning. They are also used in medical imaging techniques, such as electrocardiograms and electroencephalograms.

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