Calculations involving different Dielectrics and Capacitors

  • #1
as2528
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TL;DR Summary: Need dielectric constant for given capacitor

Given a 7.4 pF air-filled capacitor, you are asked to convert it to a capacitor that can store up to 7.4 mJ with a maximum potential difference of 652 V. Which dielectric in Table 25-1 should you use to fill the gap in the capacitor if you do not allow for a margin of error?

I did the following:

C=(k*e*A)/d and Q=C*V=>C=Q/V

So:

Q/V=(k*e*A)/d=>k=Q/V*d/(A*e)=>k=7.4*10^-6/652*7.4*10^-12=>k=8.7616*10^-20

The answer is 4.7, and uses the potential between capacitors formula. Why are the formulas I used wrong? It seems to me like it makes sense.
 
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  • #2
as2528 said:
the potential between capacitors formula
What does that mean? What is Table 25-1?
 
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  • #3
In addition to what @hutchphd said:

Is the energy really 7.4mJ (millijoules) or did you mean 7.4μJ (microjoules)?

What formula relates energy (not charge) stored to C and V?
 
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  • #4
hutchphd said:
What does that mean? What is Table 25-1?
Table 25-1 was a table showing a few dielectric constants along with the materials they corresponded to. I was supposed to get 4.7 from the question, which was the part I failed on. On the table it said Pyrex was corresponding to that kappa.
 
  • #5
Steve4Physics said:
In addition to what @hutchphd said:

Is the energy really 7.4mJ (millijoules) or did you mean 7.4μJ (microjoules)?

What formula relates energy (not charge) stored to C and V?
That was u=.5c*v^2. So I calculated with charge which causes the error?
 
  • #6
as2528 said:
That was u=.5c*v^2. So I calculated with charge which causes the error?
And it was the microjoules.
 
  • #7
as2528 said:
That was u=.5c*v^2.
That's the correct formula. But as far as I can see, you didn't use it.

as2528 said:
So I calculated with charge which causes the error?
The charge (Q) is not needed. Try this:

Step 1: With the dielectric present, U(energy stored) =7.4μJ when V =652V. Use the formula U=½CV² to find C (the required capaicitance with the dielectric in place).

Step 2: Note that without the dielectric, the capacitance is 7.4pF. Use this and your result from Step 1 to find the dielectric constant.
 
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  • #8
Steve4Physics said:
That's the correct formula. But as far as I can see, you didn't use it.The charge (Q) is not needed. Try this:

Step 1: With the dielectric present, U(energy stored) =7.4μJ when V =652V. Use the formula U=½CV² to find C (the required capaicitance with the dielectric in place).

Step 2: Note that without the dielectric, the capacitance is 7.4pF. Use this and your result from Step 1 to find the dielectric constant.
Thank you! This cleared it up for me.
 
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1. How do I calculate the capacitance of a parallel plate capacitor with different dielectric materials?

The capacitance of a parallel plate capacitor with different dielectric materials can be calculated using the formula C = ε₀A/d, where ε₀ is the permittivity of free space, A is the area of the plates, and d is the distance between the plates. The permittivity of the dielectric material should be substituted for ε₀ in the formula.

2. What is the effect of using a dielectric material on the capacitance of a parallel plate capacitor?

The use of a dielectric material in a parallel plate capacitor increases its capacitance. This is because the dielectric material has a higher permittivity than air, which reduces the electric field between the plates and allows more charge to be stored.

3. How do I calculate the electric field between the plates of a parallel plate capacitor with different dielectric materials?

The electric field between the plates of a parallel plate capacitor with different dielectric materials can be calculated using the formula E = V/d, where V is the potential difference between the plates and d is the distance between the plates. The permittivity of the dielectric material should be substituted for ε₀ in the formula.

4. How does the dielectric constant affect the capacitance of a parallel plate capacitor?

The dielectric constant, also known as the relative permittivity, is a measure of how well a dielectric material can store electric charge. A higher dielectric constant means a higher capacitance for a parallel plate capacitor, as it allows more charge to be stored between the plates.

5. Can I mix different dielectric materials in a parallel plate capacitor?

Yes, it is possible to mix different dielectric materials in a parallel plate capacitor. The overall capacitance will be a combination of the individual capacitances of each material, calculated using the formula C = ε₀A/d. However, it is important to note that the electric field may not be uniform between the plates if the dielectric materials have different permittivities.

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