Capacitor: Difference between revisions
No edit summary |
|||
| Line 22: | Line 22: | ||
<math>\ E=\frac{Q/A}{\epsilon_0 }</math> One plate has charge <math>\ +Q</math> and other plate has charge <math>\ -Q</math>; each plate has area A; Direction is perpendicular to the plates. Assumption: separation between capacitor is very small compared to the area of a plate. | <math>\ E=\frac{Q/A}{\epsilon_0 }</math> One plate has charge <math>\ +Q</math> and other plate has charge <math>\ -Q</math>; each plate has area A; Direction is perpendicular to the plates. Assumption: separation between capacitor is very small compared to the area of a plate. | ||
| Line 29: | Line 28: | ||
<math>\ E_{fringe}=\frac{Q/A}{2\epsilon_0 }(\frac{s}{R})</math> <math>\ s</math> is the separation between plates; <math>\ R</math> is the radius of plate | <math>\ E_{fringe}=\frac{Q/A}{2\epsilon_0 }(\frac{s}{R})</math> <math>\ s</math> is the separation between plates; <math>\ R</math> is the radius of plate | ||
Step 1. Cut up the charge distribution into pieces and draw <math> \Delta \vec{E}</math> | |||
Approximate electric field of a uniformly charged disk is <math>\ E=\frac{Q/A}{2\epsilon_0 }[1-\frac{s}{R})]</math> or <math>\ E=\frac{Q/A}{2\epsilon_0 }</math> | |||
==Examples== | ==Examples== | ||
Revision as of 16:26, 19 November 2015
Short Description of Topic
This page is all about the Electric Field due to a Point Charge.
Electric Field
Electric Field of two uniformly charged disks: A Capacitor
The Electric Field of a Capacitor can be found by the formula:
Electric field near the center of a two-plate capacitor
[math]\displaystyle{ \ E=\frac{Q/A}{\epsilon_0 } }[/math] One plate has charge [math]\displaystyle{ \ +Q }[/math] and other plate has charge [math]\displaystyle{ \ -Q }[/math]; each plate has area A; Direction is perpendicular to the plates. Assumption: separation between capacitor is very small compared to the area of a plate.
Fringe Field (just outside the plates near center of disk)
[math]\displaystyle{ \ E_{fringe}=\frac{Q/A}{2\epsilon_0 }(\frac{s}{R}) }[/math] [math]\displaystyle{ \ s }[/math] is the separation between plates; [math]\displaystyle{ \ R }[/math] is the radius of plate
Step 1. Cut up the charge distribution into pieces and draw [math]\displaystyle{ \Delta \vec{E} }[/math]
Approximate electric field of a uniformly charged disk is [math]\displaystyle{ \ E=\frac{Q/A}{2\epsilon_0 }[1-\frac{s}{R})] }[/math] or [math]\displaystyle{ \ E=\frac{Q/A}{2\epsilon_0 } }[/math]