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Homework Statement
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The figure [attached] below shows a variable "air gap" capacitor for manual tuning. Alternate plates are connected together; one group of plates is fixed in position, and the other group is capable of rotation. Consider a capacitor of n = 4 plates of alternating polarity, each plate having area A = 1.60 cm^{2} and separated from adjacent plates by distance d = 4.35 mm. What is the maximum capacitance of the device?
Homework Equations
$$C=\frac {ε_0A} d$$
$$\frac 1 {C_{eq}}=\sum_{j=1}^n \frac 1 {C_j}$$
The Attempt at a Solution
I'm not entirely sure if the capability of the green plates to rotate plays a role here so I just ignored it. I treated each green/blue pair of plates as an individual capacitor.
After that, I plugged in the appropriate values. $$C= \frac {\left( 8.85 \cdot 10^{12} \right) \left( π \frac {.00016^2} 2 \right)} {4.35 \cdot 10^{3}}=8.18 \cdot 10^{17}$$
The capacitors are connected in series, therefore
$$\frac 1 {C_{eq}}=\frac 4 {8.18 \cdot 10^{17}}$$
$$C_{eq}=\frac {8.18 \cdot 10^{17}} 4=2.045 \cdot 10^{17}$$
This is actually a problem from an old assignment so I can't see if it's correct. I have a feeling this is all wrong. Any help would be appreciated!
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