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

1. Apr 25, 2016

### Kernul

1. The problem statement, all variables and given/known data
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.

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

3. 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?

2. Apr 25, 2016

### TSny

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. Apr 25, 2016

### Kernul

Why is it indeterminate? Shouldn't $E$ be the maximus on the surface?

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. Apr 25, 2016

### TSny

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.
Shoudn't the $r$ on the left side be $r_0$?
At this point, use the fact that $\int {dS} = S$ and just substitute the correct expression for S.
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. Apr 25, 2016

### Kernul

So simply $S = 2 \pi r_0 l$?
Then I have $E_0 = \frac{\sigma}{\epsilon_0}$. Is it right?

6. Apr 25, 2016

### TSny

Think about whether you should have $r$ or $r_0$ in this expression.

7. Apr 25, 2016

### Kernul

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. Apr 25, 2016

### TSny

Yes, that's right.

9. Apr 25, 2016

### TSny

Of course, there is also the contribution from the line charge.

10. Apr 25, 2016

### Kernul

Which is this $\frac{1}{2 \pi \epsilon_0} \frac{\lambda}{r_0}$, right?

11. Apr 25, 2016

Yes.