Parallel plate capacitor problem

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Homework Help Overview

The discussion revolves around a parallel plate capacitor problem involving various configurations, including changes in plate spacing, charge, and dielectric materials. Participants are exploring the implications of these changes on the charge and voltage of the capacitor.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants are examining the relationships between charge, voltage, and capacitance using the formula Q = CV. There are attempts to calculate the charge after various changes to the capacitor's configuration. Questions arise regarding the validity of the charge calculations, particularly when the area of the plates is increased.

Discussion Status

Some participants have provided guidance on the principles governing charge conservation in capacitors, noting that the charge on the plates cannot change without an external influence. Multiple interpretations of the problem are being explored, particularly concerning the effects of changing plate area on charge density.

Contextual Notes

There is an assumption that the capacitor is initially disconnected from the battery when returning to its original spacing, which may affect the charge calculations. The dielectric constant of the fluid introduced is also a point of discussion, as it alters the voltage across the capacitor.

strawberrysk8
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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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Please show your work and where you get stuck so that we can help you.
 


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?
 


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).
 

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