# Double Paralell Plate Capacitor with Dielectric

• bananabandana
In summary, Homework Equations: The Attempt at a Solution includes a result for part a) - which is correct - and a problem with part b) - with the dielectrics. The solution is to use the electric field and calculate the capacitance.
bananabandana

## 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.

Thanks!

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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

CyanaLi said:
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##?

bananabandana said:
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##?
bananabandana said:
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.

CyanaLi said:
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?

bananabandana said:
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.

## 1. What is a double parallel plate capacitor with dielectric?

A double parallel plate capacitor with dielectric is a type of capacitor that consists of two parallel plates separated by a dielectric material. The plates are electrically conductive and the dielectric material is an insulator. This type of capacitor is used to store electric charge.

## 2. How does the dielectric material affect the capacitance of a double parallel plate capacitor?

The presence of a dielectric material between the plates of a double parallel plate capacitor increases the capacitance. This is because the dielectric material reduces the electric field between the plates, allowing for more charge to be stored on the plates.

## 3. What are the factors that affect the capacitance of a double parallel plate capacitor with dielectric?

The capacitance of a double parallel plate capacitor with dielectric is affected by the distance between the plates, the surface area of the plates, the dielectric constant of the material between the plates, and the permittivity of free space.

## 4. How does the presence of a dielectric material affect the voltage and charge on a double parallel plate capacitor?

The presence of a dielectric material between the plates of a double parallel plate capacitor reduces the voltage across the plates for a given charge. This is because the dielectric material increases the capacitance, which is the ratio of charge to voltage.

## 5. What are some common applications of a double parallel plate capacitor with dielectric?

Double parallel plate capacitors with dielectric are commonly used in electronic circuits to store energy and filter out unwanted frequencies. They are also used in power factor correction, energy storage systems, and as sensors in various devices.

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