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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Related to Calculations involving different Dielectrics and Capacitors

What is a dielectric, and how does it affect a capacitor?

A dielectric is an insulating material that, when placed between the plates of a capacitor, increases its capacitance by reducing the electric field within the capacitor. This occurs because the dielectric becomes polarized in the presence of an electric field, which reduces the effective field and allows the capacitor to store more charge for a given voltage.

How do you calculate the capacitance of a capacitor with a dielectric material?

The capacitance of a capacitor with a dielectric material is given by the formula \( C = \kappa \cdot \epsilon_0 \cdot \frac{A}{d} \), where \( \kappa \) is the dielectric constant of the material, \( \epsilon_0 \) is the permittivity of free space, \( A \) is the area of one plate, and \( d \) is the distance between the plates. The dielectric constant \( \kappa \) represents how much the dielectric material increases the capacitance compared to a vacuum.

What is the dielectric constant, and how is it determined?

The dielectric constant, also known as the relative permittivity, is a dimensionless number that describes how much a dielectric material can increase the capacitance of a capacitor compared to a vacuum. It is determined experimentally by measuring the capacitance of a capacitor with and without the dielectric material and taking the ratio of the two values. The dielectric constant is specific to each material and typically ranges from 1 (for a vacuum) to several thousand (for materials like barium titanate).

How does the introduction of multiple dielectrics affect the capacitance of a capacitor?

When multiple dielectric materials are introduced between the plates of a capacitor, the overall capacitance can be calculated by considering each dielectric layer separately and then combining their effects. If the dielectrics are arranged in series (one after another), the reciprocal of the total capacitance is the sum of the reciprocals of the individual capacitances. If they are arranged in parallel (side by side), the total capacitance is the sum of the individual capacitances. The equivalent capacitance can be more complex depending on the specific configuration of the dielectrics.

How do you calculate the energy stored in a capacitor with a dielectric?

The energy stored in a capacitor with a dielectric is given by the formula \( U = \frac{1}{2} C V^2 \), where \( C \) is the capacitance with the dielectric material and \( V \) is the voltage across the capacitor. Since the dielectric increases the capacitance, the energy stored for a given voltage will be higher compared to a capacitor without a dielectric. This formula shows that the energy stored is directly proportional to the capacitance and the square of

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