Dielectric with capacitor

In summary, the conversation discusses a parallel-plate air capacitor that is charged to a voltage of 50.0 V and then has a dielectric material with a dielectric constant of 4.80 inserted between the plates. The question asks how much additional charge will flow from the battery onto the positive plate. The discussion also includes calculations for capacitance and charge using equations and different values. Eventually, the issue of a miscalculation is addressed and resolved.
  • #1
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A parallel-plate air capacitor of area A= 21.0 cm2 and plate separation d= 3.20 mm is charged by a battery to a voltage 50.0 V. If a dielectric material with κ = 4.80 is inserted so that it fills the volume between the plates (with the capacitor still connected to the battery), how much additional charge will flow from the battery onto the positive plate?

C= ε0A/d
Q=C(dV)
C= κC_0_
Q= κQ_0_

Using the first equation, I found the capacitance for the vacuum-insulated capacitor. Then I found the change in capacitance using C=κC_0_. I got C_0_(vacuum-insulated capacitance) to be 7.29e-9 F and got C(dielectric-insulated capacitance) to 3.5e-8. I know that while the capacitor is hooked up to the battery, the potential difference(dV) does not change. So I used Q= C(dV) to find the charge on the plates with dielectric in between which was 1.75e-6 C. I also found the Q_0_(vacuum-insulated charge) to be 3.64e-7 C. I took the difference of the two and got 1.38e-6 C. I thought this was the answer but it was wrong. The problem asks, "how much additional charge will flow from the battery onto the positive plate?" though. I feel I'm missing something but I don't know what.
 
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  • #2
Looks like you understand the problem very clearly.
But there may be something wrong with the capacitance calcs:
I get C = ε₀A/d = 8.854E-12*21E-4 / 3.2E-3 = 5.81 E-12
which is more than a thousand times smaller than your value.
Mind you, I'm struggling a bit with calculations these days.
 
  • #3
I think i figured out my problem. My was miscalculating my ε_0_ but I got it now. Thank you!
 
  • #4
Most welcome!
 
  • #5


It seems like you have correctly calculated the change in capacitance and the resulting change in charge on the plates. However, the problem is asking for the additional charge that flows from the battery onto the positive plate. This means you need to subtract the initial charge on the plates (3.64e-7 C) from the final charge on the plates (1.75e-6 C). This will give you the additional charge that flowed from the battery onto the positive plate, which is 1.38e-6 C in this case.
 

1. What is a dielectric?

A dielectric is a material that does not conduct electricity, but can be polarized by an electric field. This means that the atoms or molecules in the material will align themselves in response to the electric field, creating a separation of charges.

2. How does a dielectric affect a capacitor?

A dielectric placed between the plates of a capacitor increases the capacitance (ability to store charge) of the capacitor. This is because the dielectric reduces the electric field between the plates, allowing more charge to be stored on the plates.

3. What is the purpose of using a dielectric in a capacitor?

The use of a dielectric in a capacitor allows for a larger amount of charge to be stored on the plates, increasing the energy storage capacity of the capacitor. It also helps to reduce the risk of electrical breakdown between the plates.

4. Can any material be used as a dielectric?

No, not all materials are suitable for use as a dielectric. The ideal dielectric should have high electrical resistance, low dielectric loss, and be able to withstand high voltages without breaking down. Some common materials used as dielectrics are air, paper, glass, and certain plastics.

5. How does the dielectric constant affect the capacitance of a capacitor?

The dielectric constant is a measure of how well a material can store electric charge. It is a factor in determining the capacitance of a capacitor, as a higher dielectric constant means a larger capacitance. This is because the higher dielectric constant allows for a greater reduction in electric field between the plates, allowing more charge to be stored on the plates.

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