# Electron's velocity

1. Mar 11, 2016

### Kernul

1. The problem statement, all variables and given/known data
Two thin cylindrical shells, respectively with radius $R_1 = 12 cm$ and $R_2 = 20 cm$ and height $D = 100 m$, are concentric and characterized by a charge density $\lambda_1 = -0.38 \frac{\mu C}{m}$, for the internal cylinder, and $\lambda_2 = 0.32 \frac{\mu C}{m}$, for the external one. An electron ($m_e = 9.1 \cdot 10^{-31} kg$ and a charge $e = 1.6 \cdot 10^{-19} C$) is released stationary from the internal cylinder's surface; what is the velocity when it reaches the other cylinder?

2. Relevant equations
Gauss Teorem
$\Phi_S (\vec E_0) = \int_S \vec E_0 \cdot d\vec S = \frac{Q}{\epsilon_0}$
and
$a = \frac{q}{m}E$

3. The attempt at a solution
So, I started by calculating the electric field on both the cylinders.
$\Phi_S (\vec E_1) = \int_S \vec E_1 \cdot d\vec S = E_1 2 \pi r D = \frac{\lambda_1 D}{\epsilon_0}$
so
$E_1 = \frac{\lambda_1}{2 \pi r \epsilon_0}$
Same thing for the other one.
$E_2 = \frac{\lambda_2}{2 \pi r \epsilon_0}$

At this point I should calculate the velocity starting from the acceleration $a = \frac{e}{m_e}E$, but I'm blocked here. The $E$ in the acceleration is the "total" electric field or it is $E_2$ since the electron is now moving inside the external cylinder?
I know there is this other equation:
$v = \sqrt{\frac{2q(V_1 - V_2)}{m}}$
But in this case I should calculate the two electric potential, which would be:
$\int_A^B \vec E_0 \cdot d\vec l = V(A) - V(B)$
with $A$ being the internal cylinder and $B$ the other one.

Last edited: Mar 11, 2016
2. Mar 11, 2016

### Staff: Mentor

Working with the potential is the easiest approach here.

What is the field inside a long cylinder with a constant surface charge? (Just the field from this surface charge)

3. Mar 11, 2016

### Kernul

It should be $E = \frac{\sigma R_2}{\epsilon_0 r}$, right?

4. Mar 11, 2016

### Staff: Mentor

Inside the cylinder?
So at r=0 you have infinite field strength?

5. Mar 11, 2016

### Kernul

I did a drawing of the exercise, just to be sure I'm getting this right.
I used Gauss Theorem to get that. I did like this:
$\Phi_s (E) = \int_S \vec E \cdot d \vec S = E 2 \pi r D = \frac{\sigma 2 \pi R_2 D}{\epsilon_0}$
then
$E 2 \pi r D = \frac{\sigma 2 \pi R_2 D}{\epsilon_0}$
$E = \frac{\sigma 2 \pi R_2 D}{\epsilon_0 2 \pi r D}$
$E = \frac{\sigma R_2}{\epsilon_0 r}$
so I think yes. Unless I got something wrong.

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6. Mar 11, 2016

### Staff: Mentor

For Gauß theorem, you need the charge enclosed by the area you are integrating over. What is the charge in a smaller cylinder inside a charged cylinder?

7. Mar 12, 2016

### Kernul

I'm guessing it should be smaller? Because the charged cylinder act like some kind of insulator to the other one inside?
About the area, I should do something like this then: $E = \int_{R_1}^{R_2} \frac{\sigma R_2}{\epsilon_0 R_1}$
With $R_1$ and $R_2$ being the two points describing the area.

8. Mar 12, 2016

### Staff: Mentor

Forget about the two cylinders for a moment, the problem is earlier.

What is the field inside a single spherical shell with a constant surface charge?

What is the field inside a single cylinder with a constant surface charge?

9. Mar 12, 2016

### Kernul

$\vec E = \frac{1}{4 \pi \epsilon_0} \frac{Q}{r^2}$
with $\sigma = \frac{Q}{4 \pi r^2}$ we have then
$\vec E = \frac{\sigma}{\epsilon_0}$

$\vec E = \frac{1}{4 \pi \epsilon_0} \frac{Q}{r^2}$
with $\sigma = \frac{Q}{2 \pi r h}$ we have then
$\vec E = \frac{\sigma h}{2 r \epsilon_0}$

Is this what you asked for?

10. Mar 12, 2016

### Staff: Mentor

Where does that formula come from and what does it calculate?

It does not calculate the electric field inside anything. The right side is also not a vector, so writing E as vector at the left side is not right.

11. Mar 12, 2016

### Kernul

Oh, I'm sorry about the vector thingy.
I know this formula:
$\vec E_0 (\vec r) = \frac {1}{4 \pi \epsilon_0} \int \frac{\sigma(x', y', z')(\vec r - \vec r')}{|\vec r - \vec r'|^3}dS'$ which is just a general way(not only for constant surface charge) and I thought that since $dq = \sigma(x, y, z) dS$ it would end up with the one I wrote ($E_0 = \frac{1}{4 \pi \epsilon_0} \frac{Q}{r^2}$). Is this not correct?

12. Mar 12, 2016

### Staff: Mentor

It is not correct.
What would r even be there? The position where you want to evaluate the field, or the radius of the sphere?

Anyway, let's stay at the sphere: if it has a radius R, what is the electric field at a radius r where r<R?
You can use Gauß law, but don't blindly plug numbers into a random formula. Think what the formula is doing when, and what the meaning of its components is.

13. Mar 12, 2016

### Kernul

$r$ is the position where I want to evaluate the field.

Okay, in $r < R$ we have a $Q = \sigma 4 \pi r^2$
using Gauß law we have
$\Phi = \frac{Q}{\epsilon_0}$
which becomes
$E 4 \pi r^2 = \frac{\sigma 4 \pi r^2}{\epsilon_0}$
$E = \frac{\sigma}{\epsilon_0}$

14. Mar 12, 2016

### Staff: Mentor

Where would that charge be? All the charge is at the spherical shell, at radius R. There is nothing at smaller radii.

15. Mar 12, 2016

### Kernul

Oh! Because we are talking about surface charge and not volume charge. Because if it was a volume charge then it would have a charge inside the sphere and so it would have $Q = \rho \frac{4}{3} \pi r^3$, right?

16. Mar 12, 2016

### Staff: Mentor

If it would be a volume charge, yes, but it is not.

17. Mar 12, 2016

### Kernul

So, going back to what you asked me, for $r < R$ the field would be $E = 0$ inside the sphere and $E = \frac{\sigma R^2}{\epsilon_0 r^2}$ on the surface, right?

18. Mar 12, 2016

### Staff: Mentor

Directly at the surface the field is not well-defined, but outside, yes.
Inside it is zero, right.

So going back to the cylinders: does the charge from the outer cylinder matter for the field inside?

19. Mar 12, 2016

### Kernul

No, it doesn't matter. Only the charge inside matter. This means I only have to calculate the field inside the outer cylinder(which it is the field outside the internal cylinder) $E_1$, but I have the linear charge density $\lambda_1$ and $\lambda_2$, not the surface charge, right?

20. Mar 12, 2016

### Staff: Mentor

You have the surface charge of the inner cylinder, but a single charged line along its symmetry axis would produce the same field.