• Support PF! Buy your school textbooks, materials and every day products Here!

Parallel Capacitor

  • #1

Homework Statement


A parallel plate capacitor has two conducting plates seperated by a vacuum. The distance is D and the area of each plate is A.
An alpha particle with mass M and charge Q is placed on the positively-charged plate, between the plates. It shoots through a small hole in the negatively charged place with speed S.
What is the magnitude of the uniform electric field between the plates?
How much charge lies on the positively-charged plate?
How much energy is stored in the capacitor's electric field?


Homework Equations


C=Q/V
V=Ed
C=(A/d)(epsilon_o)


The Attempt at a Solution



For the first question, I can find the capacitence, C=(epsilon_0)(A/D), but I'm not sure where to go from here.
For the second question, I could find Q using C=Q/V and use V=Ed but I need E from the first question that I don't know how to find!

I'm not sure where the alpha-particle comes in to play!?

Thank you!
 

Answers and Replies

  • #2
53
0
You are very right about the fact that E is needed. All of your logic is great up to that point. So with problems like these, take a step back and ask; How could I possibly find E (or V).

The first thing you should do is review or find out what an alpha particle is. An alpha particle is a helium atom that has been stripped of its electrons. This means that it has a 2+ charge.

Now we know when this charged particle is placed into the capacitor, it will experience a force from the electric field and start to gain velocity. The problem tells us how fast it is moving when it reaches the other side. Try and find a way to relate the velocity obtained by the particle to the electric field or voltage between the plates. *Cough!*Use Energy*Cough!*
 
  • #3
Thank you Hellabyte! :)

If I find V, then I know how to find E.
So, V=Work/charge
V=(0.5mv^2)/q
thus V=(0.5MS^2)/Q

Is this correct? I think it is, and from here I know how to do all the problem...assuming it is correct!?
 
  • #4
53
0
Yes exactly. Remember(or learn) that qV is the kinetic energy gained by a charged particle that goes through a potential difference of V. Try and draw some parallels to it and the gravity potential-kinetic energy problems that you probably remember so fondly, they're completely analogous.
 
  • #5
Thank you so so so much! What great help!
 
  • #6
53
0
No problem :)
 

Related Threads for: Parallel Capacitor

  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
5
Views
3K
  • Last Post
Replies
1
Views
874
  • Last Post
Replies
7
Views
2K
  • Last Post
Replies
6
Views
1K
  • Last Post
Replies
3
Views
1K
Replies
0
Views
1K
  • Last Post
Replies
1
Views
2K
Top