Bohr's calculation of stopping power

[EucharisCriss]

so, hey...I'm pretty new to these forums and just needed help understanding a textbook that I'm reading.

I'll give you all the background, but I'm not sure if it is necessary.

They begin by considering a heavy particle of mass M and velocity v passing through a material with an atomic electron at distance b from the particle trajectory. the electron is assumed to be free and initially at rest. Furthermore, it is assumed that the electron only moves slightly during the interaction with the heavy particle. After the collision, it is assumed that the incident heavy particle continues essentially as before (because M>>me)

the impulse is calculated:
I=integral(Fdt)=e * integral(E_perpendicular)$$\frac{dt}{dx}$$dx = e integral ( E_perpendicular $$\frac{dx}{v}$$

They claim that only E_perpendicular enters because of symmetry. I am a little confused about this actually. I kind of reasoned it out by thinking of the work function (which would give the change in energy for the particle, right) W=integral(F . dx )=integral (eE . dx) = integral( e*E_perpendicular*dx)...(their next to last step seems kind of like this...)...but I think I'm just trying to make up something to understand it...so if anyone could tell me what symmetry they mean, I'd appreciate it.

Also, they go on to calculate the integral( E_perpendicular dx) using Gauss' Law over an infinitely long cylinder centered on the particle trajectory. They claim that:

integral ( E_perpendicular *2*pi*b*dx)=4*pi*z*e

I just don't know where the 4*pi is coming from on the right side (and where epsilon is..) and would really appreciate it if someone could explain this to me.

I hope this is an okay place to post this. It isn't homework, I'm just reading it...

I tried to draw the diagram from the book and it is attached (i hope).

I just feel like I must be missing something major

 

[tiny-tim]

Welcome to PF!

Hi EucharisCriss! Welcome to PF!

(have an integral: ∫ and a pi: π and an epsilon: ε )

They claim that only E_perpendicular enters because of symmetry. I am a little confused about this actually.

I think the symmetry is because the effect before M reaches the nearest point will be the reflection of the effect after, so the total effect will be perpendicular.

Also, they go on to calculate the integral( E_perpendicular dx) using Gauss' Law over an infinitely long cylinder centered on the particle trajectory. They claim that:

integral ( E_perpendicular *2*pi*b*dx)=4*pi*z*e

I just don't know where the 4*pi is coming from on the right side (and where epsilon is..) and would really appreciate it if someone could explain this to me.

4πε₀ is the factor you always need to include when you use SI units.

(And ε₀ is the permittivity of the vacuum)

See the PF Library on Coulomb's law for details.

 

[EucharisCriss]

No, I mean I know Coulomb's law, but the four pi seems to come out of nowhere when you consider that Gauss' law:

ε0∫E . dA = qenc

I mean, Coulomb's law can be derived from this by considering a point charge with a spherical gaussian surface centered on the particle.
ε0∫E . dA = qenc
ε0∫E dA = qenc (because E and dA are at 0 degree angle when charge is positive)
ε0EA = qenc
In this case A=4πr^2
so this is where the 4πε0 in Coulomb's law comes from...

However, in their version of gauss' law, they have no ε0 and an extra 4π on the side side of the enclosed charge (the enclosed charge being ze).
They have changed dA to 2πbdx (2πb being the circumference of the circle and dx being the length of the differential cylindrical shell).
Basically, the left side of this equation makes sense while the existence of the 4π (and the lack of an ε0) on the right sense seems to come from nowhere for me.

What I was expecting to see:
rightside=ze/ε0
What they have:
rightside=4πze

ε0 is slightly related to 4π because c=1/sqrt(ε0μ0)
μ0=4π * 10 ^-7


I would agree with you about the symmetry argument, but it was my thought that the charged particle ze is on a collision path with the electron. I should get the book out, but I'm kind of lazy right now.

 

[tiny-tim]

… However, in their version of gauss' law, they have no ε0 and an extra 4π on the side side of the enclosed charge (the enclosed charge being ze). …

Hi EucharisCriss!

My guess is that the book isn't using SI units.