# electron between charged plate question

Discussion in 'MCAT Study Question Q&A' started by dorjiako, Aug 16, 2011.

1. ### dorjiako 2+ Year Member

Joined:
Dec 9, 2010
Messages:
71
0
An electron starts from rest at tone plate of a parallel-plate capacitor and accelerates to the other plate. The plate separation is 2mm, and it takes 1ms for the electron to travel from one side to the other. What is the capacitance of this capacitor if there are 2C of charge stored on the plates? m = 9.11 x 10^-31kg, e=1.6 x 10^-19C.

Joined:
Aug 9, 2011
Messages:
21
0
Status:
Pre-Medical
3. ### Bumbl3b33 Removed

Joined:
Feb 12, 2011
Messages:
522
1
Status:
Medical Student
Unlike the person above me, I'll try to help you. I haven't done the calculation to see if this is right, but here is how I would approach it:

First and foremost, understand that capacitance is how much ENERGY is stored on you capacitor. If you want more help with that, http://en.wikipedia.org/wiki/Capacitance The minute you see energy and objects moving the first thing going in my head is always KE=PE at the end. Thus, kinetic energy gained by the electron as it accelerates is the energy provided by the capacitor (this is the easiest way for me to think about it, sorry). So, I'm guessing they told us the distance and time the e- took to travel so we can figure out the final velocity. Why do we need that? So we can plug it into KE=1/2mv^2.

So, you can use kinematics to get the final velocity of the electron as it hits the positive plate, right? Displacement=1/2at^2 so you can get acceleration, and then with that acceleration you can get the final velocity by doing finalvelocity^2=2(acceleration)(displacement)...in both cases, the Vinitial is disregarded since it's at rest. Now plug that into KE equation.

Now, that's our KE, it has to equal to PE. The PE for a capacitor is given by 1/2QV...when you rearrange it by plugging in V=Q/C (from Q=CV) for the V in 1/2QV, you get 1/2(Q^2)/C. So since all the potential energy of the capacitor was used to get to the positive plate, we can set KE=PE=0.5Q^2/C. Now, we know KE, we know Q, the charge of 2 coulombs, and all you have to do is solve for C and that should be the answer. Remember to watch your units when doing the kinematics and stuff. Let me know if this is not right!

4. ### justhanging 2+ Year Member

Joined:
May 22, 2010
Messages:
335
2
Status:
Pre-Pharmacy
You have to set K.E = Vq which is the work done by the electric field of the capacitor to move the electron across. 1/2QV is the energy stored in the capacitor. Another way you could do is that once you find out acceleration from kinematics set it equal to the net force which is just the force of the electric field from the capacitor.

m(electron) a(kinamatics) = Eq(electron)

Find out E then use V = Ed(distance between plates)

With V you could then use Q = CV

5. ### Bumbl3b33 Removed

Joined:
Feb 12, 2011
Messages:
522
1
Status:
Medical Student
He's right. I don't know what I was thinking....my bad! PE=qV is for one charge, like in your scenario.

From an old forum response:
"Let's assume that you understand that PE=qV for one charge. Now, for condenser you can move one small charge(q) from plate to plate and work will be qV also. However, when you move the next charge(q2) the V is less, since you total charge on both plates are now Q-q. and so on. When you move the "last" charge V is almost zero. To find the total PE you need to sum all these works. (Integrate) or take the average which will be 0.5*QV (Note, I use the capital Q in this formula) "

We’ve been on the Internet for over 20 years doing just one thing: providing career information for free or at cost. We do this because we believe that the health education process is too expensive and too competitive.

We believe that all students deserve the same access to high quality information. We believe that providing high quality career advice and information ensures that everyone, regardless of income or privilege, has a chance to achieve their dream of being a doctor.