Energy Density and Electric Field: Difference between revisions
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<math>ΔU=\frac{1}{2}\epsilon_0{(\frac{Q/A}{\epsilon_0})^2}AΔs </math> | <math>ΔU=\frac{1}{2}\epsilon_0{(\frac{Q/A}{\epsilon_0})^2}AΔs </math> | ||
The expression in parenthesis that we are squaring is the same as the electric field inside the capacitor. Substituting, we get: | The expression in parenthesis that we are squaring is the same as the electric field inside the capacitor. Substituting, we get: Field Energy Density= | ||
<math> | <math>\frac{ΔU}{Δ(volume)}=\frac{1}{2}\epsilon_0E^2 </math> |
Revision as of 17:34, 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]\displaystyle{ ΔU=\frac{1}{2}\epsilon_0{(\frac{Q/A}{\epsilon_0})^2}AΔs }[/math]
The expression in parenthesis that we are squaring is the same as the electric field inside the capacitor. Substituting, we get: Field Energy Density= [math]\displaystyle{ \frac{ΔU}{Δ(volume)}=\frac{1}{2}\epsilon_0E^2 }[/math]