Energy Density and Electric Field: Difference between revisions

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<math>F=Q\frac{Q/A}{2\epsilon_0} </math>
<math>F=Q\frac{Q/A}{2\epsilon_0} </math>


Now, suppose that you pull the positive plate of the capacitor slowly away from the negative plate, exerting a force that is just barely larger than the one exerted by the other plate. This work that you do, moving the plate a distance \deltas, increases the potential energy.
Now, suppose that you pull the positive plate of the capacitor slowly away from the negative plate, exerting a force that is just barely larger than the one exerted by the other plate. This work that you do, moving the plate a distance Δs, increases the potential energy:
 
<math>ΔU=W=Q\frac{Q/A}{2\epsilon_0}Δs</math>
 
For the sake of simplification, working out the math, we get this representation:
 
<math>ΔU=\frac{1}{2}\epsilon_0{\frac{Q/A}{\epsilon_0}^2}AΔs

Revision as of 17:28, 30 November 2015

claimed by Samir Nileshwar

Main Idea

This section takes an alternative view and treats electric fields as if they have energy stored in them. To learn how this works, consider moving one plate of a capacitor. For a very small gap, the force of one capacitor plate on another capacitor plate is charge Q times the field made by the other plate:

[math]\displaystyle{ E_{one plate}=\frac{Q/A}{2\epsilon_0 } }[/math]

Then the force on one plate is:

[math]\displaystyle{ F=Q\frac{Q/A}{2\epsilon_0} }[/math]

Now, suppose that you pull the positive plate of the capacitor slowly away from the negative plate, exerting a force that is just barely larger than the one exerted by the other plate. This work that you do, moving the plate a distance Δs, increases the potential energy:

[math]\displaystyle{ ΔU=W=Q\frac{Q/A}{2\epsilon_0}Δs }[/math]

For the sake of simplification, working out the math, we get this representation:

<math>ΔU=\frac{1}{2}\epsilon_0{\frac{Q/A}{\epsilon_0}^2}AΔs