Double Paralell Plate Capacitor with Dielectric

  • #1

Homework Statement


Please see attached.

Homework Equations




The Attempt at a Solution


I get a result of ## \alpha =\frac{1}{2} ## for part a) - which I think is correct.

I'm stuck however on part b) - with the dielectrics. I have that the field in the region where the dielectric is ##E_{1}## is:

$$ E_{1} = \frac{Q}{2\epsilon_{0}\epsilon_{r}A}(1+\beta) $$

Similarly, the field where this is no dielectric is given by:

$$E_{2} = \frac{Q}{2\epsilon_{0}A}(2-\beta) $$

This just comes out of superposing the field due to the positive ##+Q## plate (## \frac{Q}{2A\epsilon_{0}}## ) and the negative plate with the correct charge - calculated using the standard result for an infinite plate (via Gauss's law).

$$V_{2} = E_{2}d $$

$$ V_{1}=E_{1}d $$

Since plates are connected:
$$ V_{1} = V_{2} $$
This implies:
$$ \frac{\beta +1}{\epsilon_{r}} = 2-\beta \implies \beta =\frac{2\epsilon_{r}-1}{1+\epsilon_{r}} $$

That leaves me with
$$ V_{1} =V_{2} = V = \frac{3Qd}{2\epsilon_{0}A} $$

This system looks to me to be two capacitors in parallel - so I try to use:
$$ \frac{1}{C_{effective}}=\frac{1}{C_{1}}+\frac{1}{C_{2}} = V \bigg( \frac{1}{Q_{1}} +\frac{1}{Q_{2}} \bigg) $$

where ##Q_{1} = \beta## ##Q_{2} = 1-\beta ## - but that does not get me the required result.

Could someone please tell me what assumption I've made that's wrong?

Thanks!
 

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Answers and Replies

  • #2
2
0
First,i detect a little mistake when you calculated the capacitance.
It's true that the capacitors are in parallel,so you sum the capacitance regularly,i.e :

C = C1 + C2

The reason why the capacitors are in parallel not just because it looked like in parallel,but because they have the same potential.

I don't really understand how you can get the answer,but i have simpler (well,at least for me) way to get the answer.

Note : the index 1 denote the capacitor with dielectric
C1 = Q1 / V
C2 = Q2 / V
(denoting that V1 = V2)

Then use the information
V1 = V2
E1 d = E2 d

substitute the value of E1 and E2

Name the charge on capacitors with dielectric as Q1 and the capacitors without dilectric as Q2
With equaling the potential difference,you can get
Q1 = Q2 εr

Then you can also make V in terms of Q1 and the other known constan

We know that
C= C1 + C2
All the charge variabels can cancel out and then you will get the answer.

If you still need help for the value of electric field;
E1 = Q1 / εo εr A
E2 = Q2 / εo A
 
  • #3
First,i detect a little mistake when you calculated the capacitance.
It's true that the capacitors are in parallel,so you sum the capacitance regularly,i.e :

C = C1 + C2

The reason why the capacitors are in parallel not just because it looked like in parallel,but because they have the same potential.

I don't really understand how you can get the answer,but i have simpler (well,at least for me) way to get the answer.

Note : the index 1 denote the capacitor with dielectric
C1 = Q1 / V
C2 = Q2 / V
(denoting that V1 = V2)

Then use the information
V1 = V2
E1 d = E2 d

substitute the value of E1 and E2

Name the charge on capacitors with dielectric as Q1 and the capacitors without dilectric as Q2
With equaling the potential difference,you can get
Q1 = Q2 εr

Then you can also make V in terms of Q1 and the other known constan

We know that
C= C1 + C2
All the charge variabels can cancel out and then you will get the answer.

If you still need help for the value of electric field;
E1 = Q1 / εo εr A
E2 = Q2 / εo A
Ah yes, that was very stupid - I should be adding the capacitance. The equation for the voltage is also wrong: it should read:
$$ V=\frac{3Qd}{2\epsilon_{0}A(1+\epsilon_{r})} $$. I made a typo. This means I now have a result:
$$C = \frac{\epsilon_{0}A(1+\epsilon_{r})}{3d}$$
So I'm out by a factor of 3.
What is your value for ##\beta##?
 
  • #4
2
0
Ah yes, that was very stupid - I should be adding the capacitance. The equation for the voltage is also wrong: it should read:
$$ V=\frac{3Qd}{2\epsilon_{0}A(1+\epsilon_{r})} $$. I made a typo. This means I now have a result:
$$C = \frac{\epsilon_{0}A(1+\epsilon_{r})}{3d}$$
So I'm out by a factor of 3.
What is your value for ##\beta##?
Ah yes, that was very stupid - I should be adding the capacitance. The equation for the voltage is also wrong: it should read:
$$ V=\frac{3Qd}{2\epsilon_{0}A(1+\epsilon_{r})} $$. I made a typo. This means I now have a result:
$$C = \frac{\epsilon_{0}A(1+\epsilon_{r})}{3d}$$
So I'm out by a factor of 3.
What is your value for ##\beta##?
I get β = εr/(1+εr)
try to check your electric field. There shouldn't be a factor ½ in the electric field between plates.
 
  • #5
I get β = εr/(1+εr)
try to check your electric field. There shouldn't be a factor ½ in the electric field between plates.
I think there should definitely be a factor of 1/2 - this is because the charge density is split between the top and bottom plates. -Gauss's law for a flat sheet?
 
  • #6
SammyS
Staff Emeritus
Science Advisor
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I think there should definitely be a factor of 1/2 - this is because the charge density is split between the top and bottom plates. -Gauss's law for a flat sheet?
Definitely not a factor of 1/2.

Two capacitors in parallel. You find an equivalent capacitance. Then the voltage across the plates is related to the sum of the charges on the two capacitors.
 

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