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.


Answer: 31.9 nC

Homework Equations





The Attempt at a Solution



C=(K[tex]\epsilon[/tex]A)/d

and I know that E=V/d


K=2.4
[tex]\epsilon[/tex]=8.85*10^-12
A=0.3
E=5.00

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

Answers and Replies

  • #2
Hootenanny
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Note that the effective electric field, [itex]E_f[/itex] can be written as,

[tex]E_f = E_0-E_p[/tex]

Where [itex]E_p[/itex] is the electric field due to the polarisation charges in the dielectric and [itex]E_0[/itex] is the electric field in the absence of a dielectric material.
 
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  • #3
is there another method of approaching this question? I dont believe we've learned this.

Could you expand your equatiion please
 
  • #4
Hootenanny
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How much physics have you done? Have you worked with Gaussian surfaces and Gauss' law before?
 
  • #5
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
Hootenanny
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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,

[tex]E = \frac{\sigma}{\kappa \varepsilon_0}[/tex]

Where [itex]\sigma[/itex] is the surface charge density and [itex]\kappa[/itex] 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
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
Hootenanny
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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
so what is the charge density?
 
  • #10
Hootenanny
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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.
 
  • #11
Is this right

Total charge on plate is q= E*E0A -- gauss' law
Plates area is, of course, 3.00
 
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  • #12
Hootenanny
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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.
 
  • #13
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
Hootenanny
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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[tex]\epsilon[/tex]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,
C=q/v
may also come in handy at some point :wink:
 
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  • #15
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 :wink:



haha yes, yes i can.
I had a feeling there was another, easier method. Thanks for you help
 
  • #16
Hootenanny
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haha yes, yes i can.
I had a feeling there was another, easier method. Thanks for you help
No problem :smile:
 
  • #17
I got the answer, but I have a question. Why do we use 3.00cm^2 instead of m^2
 
  • #18
Hootenanny
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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 [itex]3cm^2 \neq 0.3m^2[/itex].
 

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