Electric Field of a cube with a rod

In summary, the conversation discusses the calculation of the electric field and flux in a scenario involving a long copper rod. The first part calculates the charge per unit length on the rod, while the second part involves determining the electric flux through a cube situated perpendicular to the rod. Gauss's law is used to calculate the flux, and it is concluded that the shape of the cube does not affect the calculation as long as the enclosed charge remains the same. Additionally, external charges do not affect the flux as they do not enter the cube.
  • #1
doggydan42
170
18

Homework Statement


The electric field at 2 cm from the center of long copper rod of radius 1 cm has a magnitude ##3 \frac{N}{C}## and directed outward from the axis of the rod. (a) How much charge per unit length exists on the copper 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?

Homework Equations


For a cylindrical symmetry: $$\phi = 2\pi rLE$$$$\phi = \frac{q}{\epsilon_0}$$$$\lambda = 2\pi r\epsilon_0 E$$
Where ##\lambda## is the linear charge density, ##\phi## is the flux, and L is the length.

I know I am missing an equation, which is the equation I am trying to find.

The Attempt at a Solution


For the first part, I solved for ##\lambda## to be ##3.34\times10^{-12}\frac{C}{m}##. I know that this must be used in the second part, but I am unsure of how to include the cube into the equation.

Thank you in advance
 
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  • #2
doggydan42 said:
but I am unsure of how to include the cube into the equation.
Are you familiar with "Gaussian surfaces"? If the cube is replaced by a rectangular box still length 5cm parallel to the rod but 10cm in the other two directions, would the flux through it be more, the same, or less?
 
  • #3
doggydan42 said:
I know that this must be used in the second part, but I am unsure of how to include the cube into the equation.
Does Gauss's law care what the shape of the Gaussian surface is?
 
  • #4
Recalls the physical interpretation of Gauss's theorem: flow is the closed charge into the surface divided by epsilon. Can you calculate these charge?

Another thing, when the closed electric charge inside the surface is the same, is important the shape of the surface to calculate the total flow?
 
  • #5
Oh. So the charge enclosed would just be ##q_{enc}=\lambda L##, where L is the length of the cube. Right? Since if I understand correctly, it doesn't matter what shape it is, as long as the charge enclosed is the same.

So if I solved this correctly, then:
$$\phi=\frac{q_{enc}}{\epsilon_0}=\frac{\lambda L}{\epsilon_0} = \frac{(3.34\times 10^{-12} \frac{C}{m})(0.05 m)}{8.85\times 10^{-12}\frac{C^2}{Nm^2}}
\\ \phi=1.89\times 10^{-2} \frac{Nm}{c}$$

If not what am I missing? Also, how would external charges affect the flux, since the rod extends past the cube?

Thank you.
 
  • #6
doggydan42 said:
how would external charges affect the flux, since the rod extends past the cube?
By symmetry, the field lines are all perpendicular to the axis of the rod, so flux from other parts of the rod do not enter the cube. Also, more generally, a line of flux from any charge external to the cube must enter and exit it an equal number of times, so produces no net flux.
 

FAQ: Electric Field of a cube with a rod

1. What is an electric field?

An electric field is a physical field that is created by electrically charged particles. It describes the force that a charged particle would experience if placed in that field.

2. How is an electric field of a cube with a rod calculated?

The electric field of a cube with a rod can be calculated using the Coulomb's law. The electric field is the force per unit charge, so it can be calculated by dividing the force on a test charge by the magnitude of that charge.

3. What factors affect the strength of an electric field?

The strength of an electric field is affected by the distance from the charged object, the magnitude of the charge, and the medium in which the field is measured.

4. How is the direction of an electric field determined?

The direction of an electric field is determined by the direction in which a positive test charge would be pushed or pulled if placed in that field.

5. How does the presence of a rod affect the electric field of a cube?

The presence of a rod in a cube can change the distribution of the electric field. The rod may act as a point charge or may create a non-uniform electric field within the cube.

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