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

1. Homework Statement

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. Homework 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

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Hootenanny
Staff Emeritus
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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.

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

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

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

Hootenanny
Staff Emeritus
Gold Member
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
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).

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?

Hootenanny
Staff Emeritus
Gold Member
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.

so what is the charge density?

Hootenanny
Staff Emeritus
Gold Member
so what is the charge density?
The [surface] charge density on a plate is the total charge on the plate divided by the plate's area.

Is this right

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

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Hootenanny
Staff Emeritus
Gold Member
Is this right

Total charge on plate is q= E*E0A -- gauss' law
Plates area is, of course, 3.00
That is only true in the absence of a dielectric material.

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

Hootenanny
Staff Emeritus
Gold Member
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
I'm sorry I think I've just dragged you the long way round. Going back to the two equations in your OP,
C=(K$$\epsilon$$A)/d

and I know that E=V/d
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:
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 haha yes, yes i can.
I had a feeling there was another, easier method. Thanks for you help

Hootenanny
Staff Emeritus
Gold Member
haha yes, yes i can.
I had a feeling there was another, easier method. Thanks for you help
No problem I got the answer, but I have a question. Why do we use 3.00cm^2 instead of m^2

Hootenanny
Staff Emeritus
You should note that $3cm^2 \neq 0.3m^2$.