Electric Field in all of space from charges along the Z-axis

  • Thread starter Kernul
  • Start date
  • #1
211
7

Homework Statement


A charge is distributed evenly along the ##z## axis with a density ##\lambda = 3 \frac{\mu C}{m}##. There is also a cylinder with radius ##r = 2## and on his side there is a superficial density charge of ##\sigma = -\left( \frac{1.5}{4 \pi} \right) \frac{\mu C}{m^2}##. Both distributions are unlimited on the ##z## axis that coincide with the axis of the system.

Homework Equations


Gauss Theorem:
##\Phi_S (\vec E_0) = \int_S \vec E_0 \cdot d\vec S = \frac{1}{\epsilon_0} \int_l \lambda (x, y, z) dl##
##\Phi_S (\vec E_0) = \int_S \vec E_0 \cdot d\vec S = \frac{1}{\epsilon_0} \int_S \sigma (x, y, z) dS##

The Attempt at a Solution


So I started first by calulating the electric field inside the cylinder with a ##r_0 < r##. Since it is only the charge on the ##z## axis creating the field, we will use ##\lambda## and integrate on the line.
##\Phi_S (\vec E_0) = \int_S \vec E_0 \cdot d\vec S = \frac{1}{\epsilon_0} \int_l \lambda (x, y, z) dl##
##\int_S \vec E_0 \cdot d\vec S = \frac{1}{\epsilon_0} \int_l \lambda dl##
##E_0 2 \pi r_0 l = \frac{1}{\epsilon_0} \int_l \lambda dl##
##E_0 = \frac{1}{2 \pi \epsilon_0} \frac{1}{r_0 l} \int_l \lambda dl##
##E_0 = \frac{1}{2 \pi \epsilon_0} \frac{\lambda}{r_0 l} \int_l dl##
##E_0 = \frac{1}{2 \pi \epsilon_0} \frac{\lambda l}{r_0 l}##
##E_0 = \frac{1}{2 \pi \epsilon_0} \frac{\lambda}{r_0}##

Then I calculated the electric field at ##r_0 = r##. Now there are both the charge on the ##z## axis and the charge on the cylinder's side ##\sigma##.
For the contribution of the linear charge we have ##\frac{1}{2 \pi \epsilon_0} \frac{\lambda}{r}##.
For the contribution of the superficial charge we have:
##\Phi_S (\vec E_0) = \int_S \vec E_0 \cdot d\vec S = \frac{1}{\epsilon_0} \int_S \sigma (x, y, z) dS##
##\int_S \vec E_0 \cdot d\vec S = \frac{1}{\epsilon_0} \int_S \sigma dS##
##E_0 2 \pi r l = \frac{1}{\epsilon_0} \int_S \sigma dS##
##E_0 = \frac{1}{2 \pi \epsilon_0} \frac{1}{r l} \int_S \sigma dS##
##E_0 = \frac{1}{2 \pi \epsilon_0} \frac{\sigma}{r l} \int_S dS##
##E_0 = \frac{1}{2 \pi \epsilon_0} \frac{\sigma}{r l} \int_{0}^{r} 2 \pi l dr##
##E_0 = \frac{1}{2 \pi \epsilon_0} \frac{\sigma 2 \pi l}{r l} \int_{0}^{r} dr##
##E_0 = \frac{1}{2 \pi \epsilon_0} \frac{\sigma 2 \pi r l}{r l}##
##E_0 = \frac{\sigma}{\epsilon_0}##
So ##E = \frac{1}{2 \pi \epsilon_0} \frac{\lambda}{r} + \frac{\sigma}{\epsilon_0}##

In the end I calculated the electric field for ##r_0 > r##. Like before, there are both the charge on the ##z## axis and the charge on the cylinder's side ##\sigma##.
For the contribution of the linear charge we have ##\frac{1}{2 \pi \epsilon_0} \frac{\lambda}{r_0}##.
For the contribution of the superficial charge we have ##\frac{\sigma}{\epsilon_0 r_0} (r_0 - r)##.
So ##E = \frac{1}{2 \pi \epsilon_0} \frac{\lambda}{r_0} + \frac{\sigma}{\epsilon_0 r_0} (r_0 - r)##

Is it all correct?
 

Answers and Replies

  • #2
TSny
Homework Helper
Gold Member
12,960
3,315
Your answer looks correct for the case ro < r.

I would not try to use Gauss' law for the special case of r0 = r. For this value of r, you are at the location of the surface charge density where the E field is indeterminate.

For the case r0 > r, you have not found the correct charge inside the Gaussian surface due to the surface charge density σ. There should not be any integration with respect to r. Since σ is constant, you have ##\int_{S} \sigma dS = \sigma S##. Be sure to draw a figure which helps you identify the surface ##S##.
 
  • #3
211
7
I would not try to use Gauss' law for the special case of r0 = r. For this value of r, you are at the location of the surface charge density where the E field is indeterminate.
Why is it indeterminate? Shouldn't ##E## be the maximus on the surface?

For the case r0 > r, you have not found the correct charge inside the Gaussian surface due to the surface charge density σ. There should not be any integration with respect to r. Since σ is constant, you have ∫SσdS=σS∫SσdS=σS\int_{S} \sigma dS = \sigma S. Be sure to draw a figure which helps you identify the surface SSS.
Why? I'm assuming a cylinder with a radius ##r_0 > r## and I want to find the electric field there. Shouldn't it be this the contribution of the surface charge:
##E_0 2 \pi r l = \frac{1}{\epsilon_0} \int_S \sigma dS##
##E_0 = \frac{1}{2 \pi \epsilon_0} \frac{1}{r l} \int_S \sigma dS##
##E_0 = \frac{1}{2 \pi \epsilon_0} \frac{\sigma}{r l} \int_S dS##
##E_0 = \frac{1}{2 \pi \epsilon_0} \frac{\sigma}{r l} \int_{r}^{r_0} 2 \pi l dr##
##E_0 = \frac{1}{2 \pi \epsilon_0} \frac{\sigma 2 \pi l}{r l} \int_{r}^{r_0} dr##
##E_0 = \frac{1}{2 \pi \epsilon_0} \frac{\sigma 2 \pi l}{r l} (r_0 - r)##
##E_0 = \frac{\sigma}{\epsilon_0 r} (r_0 - r)##
 
  • #4
TSny
Homework Helper
Gold Member
12,960
3,315
Why is it indeterminate? Shouldn't ##E## be the maximus on the surface?
I think you'll see why E is undefined for points on the surface of the cylinder once you have worked out E for the inner and outer regions.
I'm assuming a cylinder with a radius ##r_0 > r## and I want to find the electric field there. Shouldn't it be this the contribution of the surface charge:
##E_0 2 \pi r l = \frac{1}{\epsilon_0} \int_S \sigma dS##
Shoudn't the ##r## on the left side be ##r_0##?
##E_0 = \frac{1}{2 \pi \epsilon_0} \frac{1}{r l} \int_S \sigma dS##
##E_0 = \frac{1}{2 \pi \epsilon_0} \frac{\sigma}{r l} \int_S dS##
At this point, use the fact that ##\int {dS} = S## and just substitute the correct expression for S.
##E_0 = \frac{1}{2 \pi \epsilon_0} \frac{\sigma}{r l} \int_{r}^{r_0} 2 \pi l dr##
It doesn't make any sense to integrate over r. You are integrating over part of the surface of the cylinder where ##r## is a fixed number, not a variable.
 
  • #5
211
7
Shoudn't the rrr on the left side be r0r0r_0?
Yeah, you're right. My bad.
At this point, use the fact that ∫dS=S∫dS=S\int {dS} = S and just substitute the correct expression for S.
So simply ##S = 2 \pi r_0 l##?
Then I have ##E_0 = \frac{\sigma}{\epsilon_0}##. Is it right?
 
  • #6
TSny
Homework Helper
Gold Member
12,960
3,315
So simply ##S = 2 \pi r_0 l##?
Think about whether you should have ##r## or ##r_0## in this expression.
 
  • #7
211
7
Gauss Theorem says that ##\Phi## of the electric field in the void ##E_0## through a closed surface ##S## is the integral of the charge inside ##S## divided by ##\epsilon_0##. The surface we are talking about in this case is the cylinder with radius ##r## while the electric field we want to find out is at radius ##r_0 > r##. So on the left, where there is ##S##, we have ##r_0##, while on the right, where there is the other ##S##, we have ##r##. This is because on the left we have the surface with radius ##r_0## and on the right the surface with the surface charge ##\sigma##, so the one at radius ##r##. So I got it wrong. I should have written ##E_0 = \frac{\sigma r}{\epsilon_0 r_0}##. Am I right?
 
  • #8
TSny
Homework Helper
Gold Member
12,960
3,315
Yes, that's right.
 
  • Like
Likes Kernul
  • #9
TSny
Homework Helper
Gold Member
12,960
3,315
Of course, there is also the contribution from the line charge.
 
  • #10
211
7
Which is this ##\frac{1}{2 \pi \epsilon_0} \frac{\lambda}{r_0}##, right?
 
  • #11
TSny
Homework Helper
Gold Member
12,960
3,315
Which is this ##\frac{1}{2 \pi \epsilon_0} \frac{\lambda}{r_0}##, right?
Yes.
 
  • Like
Likes Kernul

Related Threads on Electric Field in all of space from charges along the Z-axis

Replies
3
Views
3K
Replies
2
Views
317
Replies
2
Views
746
Replies
5
Views
5K
Replies
1
Views
5K
Replies
1
Views
8K
Replies
0
Views
6K
Replies
11
Views
594
Replies
22
Views
1K
Top