Electrodynamics, Need help with problem

In summary, the electric field inside the cylindrical volume is negative everywhere except at the axis of symmetry. The potential inside the cylinder is negative, and the work needed to take a unit test charge from the axis out to the wire surface and then out some more to infinity is 2∏klR^6 / 6 εo.
  • #1
Stan12
20
0

Homework Statement



An infinitely long cylindrical volume of radius R contains a charge density ρ(s)=ks4 where k is a constant and s is the distance from the axis of the cylinder. Note that this is NOT a constant density.

a) Find the electric field everywhere in space.

b) From your result in part a) find the electrostatic potential inside the cylinder, assuming that the potential vanishes along the axis of symmetry, V(s)|[s=0]=0

Homework Equations



∫E*dA

closed surface : ∫E*dA = Qenc /εo

Qenc = ∫ρ dτ

The Attempt at a Solution



Qenc = k ∫∫∫ s'^4 s' ds'dz'dθ'

it came out to be 2∏klR^6 / 6 εo

flux = kR^5 / 6εo

I'm not sure this is correct so I cannot continue to part b)
 
Last edited:
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  • #2
You have (apparently) assumed a unit length of your wire. In your calculations I would have assumed an arbitrary length L intead. Why? Because that way you can check your dimensions in all your terms as you proceed in your calculations.

But, for unit length of wire, your integral is correct. And so is your calculation of flux E(s).

So you can go on. Realize of course that E(s) = what you came up with is valid only for 0 <= s <= R.

So potential is defined as zero on the axis. What is the work needed to take a unit test charge from the axis out to the wire surface and then out some more to infinity? (You are moving the test charge in a radial direction).
 
  • #3
Why did your flux lose an R in the numerator? Remember that flux= Q/(e_0). It should be your net charge enclosed and not that divided by R.

Edit: Perhaps you wrote flux meaning the field?

In that case, OP did NOT assume unit length- the L's cancel when you set the flux equal to Q/epsilon, because each involves an L. If he had an L left somewhere the units would be off.
 
  • #4
So with the E I can find the potential but plugging it to this,
V(r) = -∫E*dr . Thank you guys!
 
  • #5
I worked it out today and change some notations, I came up with E(r) = kr^5/6εo for 0<= r <= R and E(r) = kR^6/6rεo for R< r.
 
  • #6
schaefera said:
Why did your flux lose an R in the numerator? Remember that flux= Q/(e_0). It should be your net charge enclosed and not that divided by R.

Edit: Perhaps you wrote flux meaning the field?

In that case, OP did NOT assume unit length- the L's cancel when you set the flux equal to Q/epsilon, because each involves an L. If he had an L left somewhere the units would be off.

Never mind. I did not see the small "l" in his integral.
 
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  • #7
Stan12 said:
So with the E I can find the potential but plugging it to this,
V(r) = -∫E*dr . Thank you guys!

You are on the right track!
 

1. What is Electrodynamics?

Electrodynamics is the branch of physics that deals with the study of electric and magnetic fields and their interactions with charged particles. It is a fundamental theory that explains the behavior of electricity and magnetism in various situations.

2. How is Electrodynamics different from Electromagnetism?

Electrodynamics is a broader term that encompasses the study of both electric and magnetic fields, while electromagnetism specifically refers to the interactions between electric and magnetic fields. In other words, electromagnetism is a subset of electrodynamics.

3. What are some real-world applications of Electrodynamics?

Electrodynamics has numerous practical applications, including the functioning of electric motors and generators, the transmission of electricity through power lines, the operation of electronic devices, and the production of electromagnetic waves used in communication technologies.

4. What are some common problems encountered in Electrodynamics?

Some common problems in Electrodynamics include calculating the electric and magnetic fields generated by various charge distributions, determining the motion of charged particles in electric and magnetic fields, and analyzing the behavior of electromagnetic waves in different mediums.

5. How can I solve problems in Electrodynamics?

The key to solving problems in Electrodynamics is to have a clear understanding of the fundamental principles and equations. It is important to practice solving various types of problems and pay attention to units and mathematical operations. It can also be helpful to seek guidance from textbooks, online resources, and peers or instructors.

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