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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 x to x + dx.
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 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.
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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