# Force to open a capacitor

Karol

## Homework Statement

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.

## Homework Equations

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

## 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?

Homework Helper

## Homework Statement

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.

## Homework Equations

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

## 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?
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}##.

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

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

Karol
Thanks