# Calculate the maximum charge that can be stored in the capacitor

1. Apr 12, 2008

### unhip_crayon

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

Each plate of a parallel plate capacitor has an area of 3.00 cm2. A dielectric material with
dielectric constant κ = 2.40 and a dielectric strength of 5.00 kV/mm is placed between the plates.
Calculate the maximum charge that can be stored in the capacitor.

2. Relevant equations

3. The attempt at a solution

C=(K$$\epsilon$$A)/d

and I know that E=V/d

K=2.4
$$\epsilon$$=8.85*10^-12
A=0.3
E=5.00

but i cant find d, which is what i probably need

2. Apr 12, 2008

### Hootenanny

Staff Emeritus
Note that the effective electric field, $E_f$ can be written as,

$$E_f = E_0-E_p$$

Where $E_p$ is the electric field due to the polarisation charges in the dielectric and $E_0$ is the electric field in the absence of a dielectric material.

Last edited: Apr 12, 2008
3. Apr 12, 2008

### unhip_crayon

is there another method of approaching this question? I dont believe we've learned this.

4. Apr 12, 2008

### Hootenanny

Staff Emeritus
How much physics have you done? Have you worked with Gaussian surfaces and Gauss' law before?

5. Apr 12, 2008

### unhip_crayon

i missed a week of school when I tore a ligament in my ankle so I missed those lectures, but I do have the notes and I know about gauss' law

E=q/E0A

Could you please just start me off, I have exams comming up

6. Apr 12, 2008

### Hootenanny

Staff Emeritus
It can be shown that the electric field of a dielectric capacitor may be written thus,

$$E = \frac{\sigma}{\kappa \varepsilon_0}$$

Where $\sigma$ is the surface charge density and $\kappa$ is the dielectric constant. If I have a little more time this evening I'll post a write up of how it is derived (or you could just look at your notes).

7. Apr 12, 2008

### unhip_crayon

that formula is nowhere in our notes...hmm. I'm going to ask a buddy of mine for the solution, but if you dont mind could you post you method too.

Surface charge density...is that a constant?

8. Apr 13, 2008

### Hootenanny

Staff Emeritus
I'm afraid until you've answered the question, forum guidelines prohibit me from posting the derivation.

Yes, once the capacitor is charged and the dielectric is inserted, the charge density is constant.

9. Apr 13, 2008

### unhip_crayon

so what is the charge density?

10. Apr 13, 2008

### Hootenanny

Staff Emeritus
The [surface] charge density on a plate is the total charge on the plate divided by the plate's area.

11. Apr 13, 2008

### unhip_crayon

Is this right

Total charge on plate is q= E*E0A -- gauss' law
Plates area is, of course, 3.00

Last edited: Apr 13, 2008
12. Apr 13, 2008

### Hootenanny

Staff Emeritus
That is only true in the absence of a dielectric material.

13. Apr 13, 2008

### unhip_crayon

first off, id like to thank you for your help, im learning a lot.

the only formulas i have for a dielectric material are:

E = 1/2(qV) = 1/2(q0)(V0/k) = E0/k

C=q/v = qk/V0 = kC0

14. Apr 13, 2008

### Hootenanny

Staff Emeritus
I'm sorry I think I've just dragged you the long way round. Going back to the two equations in your OP,
Can you think of a way of eliminating d from the first equation using the second?

And this equation,
may also come in handy at some point

Last edited: Apr 13, 2008
15. Apr 13, 2008

### unhip_crayon

haha yes, yes i can.
I had a feeling there was another, easier method. Thanks for you help

16. Apr 13, 2008

### Hootenanny

Staff Emeritus
No problem

17. Apr 13, 2008

### unhip_crayon

I got the answer, but I have a question. Why do we use 3.00cm^2 instead of m^2

18. Apr 14, 2008

### Hootenanny

Staff Emeritus
I'm not sure what you mean, it doesn't matter which units one uses provided one converts them all to the same type. However, I noticed in your OP that,
You should note that $3cm^2 \neq 0.3m^2$.