Calculate the maximum charge that can be stored in the capacitor

[unhip_crayon]

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.


Answer: 31.9 nC

Homework Equations

The Attempt at a Solution

C=(K\epsilonA)/d

and I know that E=V/d


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

but i can't find d, which is what i probably need

 

[Hootenanny]

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.

 

[unhip_crayon]

is there another method of approaching this question? I don't believe we've learned this.

Could you expand your equatiion please

 

[Hootenanny]

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

 

[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

 

[Hootenanny]

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).

 

[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 don't mind could you post you method too.

Surface charge density...is that a constant?

 

[Hootenanny]

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.

 

[unhip_crayon]

so what is the charge density?

 

[Hootenanny]

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.

 

[unhip_crayon]

Is this right

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

 

[Hootenanny]

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.

 

[unhip_crayon]

first off, id like to thank you for your help, I am 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]

first off, id like to thank you for your help, I am 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\epsilonA)/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,

C=q/v

may also come in handy at some point

 

[unhip_crayon]

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]

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

No problem

 

[unhip_crayon]

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

 

[Hootenanny]

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

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,

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.
[...]
A=0.3

You should note that 3cm^2 \neq 0.3m^2.