# Parallel Plate Electric Fields

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• Dec 30th 2009, 11:18 AM
mikewhant
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!

Thanks in advance.

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
• Dec 30th 2009, 07:57 PM
sym0110
Quote:

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!

Thanks in advance.

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
• Jan 1st 2010, 10:20 AM
mikewhant
Thank you, glad to see I'm doing it correctly.

Happy new year!