Variable Capacitance of a Radio Dial

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The discussion revolves around calculating the charge on a variable capacitor after it is disconnected from a battery and its capacitance is altered. Initially, the capacitor's capacitance ranges from 100 to 350 pF, and it is charged to 130 V before being disconnected. The key point is that charge is conserved when the dial is turned, meaning the charge at 180° remains the same when the capacitance is reduced to 100 pF at 0°. Participants emphasize the importance of recognizing that while energy is conserved, the work done in adjusting the dial affects the system. Ultimately, the correct approach involves using the relationship Q=CV to find the charge at the new capacitance setting.
matthew1991
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Homework Statement



The capacitance of the variable capacitor of a radio can be changed from 100 to 350 pF by turning the dial from 0° to 180°. With the dial set at 180°, the capacitor is connected to a 130 V battery. After charging, the capacitor is disconnected from the battery and the dial is turned to 0°. What is the charge on the capacitor now?
K air= 1.000
K polystyrene= 2.3

Homework Equations



Q=C/V
V/V(sub)0=K
C=kCsub0=k*Epsilon naught*(A/D)
E=Q^2/2C=V^2*C/2=Q*V/2

The Attempt at a Solution



I tried using conservation of energy to find charge by using V^2*C/2=Q6^2/2C, but it was't the right answer.
there is a hint given that says think about what quantity is conserved. I thought energy was always conserved?
 
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matthew1991 said:
I tried using conservation of energy to find charge by using V^2*C/2=Q6^2/2C, but it was't the right answer.
there is a hint given that says think about what quantity is conserved. I thought energy was always conserved?

Energy is always conserved, but did you account for the work you had to do in order to turn the dial?

The hint was a good one though. Question: where can the charge go when you turn the dial if the capacitor is disconnected?
 
I made this way more complicated than it needed to be. Charge is always conserved, so Q=CV at 180 deg is the same Q still at 0 deg. I had pF as E-9 instead of E-12
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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