Solve Parallel Capacitor Homework: Electric Field, Charge & Energy

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The discussion centers around solving a homework problem involving a parallel plate capacitor and an alpha particle. Key points include the need to determine the electric field between the plates, the charge on the positively charged plate, and the energy stored in the capacitor. Participants emphasize the importance of understanding the relationship between the alpha particle's velocity and the electric field or voltage, suggesting the use of energy equations. The conversation highlights the connection between electric potential energy and kinetic energy for charged particles. Ultimately, the focus is on deriving the necessary equations to solve the problem effectively.
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Homework Statement


A parallel plate capacitor has two conducting plates separated 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 into play!?

Thank you!
 
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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!*
 
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!?
 
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.
 
Thank you so so so much! What great help!
 
No problem :)
 
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