# Homework Help: Parallel plate capacitor help

1. Aug 3, 2009

### sw1mm3r

1. The problem statement, all variables and given/known data

An isolated parallel-plate capacitor (not connected to a battery) has a charge of Q=1.4x10^-5 C. The separation between the plates initially is d=1.2 mm, and for this separation the capacitance is 3.1x10^-11 F. Calculate the work that must be don to pull the plates apart until their separation becomes 4.5 mm if the charge on the plates remains constant. The capacitor plates are in a vacuum.

2. Relevant equations

u=(1/2)E^2(epsilon,zero)
W=U2-U1
U=(1/2)(Q^2)/C

3. The attempt at a solution

CAN I HAVE SOME HELP?

Last edited by a moderator: Aug 4, 2009
2. Aug 3, 2009

### Dick

You will probably also need a formula for the relation between the capacitance between the two plates and the distance between them, right? Then use the potential energy equation. And leave the CAPS LOCK off, ok?

3. Aug 3, 2009

### sw1mm3r

ok i have been tryin to figure this out for the past hour and got a work of 4.7x10^-10 J would that be correct?
i used

C=(C*E)/(d)
U=(1/2)(Q^2)/C)
W=U2-U1

Would that be right??

4. Aug 3, 2009

### sw1mm3r

or would i have to use

C=Q/Ed for capacitance

5. Aug 3, 2009

### sw1mm3r

nevermind i made a huge mistake in my math

6. Aug 3, 2009

### sw1mm3r

ok after i fixed my math i get an answer of -539.07 J

7. Aug 3, 2009

### sw1mm3r

is that correct? Or is my method correct?

8. Aug 3, 2009

### rl.bhat

In the first case
C1 = εοA/d1
In the second case

C2 = εοA/d2
Substitute the values of d1 and d2.
C1 is given. Find C2.
Find the energies in C1 and C2 and find the difference.

9. Aug 3, 2009

### Dick

Your capacitance is the only thing that changes. C_after=C_before*(1.2/4.5). Both of your energy numbers seem somewhat off to me. Isn't (1/2)*(1.4E-5*coulomb)^2/(3.1E-11*farad)=3.16J? Or is my list of constants off?

10. Aug 3, 2009

### sw1mm3r

i thought you had to convert the mm to m

11. Aug 3, 2009

### sw1mm3r

idk maybe i missed something in lecture but can you explain why C after=C before*(1.2/4.5)

12. Aug 4, 2009

### Dick

You don't have to convert the distance to any particular units. The capacitance is proportional to 1/D. You can't directly use the e0*A/D formula because you don't know A. But you do know A is the same at both distances. Take a ratio.

13. Aug 4, 2009

### sw1mm3r

But i thought the question was asking for work not the capacitance? and I did not use that formula of e0*A/D cuz of the same reason we have no area which is why i used

C=(C*E)/(d)... To find capacitance at 4.5 mm..... Because they gave us capacitance at 1.5 mm in the problem

U=(1/2)(Q^2)/C)... to find potential energy at 1.5 mm and at 4.5 mm

W=U2-U1... to find the work done

Is what I did incorrect??, because we do not get the same answers.... Idk i may be wrong but what your doing seems like your just solving for capacitance after and not for the work

14. Aug 4, 2009

### Dick

Yes, you need to find the capacitance at 4.5 mm. Then you can just take the difference of the potential energies. But how are you doing that? C=C*E/d doesn't even make sense. And what do you get for the potential energy at 1.5 mm?

15. Aug 4, 2009

### sw1mm3r

the problem gives you the capacitance at 1.5 so using U=(1/2)(Q^2)/C.... I get U=3.16
and U at 4.5 mm gives me U=195.93
then taking the difference gives me Work=192.76 J

and C=C*E/d is an equation our prof gave us, it gives the same values if you use C=Q/Ed

16. Aug 4, 2009

### Dick

I get the C at 4.5mm is 1/3 of C at 1.5mm. I still don't get C=C*E/d. There are two C's in there.

17. Aug 4, 2009

### sw1mm3r

what formula for C are you using?
What U at 4.5 mm are you getting?

yea that formula uses C to get a new C i think.... our prof gave it to use in class.... it gives the same values as C=Q/Ed

18. Aug 4, 2009

### Dick

I'm using C=epsilon*A/d. If d changes from 1.5mm to 4.5mm C is divided by 3. If that formula gives you a new C somehow, then what do you get for C at 4.5mm? Is E electric field? You don't know that either, do you?

19. Aug 4, 2009

### sw1mm3r

How can you use the C formula if you dont know area

I found the electric field by using E=(K*Q)/d^2

C at 4.5 is 5.00x10^-13

20. Aug 4, 2009

### Dick

I don't need to know area, I just need to know A doesn't change. Only d changes by a factor of 3. Look back at rl.bhat's post. E=(K*Q)/d^2 isn't right. E is proportional to charge density (charge per area). d is the separation between the plates. It doesn't have anything to do with charge density.

21. Aug 4, 2009

### sw1mm3r

ok that makes sense so then in solving for the capacitance for 4.5 mm i would neglect A because its constant??

22. Aug 4, 2009

### Staff: Mentor

Yes.

23. Aug 4, 2009

### sw1mm3r

so then just to make sure i would have

C=epsilon/d only

24. Aug 4, 2009

### Staff: Mentor

No, that is not a valid equation. But when you write the equation for the capacitance change or ratio, the areas A cancel out of that equation.

25. Aug 4, 2009

### sw1mm3r

Ok now i am super confused do you think someone could give me the value of the capacitance at 4.5 mm, cuz now i dont know how to solve for it