Double Paralell Plate Capacitor with Dielectric

[bananabandana]

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!

 

[CyanaLi]

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

 

[bananabandana]

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 $\beta$?

 

[CyanaLi]

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.

 

[bananabandana]

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?

 

[SammyS]

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.