Capacitance of a variable capacitor

[astenroo]

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?

 

[cupid.callin]

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

 

[astenroo]

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?

 

[cupid.callin]

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"

 

[astenroo]

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"

Well, if they are in parallell, then the total capacitance will be the sum of each "sub-capacitor".

 

[gneill]

Find the total area of overlap for + and - faces. Then do the obvious.

 

[astenroo]

Find the total area of overlap for + and - faces. Then do the obvious.

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.

 

[gneill]

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.

 

[cupid.callin]

Yes they do have fields till they have some effective area acting as capacitor

 

[astenroo]

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.