What is the Work Required to Increase Separation of a Parallel Plate Capacitor?

[sw1mm3r]

Homework Statement

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.

Homework Equations

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

The Attempt at a Solution

CAN I HAVE SOME HELP?

 

[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?

 

[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??

 

[sw1mm3r]

or would i have to use

C=Q/Ed for capacitance

 

[sw1mm3r]

nevermind i made a huge mistake in my math

 

[sw1mm3r]

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

 

[sw1mm3r]

is that correct? Or is my method correct?

 

[rl.bhat]

is that correct? Or is my method correct?

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

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

 

[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?

 

[sw1mm3r]

i thought you had to convert the mm to m

 

[sw1mm3r]

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

 

[Dick]

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

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.

 

[sw1mm3r]

But i thought the question was asking for work not the capacitance? and I did not use that formula of e0*A/D because 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

 

[Dick]

But i thought the question was asking for work not the capacitance? and I did not use that formula of e0*A/D because 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

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?

 

[sw1mm3r]

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?

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

 

[Dick]

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

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.

 

[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

 

[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?

 

[sw1mm3r]

How can you use the C formula if you don't know area

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

C at 4.5 is 5.00x10^-13

 

[Dick]

How can you use the C formula if you don't know area

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

C at 4.5 is 5.00x10^-13

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.

 

[sw1mm3r]

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

 

[berkeman]

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

Yes.

 

[sw1mm3r]

so then just to make sure i would have

C=epsilon/d only

 

[berkeman]

so then just to make sure i would have

C=epsilon/d only

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.

 

[sw1mm3r]

Ok now i am super confused do you think someone could give me the value of the capacitance at 4.5 mm, because now i don't know how to solve for it

 

[Dick]

Ok now i am super confused do you think someone could give me the value of the capacitance at 4.5 mm, because now i don't know how to solve for it

You don't 'ignore the A' you take the ratio of the two equations c1=e*A/d1 and c2=e*A/d2 and conclude c1/c2=d2/d1. Because the e and the A cancel. c2=(3.1x10^-11 F)/3.

 

[sw1mm3r]

ok i think i figured out on my own with a different easier method... where c2=8.26x10^-12

and
W=8.69 J

 

[Dick]

ok i think i figured out on my own with a different easier method... where c2=8.26x10^-12

and
W=8.69 J

It's nice that you've found an easier method. Your answer is still wrong.

 

[sw1mm3r]

well by using your formula you do not get c2 being c1/3... you get c2=8.26x10^-12... and w=8.69 J

But if its wrong can you give me the right answers as i have been grinding my gears on this problem??

 

[sw1mm3r]

You don't 'ignore the A' you take the ratio of the two equations c1=e*A/d1 and c2=e*A/d2 and conclude c1/c2=d2/d1. Because the e and the A cancel. c2=(3.1x10^-11 F)/3.

thats the formula i was talking about

 

[Dick]

thats the formula i was talking about

c2=(3.1x10^-11 F)/3 doesn't give you c2=8.26x10^-12F.

 

[sw1mm3r]

no look c1/c2=d2/d1

(3.1x10^-11)/c2=.0045/.0012

c2=8.26x10^-12 so c2 is obviously not c1/3

 

[Dick]

OOhhhh. You are talking about c2=(1.2/4.5)*c1. The 1.2 was a typo. There's no '1.2' in the problem anywhere. I apologize for that. It should have been 1.5. You really haven't understood anything of this, have you?

 

[sw1mm3r]

1.2 is in the problem its 1.2 mm... and i have understood a good chunk of stuff i just got confused on the last part

 

[Dick]

Hah. You're right. I substituted the 1.5 for 1.2 way back. Apparently I'm the one who's confused. 34 posts and you lose track. Sorry again. I think you've got it right.