Question

From Fishbane‘s Physics 3rd edition.

A beam of electrons is accelerated by passing it through a region between two large charged parallel plates, 2cm apart.
Calculate the charge density on the plates if the electrons accelerate from 1.4x10^6 m/s to 3.0x10^7 m/s between plates.

Answer 1

Ok, so despite the large velocities, we can still work in the non-relativistic regime here (accurate to within half a percent).

The increase in kinetic energy is

\[\Delta KE = \frac12 m_e ((3\times10^7~\text{ms}^{-1})^2-(1.4\times10^6~\text{ms}^{-1})^2)\]

You can work this out in Joules then divide by the elementary charge to get the potential difference between the two plates in Volts. Or you can work out the change in kinetic energy in electronvolts if you want.

Either way this gives you a voltage between the plates of 2.553 kV.
Divide by separation of the plates to give electric field strength (since electric field is the gradient of potential, and these are parallel plates).

Then I think you just divide by permittivity (use \(\epsilon_0\) assuming the system is in vacuum) to get charge density. It gives you the right dimensions (charge per unit area) anyway.

Answer 2

But I can‘t get that result.
I was wondering if the kinetic energy would be \[\frac{ 1 }{ 2 } \times m _{e} \times \left( v _{f} - v _{i}\right)^{2}\]

Answer 3

No, energy adds linearly.
\[\Delta E = E_f -E_i = \frac12m_ev_f^2 - \frac12m_ev_i^2 = \frac12m_e(v_f^2-v_i^2)\]

I should have said multiply by permittivity at the end. Oops

Answer 4

This is what the calculation should be (with numerical result):
http://www.wolframalpha.com/input/?i=1%2F2+vacuum+permittivity+electron+mass+%28%283e7+m%2Fs%29^2+-+%281.4e6+m%2Fs%29^2%29%29+%2F+%28electron+charge+2+cm%29

Answer 5

I was doing:

\[\frac{ 1 }{ 2 } \times 9.1 \times 10^{-31} \times \left( 3 \times 10^{7} - 1.4 \times 10^{6} \right)^{2}\]

and next, I divided the result by the charge of the electron e=1,6022X10ˆ(-19) C;
afterwords, I divided by the 0.02m, that is the distance between plates and then by the permitivity.

Answer 6

Thanks a lot.
I was dividing by the permitivity, instead of multiplying.

Answer 7

very nice resource, the Wolfram Alpha.

Double thank you a lot!!

Answer 8

I've a doubt.
Once found the kinetic energy, dividing by the distance between the plates, will return the electric field of the two plates?
And why divide, also, by the charge?
I'm trying to follow your thinking, and connecting the kinetic energy to the electric field. I'm digging the Physics book for the link between them.

Answer 9

Divide kinetic energy by elementary charge to get potential difference (voltage).

Divide voltage by distance to get electric field.