Parallel plate capacitor problem

In summary, the conversation discusses a parallel plate capacitor with changing plate spacing and dielectric material. The initial charge on the positive plate is 10.5 nC, but it increases to 16.2 nC when connected to a 1.51 V battery. When disconnected, the charge remains the same but the voltage increases to 16.7 V. When the dielectric material is changed, the voltage decreases to 3.297 V. Finally, when the area of overlap increases, the charge decreases but remains the same.
  • #1
strawberrysk8
27
0
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?
 
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  • #2


Please show your work and where you get stuck so that we can help you.
 
  • #3


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?
 
  • #4


Trick question. The charge on the plates can't change without an external source to drive it. The charge stays the same when the area increases (although the charge density decreases since the area increases).
 

Related to Parallel plate capacitor problem

What is a parallel plate capacitor?

A parallel plate capacitor is an electronic component that consists of two parallel conductive plates separated by a dielectric material. It stores electrical energy in the form of an electric field between the plates.

How do you calculate the capacitance of a parallel plate capacitor?

The capacitance of a parallel plate capacitor can be calculated using the formula C = ε0A/d, where C is the capacitance, ε0 is the permittivity of free space, A is the area of the plates, and d is the distance between the plates. Alternatively, it can also be calculated using the formula C = Q/V, where Q is the charge on the plates and V is the potential difference between the plates.

What is the relationship between capacitance and plate area in a parallel plate capacitor?

The capacitance of a parallel plate capacitor is directly proportional to the area of the plates. This means that as the area of the plates increases, the capacitance also increases, and vice versa.

What is the effect of changing the distance between the plates in a parallel plate capacitor?

The distance between the plates in a parallel plate capacitor has an inverse relationship with the capacitance. This means that as the distance between the plates decreases, the capacitance increases, and vice versa. This is because a smaller distance between the plates allows for a stronger electric field and therefore a higher capacitance.

How does the dielectric material between the plates affect the capacitance in a parallel plate capacitor?

The dielectric material between the plates in a parallel plate capacitor affects the capacitance by increasing it. This is because the dielectric material has a higher permittivity than air, which allows for a stronger electric field and therefore a higher capacitance.

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