Circuit with a parallel plate capacitor

In summary, we are given a parallel plate capacitor with a certain plate area and spacing. The capacitor is initially filled with air and has a certain charge on its plates. The spacing is then reduced and the capacitor is connected to a battery, causing a change in charge. After being disconnected from the battery, the plates are returned to their initial spacing and a different fluid is added between them, resulting in a change in voltage. Finally, the area of overlap between the plates is increased, leading to a question about the charge on the positive plate of the capacitor. The equations used to calculate the various values are also provided.
  • #1
strawberrysk8
27
0

Homework Statement



1.) Consider a parallel plate capacitor with plates of area 1.36 m2 whose plates are 1.24 cm apart. The gap between the plates is filled with air (assume that κair is unity) and the positive plate has a charge of 10.5 nC on it while the negative plate has a negative charge of equal magnitude on it.

The capacitor has its plate-spacing reduced to 1.12 mm, and the plates are connected to a 1.51 V battery. So that the charge on the positive side of the capacitor is now 16.2nC.

The capacitor is disconnected from the battery and the plates are returned to their initial spacing. So that the voltage across the capacitor is now 16.7V.

The space between the plates is now filled with a fluid with a dielectric constant of 5.07.
So that the voltage across the capacitor is now 3.297V.

The area of overlap between the capacitor plates is increased to 2.01 m2.

What is the charge on the positive plate of the capacitor?


Homework Equations





The Attempt at a Solution



V = Q/C
C = AE/d

Q = CV = AEV/d
d = 0.00112m
V = 1.51V
A = 1.36
Q = 16.2nC

V = Q/C = Q/ (AE/d) = Qd/AE
d = 0.0124m
Q = 16.2nC
A = 1.36m2
V = 16.7V

V = V(0)/k
V(0) = 16.7V
k = 5.07
V = 3.297V

Q = CV = AEV/d
A = 2.01m2
V = 3.297V
d = 0.0124m
Q = 4.73nC
BUT Q IS WRONG? WHY IS IT WRONG?
 
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  • #2


The capacitor is disconnected from the battery and the plates are returned to their initial spacing.

Where does charge come from after the battery is disconnected?
 
  • #3


i guess there is no new charge so is the charge still 16.2nC?
 
  • #4


if it is...why is V = Q/C = 11.2V? If C = AE/d = 2.01(8.85*10^-12)/0.0124
According to the answer key it is supposed to be 2.23V using the formula initial VC = final VC.
 
  • #5


strawberrysk8 said:
i guess there is no new charge so is the charge still 16.2nC?

Yes. That's what it looks like to me.
 
  • #6


strawberrysk8 said:
if it is...why is V = Q/C = 11.2V? If C = AE/d = 2.01(8.85*10^-12)/0.0124
According to the answer key it is supposed to be 2.23V using the formula initial VC = final VC.

Wasn't the question just how much charge?

The voltage will drop when the area increases if the charge is the same.
 
  • #7


oh okay. thank you so much!
 

1. What is a parallel plate capacitor?

A parallel plate capacitor is a type of capacitor that consists of two conductive plates separated by a dielectric material, such as air or a non-conducting material. It is used to store electrical energy by creating an electric field between the plates.

2. How does a parallel plate capacitor work?

A parallel plate capacitor works by storing electrical charge on the two plates, creating an electric field between them. When a voltage is applied, one plate becomes positively charged and the other becomes negatively charged. The electric field between the plates stores the energy, which can then be released when the capacitor is discharged.

3. What is the capacitance of a parallel plate capacitor?

The capacitance of a parallel plate capacitor is determined by the area of the plates, the distance between them, and the dielectric constant of the material between the plates. It is represented by the equation C = εA/d, where C is the capacitance, ε is the dielectric constant, A is the area of the plates, and d is the distance between them.

4. How is a parallel plate capacitor used in circuits?

A parallel plate capacitor is used in circuits to store electrical energy and to smooth out fluctuations in voltage. It can also be used in filters and frequency-selective circuits, as well as in timing circuits and oscillators.

5. What factors affect the capacitance of a parallel plate capacitor?

The capacitance of a parallel plate capacitor is affected by the area of the plates, the distance between them, and the type of dielectric material used. It also depends on the voltage applied and the temperature of the capacitor. Additionally, the shape and orientation of the plates can also impact the capacitance.

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