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I found this problem while self-studying Electricity and Magnetism, and I want to know if my solution is rigorous.

Show that the plates of a plane capacitor attract each other mutually with a force equal to

[tex]F=\frac{q^2}{2\epsilon_0 A}[/tex]

Obtain this result by calculating the work required to increase the separation between the plates from

Energy stored in a capacitor with electric field

[tex]U=\frac{1}{2}\epsilon_0 E^2 Ax[/tex]

The energy

[tex]U=\frac{1}{2}\epsilon_0 E^2 Ax[/tex]

The electric field

[tex]U=\frac{1}{2}\frac{q^2 x}{\epsilon_0 A}[/tex]

The work

[tex]dU=\vec{F}\cdot d\vec{x}[/tex]

where [itex]\vec{F}[/itex] is the force applied by the external agent.

Because in this situation force and displacement are in the same direction, it may be rewritten as:

[tex]dU=Fdx[/tex]

Isolating the force:

[tex]F=\frac{dU}{dx}[/tex]

Thus:

[tex]F=\frac{d}{dx}\frac{1}{2}\frac{q^2 x}{\epsilon_0 A}[/tex]

[tex]F=\frac{1}{2}\frac{q^2}{\epsilon_0 A}[/tex]

Thank you in advance.

## Homework Statement

Show that the plates of a plane capacitor attract each other mutually with a force equal to

[tex]F=\frac{q^2}{2\epsilon_0 A}[/tex]

Obtain this result by calculating the work required to increase the separation between the plates from

*x*to*x*+*dx*.## Homework Equations

Energy stored in a capacitor with electric field

*E*, plate area*A*and distance*x*between the plates:[tex]U=\frac{1}{2}\epsilon_0 E^2 Ax[/tex]

## The Attempt at a Solution

The energy

*U*stored in a capacitor can be viewed as the work that needs to be done by an external agent to separate its plates by a distance*x*:[tex]U=\frac{1}{2}\epsilon_0 E^2 Ax[/tex]

The electric field

*E*can be rewritten as [itex]q/(\epsilon_0 A)[/itex], where*q*is the magnitude of the charge on each plate. Thus, the work can be rewritten as:[tex]U=\frac{1}{2}\frac{q^2 x}{\epsilon_0 A}[/tex]

The work

*dU*done by the external agent to separate the plates from a separation*x*to a separation*x*+*dx*is:[tex]dU=\vec{F}\cdot d\vec{x}[/tex]

where [itex]\vec{F}[/itex] is the force applied by the external agent.

Because in this situation force and displacement are in the same direction, it may be rewritten as:

[tex]dU=Fdx[/tex]

Isolating the force:

[tex]F=\frac{dU}{dx}[/tex]

Thus:

[tex]F=\frac{d}{dx}\frac{1}{2}\frac{q^2 x}{\epsilon_0 A}[/tex]

[tex]F=\frac{1}{2}\frac{q^2}{\epsilon_0 A}[/tex]

Thank you in advance.

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