# Homework Help: Change in energy stored on a capacitor when dielectric is changed.

1. Jun 28, 2013

### Final_HB

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

the separation of a parallel plate capacitor is 2mm with a potential difference between them of 45V.
(i) Find the electric field between the plates.
(ii)Find the charge per unit area of the plates if the permittivity of the dielectric medium is 20 $\epsilon_0$
(iii) Find the capacitance of each plate if the area is .13m2
(iv) If the dielectric medium between the plates is removed, so air occupies the region between the plates, find the resulting change in energy stored on the capacitor if (a) the plates are kept electrically isolated, and (b) If the plates are connected to the battery, such that the potential difference between them remains at 45V as the dielectric is removed.
(v) If the medium is only partially removed, such that half the area has the original dielectric and the other half has only air separating the plates, find the capacitance of this capacitor.

2. Relevant equations
Capacitance of a parallel capacitor: C= $\frac{\epsilon_0 A}{d}$
Charge per unit area of capacitor: E= $\frac{σ}{\epsilon_0}$
$\rightarrow$ σ=E ε 0
Electric field between plates: E=$\frac{V}{d}$

The question says we may assume the permitivity of air is the same as a vaccum, ε
0 = 8.85x10-12

3. The attempt at a solution
The first three parts are fine. I think :tongue: Ill run through them just to be sure
(i) E= $\frac{V}{d}$
E=$\frac{45}{.002}$
E=22500 (units im not sure of for an electric field )

(ii) E=$\frac{σ}{(20)(\epsilon_0)}$ We re arrange to get σ (charge/area) on its own.
so: σ= E ε0
σ= (22500) (1.17x10^-10)
σ= 2.63x10-6 C/m2

(iii)
C=$\frac{\epsilon_0 A}{d}$
C=$\frac{1.17x10-10 x .13}{.002}$
C= 7.605-9 F

And here is where it gets crazy

I know that the energy stored for a capacitor is $\frac{1}{2}$ CV2 so since its an energy change, we find what it is originally first.

The energy stored at first then is :
E=$\frac{1}{2}$ (C)(V2 )
E=(.5)(7.605-9) (45)2
E=7.7x10-6 joules.

Since the dielectric has changed, C changes.
New C is just= 5.7525x10-10

Since no current is flowing in (a) The change would be -7.710-6 J (no current, no charge... right?)

And for b) Charge is gonna be
(.5)(5.7525x10-10) (45)2
E=5.8244x10-7
So change in charge is:
7.11756-6
That makes sense to me anyways :tongue:
And ive spend an hour trying to work out a way to start the last part, with no luck at all.

Thanks in advance

Last edited: Jun 28, 2013
2. Jun 28, 2013

### Staff: Mentor

What are the units of V? What are the units of d? So what are the units of V/d?
Check your math for the final value of σ. Looks like a calculator finger problem
Again, looks like a math error crept in during your calculation of the capacitance. Redo it.
Also, I suggest that you hang onto a couple of extra digits for intermediate values in order to keep rounding errors in check.
You'll want to sort out your issues above before plunging on. But I'll give you some hints that might pan out for you. First, consider what remains constant when you remove the dielectric in part (iv)(a) versus part (iv)(b). There's another expression for the energy stored in a capacitor that involves the other thing that can remain constant...

3. Jun 28, 2013

### Final_HB

Ah yes of course :tongue:
i)Electric field is Volts/meter, yes?
ii)σ is apparently 3.98x10-6 does that look better?:shy:
iii) I left out the 'x10' bit by accident so the actual C is 7.605x10-9 unless that still looks wrong.

The only things that remain constant are the separation between the plates, the area of the plates in part a.

In part b, the 45V is also constant.

As for a formula relating those things that brings me back to energy... im lost :rofl:

4. Jun 28, 2013

### Staff: Mentor

Yes
yes
Still looks wrong to me... You're calculating $C = 20 \epsilon_o \frac{A}{d}$, right?
In part (a), something else is constant so long as the capacitor is isolated from the battery (or any other connections)

Hint: What's a formula for the voltage across a capacitor?

5. Jun 28, 2013

### Final_HB

Okay.... as soon as I do it out on the calculator as you have it written, i get an answer an order of magnitude higher :tongue: C=1.505x10-8 Are we getting somewhere?

Since its electrically isolated, Charge is constant. Cause there is no volts going into the circuit or going out, charge stays the same. Is my thinking right?

Formula for Volts across a capacitor comes from C=Q/V.
which becomes V=Q/C.

6. Jun 28, 2013

### Staff: Mentor

Absolutely. That's the right value.
Almost. Charge is constant because there's no external connection by which charge can enter or leave. Volts don't "go in" or "go out" of a circuit, charge does. Volts is the potential difference.
Right. So what happens when you substitute that expression for volts into your expression for energy stored on a capacitor? How does the stored energy depend upon charge?

7. Jun 29, 2013

### Final_HB

Eventually we get somewhere :tongue:

Energy is 1/2 (C)(V)2 orginally.

So new forumla after subbing is:
E=1/2 (C)(Q/C)2
E=$\frac{1}{2}$ (C)($\frac{Q^2}{C^2}$)

Which cancels down to:
E=$\frac{Q^2}{2C}$

8. Jun 29, 2013

### Staff: Mentor

Correct. So now you have a handy expression for finding the energy stored on a capacitor when you are given the charge (or the change in energy, such as in this case where an isolated capacitor has a fixed charge but the capacitance is changed by removing its dielectric...)

9. Jun 29, 2013

### Final_HB

so since... we can get the charge on the capacitor originally, do we just put it into that little formula and get a value for E?
Charge could be gotten from charge per unit area formula.
σ=Q/A
so... Q=Aσ
$\rightarrow[itex] Q= 5.174x10-7 And our new capacitance is C=5.7525x10-10 if im right so far, E=[itex]\frac{(5.174x10^-7)^2}{2(5.7525x10^-10)}$
E=2.32x10-4

10. Jun 29, 2013

### Staff: Mentor

Yes, that looks good.

11. Jun 29, 2013

### Final_HB

Beautiful

So for part b) Same story?
Do i just sub into 1/2CV2 ?

12. Jun 29, 2013

### Staff: Mentor

Yes. Remember that you want to determine the change in energy when the dielectric is removed.

13. Jul 30, 2013

### Final_HB

so... we get the original charge before any change in dielectric, and find the new charge and do a little subtraction?

Im REALLY sorry for being a month late to do this... my internet shut down and we need to get a replacement

14. Jul 30, 2013

### Staff: Mentor

You can do that. Although, if the capacitor's potential difference is being held constant by the battery then you can use the (1/2)CV2 formula for the energy; The capacitance changes when the dielectric is removed.

15. Jul 30, 2013

### Final_HB

short and sweet way it is

E=1/2CV2

E=$\frac{1}{2}$ (5.7525x10-10 ) (45)2
E=5.82x10-7 V/M

Right?

16. Jul 30, 2013

### Staff: Mentor

That would be the total energy stored on the capacitor when the dielectric is removed, but the units you've specified are not correct. A volt is a joule/coulomb, and you've divided that by M (which I assume to mean m for meter). That does not yield joules (energy).

So, that's the total energy but you're looking for the change in energy...

17. Jul 30, 2013

### Final_HB

Sorry... Had electric field units in my head.

The original energy in the system would be found the same way.
And it comes up as 1.52x10-5

So energy on the plates has increased by 1.4618x10-5 Joules.

18. Jul 30, 2013

### Staff: Mentor

The original energy you've got there looks a bit high. Could be your capacitor value. Looking back over your posts I see where you claimed the capacitance with the dielectric in place to be C=1.505x10-8 F, and I agreed with that. I should have taken a closer look at the value, because it's missing a digit

Could you, please, calculate one more time the value of the capacitance when the dielectric is in place? Then revisit the initial energy figure. Does the energy in the capacitor increase or decrease when you remove the dielectric?

19. Aug 1, 2013

### Final_HB

And we were just getting places

am... it comes up as 1.1505x10-8

so energy stored is :
1/2CV2= 1.165x10--5 (not a massive change)

20. Aug 1, 2013

### Staff: Mentor

So now you have values for the energy on the capacitor while the dielectric is in place and for when it is removed...

21. Aug 1, 2013

### Final_HB

Beautiful

So the energy has decreased by 3.55x10-6

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