Capacitance of a variable capacitor

AI Thread Summary
The discussion focuses on calculating the capacitance of a variable capacitor made of seven semicircular plates. Participants confirm that the capacitor can be treated as multiple capacitors in parallel, with each pair of plates acting as separate capacitors. The effective area of the plates changes with rotation, affecting the capacitance. It is established that all facing surfaces with opposite charges contribute to the electric field, confirming the presence of a field between adjacent plates. The conversation concludes with appreciation for the assistance provided in understanding the problem.
astenroo
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Homework Statement


Hi again!

I've been handed an assignment in which we are to find the capacitance of a variable capacitor consisting of seven semicircular plates with a radius of 2 cm spaced 1 mm from each other. Every other plate is connected to a common +terminal, and every other to a common -terminal. We are to find the capacitance at different angles.


Homework Equations



Area of a circle sector is A=\frac{\alpha}{360\circ}\pi r^{2}
The capacitance C=\frac{\epsilon_{0}A}{d}
\alpha = angle
and d=distance between plates
C for parallell coupling is C=C1+C2+C3+C4+C5+C6

The Attempt at a Solution


Now, my intuition tells me that this variable capacitor could be treated as six capacitors in parallell. Is my intuition right or did I make serious blunder? Calculating the areas is straight forward, and iff it can be treated as a parallell coupling of capacitors, then that part is also straight forward. My actual problem statement is, can a variabel capacitor be treated as parallell coupled capacitor?
 
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well i couldn't properly understand what is your question but i guess as the plates are semicircular .. rotating them along the center will change effective area, so A dec and therefore C also dec.

So you can alter the capacitance
 
cupid.callin said:
well i couldn't properly understand what is your question but i guess as the plates are semicircular .. rotating them along the center will change effective area, so A dec and therefore C also dec.

So you can alter the capacitance

Yes I understand how the area affects capacitance. But, in a variable capacitor, can each pair of "+- plates" be considered as a separate capacitor? If so, then they are in parallell, aren't they?
 
astenroo said:
Yes I understand how the area affects capacitance. But, in a variable capacitor, can each pair of "+- plates" be considered as a separate capacitor? If so, then they are in parallell, aren't they?

Yes ... why not ... and will the fact that they are in parallal affect them?

check this: http://en.wikipedia.org/wiki/Variable_capacitor#Mechanically_controlled"
 
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Find the total area of overlap for + and - faces. Then do the obvious.:wink:
 
gneill said:
Find the total area of overlap for + and - faces. Then do the obvious.:wink:

Ok, and then I suddenly began thinking:

Let's say that plates 1, 3, 5 and 7 are + and 2, 4 and 6 are -. Do I have a field between plates 1&2 and 2&3 etc? I believe I have a field, but I am not completely sure.
 
astenroo said:
Ok, and then I suddenly began thinking:

Let's say that plates 1, 3, 5 and 7 are + and 2, 4 and 6 are -. Do I have a field between plates 1&2 and 2&3 etc? I believe I have a field, but I am not completely sure.

All facing surfaces with opposite charges count. That includes both sides of a single plate if they're both facing oppositely charged other plates.
 
Yes they do have fields till they have some effective area acting as capacitor
 
  • #10
gneill said:
All facing surfaces with opposite charges count. That includes both sides of a single plate if they're both facing oppositely charged other plates.

Thank you very much for your help. Goes for you as well cupid.
 
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