Calculating Capacitance of Isolated Capacitor with Conductor Inserted

AI Thread Summary
The discussion revolves around calculating the capacitance of an isolated capacitor after inserting a conductor that divides the space between the plates. The initial capacitance is given as 1 µF with a charge of 29 µC. The conductor reduces the effective distance between the plates, suggesting an increase in capacitance. However, the participants clarify that the configuration resembles capacitors in series rather than parallel, indicating the need to use the formula 1/C = 1/C1 + 1/C2 for the calculation. This insight is crucial for correctly determining the new capacitance value.
voelkner
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Homework Statement



An isolated capacitor with capacitance C = 1 µF has a charge Q = 29 µC on its plates

A conductor is inserted into the capacitor with thickness of the conductor is 1/3 the thickness of the capacitor and is centered in between the plates of the capacitor.

What is the capacitance of the capacitor with the conductor in place?

Homework Equations



C = Q/V

V = E*d

The Attempt at a Solution



I've been trying this problem for hours. I know that since the capacitor is a conductor it makes the distance between the two plates smaller which means that the capacitance should therefore increase. I thought that since we now had two distances that were each 1/3 the original distance the capacitance would increase by a factor of 6 however this answer does not work. Can anyone help me?!?
 
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voelkner said:
I thought that since we now had two distances that were each 1/3 the original distance the capacitance would increase by a factor of 6 however this answer does not work. Can anyone help me?!?

Hi voelkner! :smile:

6 = 3 + 3 … isn't that for capacitors in parallel?

these capacitors are in series. :smile:
 
Would it then be 1/C?
 
voelkner said:
Would it then be 1/C?

1/C = 1/C1 + 1/C2 :wink:
 
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