# Homework Help: Force to open a capacitor

1. Dec 10, 2014

### Karol

1. The problem statement, all variables and given/known data
A parallel plates capacitor has charge Q and the plates are at distance x. they are separated incremental distance dx. what is the force F to separate them.

2. Relevant equations
The capacity: $C=\varepsilon\frac{A}{x}$
The work done to charge a capacitor: $W=\frac{1}{2}\frac{Q^2}{C}$

3. The attempt at a solution
The differential of the capacity:
$$dC=-\frac{\varepsilon A}{x^2}dx$$
The differential of the work:
$$dW=-\frac{Q^2}{2C^2}dC=\frac{\varepsilon A Q^2}{2C^2}\frac{1}{x^2}dx$$
$$F\cdot dx=dW=\frac{\varepsilon A Q^2}{2C^2}\frac{1}{x^2}dx$$
$$F=\frac{\varepsilon A Q^2}{2C^2}\frac{1}{x^2}$$
It should be $F=\frac{Q^2}{2\varepsilon A}$
Is it reasonable that the force depends on the distance x, as in my result?

2. Dec 10, 2014

### ehild

It is correct for an ideal capacitor.
The derivation could have been a simpler by using that the force between the plate is the negative gradient of the potential energy.
potential energy of the capacitor with charge Q: $U=-\frac{Q^2}{2C}=-\frac{Q^2x}{2 \varepsilon_0 A}$
$F=-\frac{dU}{dx}=\frac{Q^2}{2\varepsilon_0 A}$.

3. Dec 10, 2014

### Karol

Thanks, it's correct, but i was instructed this way by the book and i want to know why my way isn't correct.
Not only x is in my result, it's totally different with the constant's placement

4. Dec 10, 2014

### ehild

It is correct, but substitute the expression for C. x will cancel.

5. Dec 10, 2014

### Karol

Thanks

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