Parallel Plates

1. Dec 23, 2008

Inertialforce

1. The problem statement, all variables and given/known data
A proton, accelerated from rest through a potential difference of 1.2 x 10^4 V, is directed at a fixed 5.0 x 10^-6 charge.

a) What is the speed of the proton as it leaves the parallel plates?

b) What is the distance d from the fixed charge when the proton is stopped?

2. Relevant equations

3. The attempt at a solution
A) For part "a" what I did was I used the principle of conservation of energy equation (-Δ Ep = Δ Ek or ΔEp = -ΔEk) and isolated for final velocity.

Note: ΔEp = ΔV x q in this question not mgh

Δ Ep = -ΔEk
Δ V x q = Eki - Ekf
Δ V x q = -1/2mvf^2
vf = √(-2ΔV x q/m)

then I plugged in all my values and the answer that I got was 1.5 x 10^6 m/s could someone please check this to see if it is correct because this is the first time that I have done these questions and I just want to see if I am doing it right before I proceed to the other questions.

B) For part "b" I do not understand the question, could someone please tell me what equations I should be using or point me in the right direction?

2. Dec 23, 2008

Defennder

Looks ok here.

Are you given the initial distance of the proton from the fixed charge when it first leaves the plates?

3. Dec 23, 2008

Inertialforce

No, I am not given a distance of any kind in this question.

4. Dec 23, 2008

Defennder

Unless I'm mistaken, I'm pretty sure you need to know the initial distance of the proton from the fixed positive charge. Otherwise all you can do is to express the answer in terms of the relative distance travelled by the proton:

ie. $$\frac{1}{r_2} - \frac{1}{r_1}$$. where r1 denotes the distance from the fixed charge and r2 the initial distance of the proton from the fixed charge when it leaves the plates.

5. Dec 23, 2008

Dick

If you aren't given a initial distance from the fixed charge, then I would assume that it's infinity. Just because there isn't much else to do. Defennder is quite right that otherwise it's ill defined.

Last edited: Dec 23, 2008
6. Dec 23, 2008

korkscrew

It seems for part b), they want you to know how close a proton having the energy found in part a) can approach the other charge. You could try integrating Coulomb's Law from infinity to "d", your unknown. This gives you another expression for the change in potential energy that you could use with the one from part a). (I am new to this forum. I have read the rules on posting, but any feedback is appreciated!)

7. Dec 31, 2008

Inertialforce

I'm not quite sure but I think we are supposed to solve this question using some of the methods you suggested. To better illustrate this question I have added an attachment of a diagram given to us along with the question to better see how to solve this question. I hope that it helps in some way.

And after seeing the diagram would I still need to use Coulomb's Law from infinity to "d"?

Attached Files:

• Parallel Plates.doc
File size:
26 KB
Views:
53
8. Dec 31, 2008

Defennder

Um, I should point out that not many people would open *.doc files for fear of viruses/trojans. Upload it as a PDF file instead.

9. Jan 2, 2009

Inertialforce

Sorry, but I do not know how to make (or change a word document into a) pdf document. If you could tell me how to do it in a few simple steps then I would be happy to change the attachment into a pdf file and re-upload it.

10. Jan 2, 2009

rootX

3rd party tools like:
pdf convertor
http://www.freepdfconvert.com/
https://www.pdfonline.com/convert_pdf.asp

.. many other

Save as pdf in Word 2007 (not in all of them)
Save as html .. other option

11. Jan 2, 2009

rootX

I think korkscrew suggestion looks good...
It just that

energy possessed by electron between the plates = final potential energy with the fixed proton/object

don't include the part a calculations (jump straight from the initial conditions)
potential energy given to proton --> kinetic (speed) [skip this one] --> final potential

12. Jan 3, 2009

Inertialforce

Thanks for all the help. I have now converted the diagram included with the question from word format (.doc) into pdf format. After looking at this diagram does it make solving the question any easier in terms of the information it gives you or no? Because when I looked at the questions and then at the diagram I found that the diagram only helped with answering part "a" of this question.

Attached Files:

• Parallel_Plates.pdf
File size:
21.5 KB
Views:
63
13. Jan 4, 2009

Staff: Mentor

As others have suggested, just assume that the fixed charge is sufficiently far away from the second plate and apply conservation of energy.

14. Jan 4, 2009

korkscrew

Nice diagram! To clarify my reasoning, I meant that for part a, you calculate the energy of the proton to find its speed. In part b, you use this energy and set it equal to your integral of Coulomb's Law so you can solve for your unknown "d" - one of your limits of integration.

More detailed clarification: The purpose of the plates is to give the proton some kinetic energy so it can approach the other positive charge. You know the voltage and you know the charge on the proton. You can find the energy of the proton using these. Since you're given the voltage, you don't need to worry about the distance between the plates. The accelerated proton leaves the plates, meaning it has some set amount of energy it can use to approach the 2nd charge. It needs this energy, otherwise, it would be repelled, right? Your distance of closest approach is "d", your unknown. Integrate Coulomb's Law with respect to r from infinity to "d", your unknown. (Watch signs b/c work is negative here, *and* you should get something in terms of 1/r.) You already found the final potential energy in part a) to get the speed by using conservation of energy. So, the change in potential energy is just final minus initial (zero), which is just your final potential energy.

Plugging everybody in, you should have a potential energy found in part a) *set equal to* a potential energy in terms of "d" from Coulomb's Law. Solve for "d".

Last edited: Jan 4, 2009