- #1

ananonanunes

- 18

- 6

- Homework Statement
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- Relevant Equations
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This circuit consis of two identical, parallel metal plates free to move, other than being connected to identical metal springs, a switch, and a battery with terminal voltage ##\Delta V##. With the switch open, the plates are uncharged, are separated by a distance ##d##, and have a capacitance of ##C##. When the switch is closed, the plates become charged and attract each other. The distance between the plates changes by a factor ##f##, after which the plates are in equilibrium between the spring forces and the attractive electric force between the plates. I am supposed to find spring constant ##k## in terms of ##C##, ##d##, ##\Delta V## and ##f##.

So what I did was equal the spring force to the electric force. The spring force I wrote as ##F_s=kdf## and the electric force as ##F_e=qE=q\frac{\Delta V}{d(1-f)}=\frac{C(\Delta V)^2}{d(1-f)}##. This gives a final result of ##k=\frac{C(\Delta V)^2}{d^2f(1-f)}##.

I know this is wrong, and that the correct answer is actually ##k=\frac{C(\Delta V)^2}{d^2f^2(1-f)}##. Since my answer is so similar to the correct one, I feel like maybe I'm just missing something simple, but I can't seem to understand where that extra ##f## could possibly come from.