Force between the plates of a plane capacitor

Click For Summary
SUMMARY

The discussion focuses on calculating the mutual attraction force between the plates of a plane capacitor, defined by the formula F = q²/(2ε₀A). The solution involves deriving this force from the work done to separate the plates, utilizing the energy stored in the capacitor, U = (1/2)ε₀E²Ax. The electric field E is expressed as q/(ε₀A), leading to the correct formulation of the force as F = (1/2)(q²/(ε₀A)). A participant identified an error in the initial substitution, which was subsequently corrected.

PREREQUISITES
  • Understanding of electric fields and forces in capacitors
  • Familiarity with the formula for energy stored in capacitors
  • Knowledge of calculus, specifically differentiation
  • Basic concepts of electrostatics and charge interactions
NEXT STEPS
  • Study the derivation of energy stored in capacitors in detail
  • Learn about the implications of electric field strength in capacitor design
  • Explore the relationship between charge, voltage, and capacitance
  • Investigate advanced topics in electrostatics, such as energy density
USEFUL FOR

Students and educators in physics, particularly those focusing on electricity and magnetism, as well as engineers involved in capacitor design and analysis.

pc2-brazil
Messages
198
Reaction score
3
I found this problem while self-studying Electricity and Magnetism, and I want to know if my solution is rigorous.

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.
 
Last edited:
  • Like
Likes   Reactions: gracy
Physics news on Phys.org
Looks like an error substituting in the q/(ϵ0A). Shouldn't you end up with an A on the bottom instead of on the top?
 
Delphi51 said:
Looks like an error substituting in the q/(ϵ0A). Shouldn't you end up with an A on the bottom instead of on the top?
Yes, there was an error; I've just corrected it.
 

Similar threads

Replies
6
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
11
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
5
Views
3K
Replies
3
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 4 ·
Replies
4
Views
999
  • · Replies 23 ·
Replies
23
Views
2K
Replies
1
Views
2K