# Thread: Parallel Plate Electric Fields

1. ## Parallel Plate Electric Fields

Hi there, again I just wanted to see if anyone could help with the answer to this past examination question as I don't have the answers so not sure if I'm doing them correctly!

An electron is liberated from the lower of two parallel plates seperated by 5mm. The upper plate has a potential of 1000v relative to the lower plate. How long does it take to reach the top plate?
(Neglect gravity)

Ok, well I thought it best to work out the field strength so I did E=V/d giving E=2.0x10^5N/C

Then I said that F=ma=qE so a=qE/m and since the chrage of electron is e, this gives a=eE/m=3.5x10^16!

Surely this is much to greater acceleration!

I then used s=ut + 1/2at^2 to get the time taken, which gives t = 3.78x10^-10.

I'm thinking these figures look way out!

Any help much appriciated.

Mike

2. Originally Posted by mikewhant
Hi there, again I just wanted to see if anyone could help with the answer to this past examination question as I don't have the answers so not sure if I'm doing them correctly!

An electron is liberated from the lower of two parallel plates seperated by 5mm. The upper plate has a potential of 1000v relative to the lower plate. How long does it take to reach the top plate?
(Neglect gravity)

Ok, well I thought it best to work out the field strength so I did E=V/d giving E=2.0x10^5N/C

Then I said that F=ma=qE so a=qE/m and since the chrage of electron is e, this gives a=eE/m=3.5x10^16!

Surely this is much to greater acceleration!

I then used s=ut + 1/2at^2 to get the time taken, which gives t = 3.78x10^-10.

I'm thinking these figures look way out!

Any help much appriciated.

Mike
You would probably notice that the speed of the electron is approaching $3*10^8$, which happens to be the upper bound.
Special Relativity states the following equation to calculate such high objects' speed:
$E^2=p^2*c^2+m^2*c^4$ where E and p are the energy and momentum of the electron respectively

3. Thank you, glad to see I'm doing it correctly.

Happy new year!