Charged Capacitor: Difference between revisions
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<br>Take the origin at the surface of the left plane, with the z-axis running to the right. We assume that each disk has a uniformly charge density (s <math>\ll</math> R). | <br>Take the origin at the surface of the left plane, with the z-axis running to the right. We assume that each disk has a uniformly charge density (s <math>\ll</math> R). | ||
<br><br>Then, the contribution of the negative capacitor is <math>E_- \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{z}{R}]</math> (to the left) and the positive capacitor is <math>E_+ \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{s-z}{R}]</math> (to the left). | <br><br>Then, the contribution of the negative capacitor is <math>E_- \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{z}{R}]</math> (to the left) and the positive capacitor is <math>E_+ \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{s-z}{R}]</math> (to the left). | ||
<br>If we add up the contributions, <math>E_{total} \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{z}{R}] + {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{s-z}{R}] \approx {\frac{Q/A}{{\epsilon}_0}} [1-\frac{s/2}{R}]</math>. Since s <math>\ll</math> R, <math>E \approx {\frac{Q/A}{{\epsilon}_0}}</math> | <br>If we add up the contributions, <math>E_{total} \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{z}{R}] + {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{s-z}{R}] \approx {\frac{Q/A}{{\epsilon}_0}} [1-\frac{s/2}{R}]</math>. Since s <math>\ll</math> R, <math>E \approx {\frac{Q/A}{{\epsilon}_0}}</math>. | ||
===A Computational Model=== | ===A Computational Model=== | ||
Revision as of 01:39, 17 April 2016
CLAIMED BY: GA HYUN OH
The Main Idea
A capacitor is when two uniformly, but oppositely (-Q and +Q), charged metal plates are held very close to each other with a separation of s.
This page is dedicated to understanding and calculating the electric field of a capacitor through definition, mathematical models, computational models, and example problems.
A Mathematical Model
The mathematical model to the electric field of a charged capacitor (near the center of the capacitor) is [math]\displaystyle{ E \approx {\frac{Q/A}{{\epsilon}_0}} }[/math], where Q is the magnitude of the plate charges and A is the area of each plates. The direction is perpendicular to the plates.
The fringe field (field located near the center of the disks but right outside of the plates) is [math]\displaystyle{ E_{fringe} \approx {\frac{Q/A}{2{\epsilon}_0}} (\frac{s}{R}) }[/math]
Derivation
Take the origin at the surface of the left plane, with the z-axis running to the right. We assume that each disk has a uniformly charge density (s [math]\displaystyle{ \ll }[/math] R).
Then, the contribution of the negative capacitor is [math]\displaystyle{ E_- \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{z}{R}] }[/math] (to the left) and the positive capacitor is [math]\displaystyle{ E_+ \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{s-z}{R}] }[/math] (to the left).
If we add up the contributions, [math]\displaystyle{ E_{total} \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{z}{R}] + {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{s-z}{R}] \approx {\frac{Q/A}{{\epsilon}_0}} [1-\frac{s/2}{R}] }[/math]. Since s [math]\displaystyle{ \ll }[/math] R, [math]\displaystyle{ E \approx {\frac{Q/A}{{\epsilon}_0}} }[/math].
A Computational Model
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