Capacitor

Short Description of Topic

This page is all about the Electric Field due to a Point Charge.

Electric Field

[Two charged plates separated by very small gap s

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

The Algorithm

Step 1. Cut up the charge distribution into pieces and find the direction of [math]\displaystyle{ \Delta \vec{E} }[/math] at each location

Approximate electric field of a uniformly charged disk [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]

At location 2, midpoint between two disks, both disks contribute electric field in the same direction. Therefore, [math]\displaystyle{ \vec{E}_{net} }[/math] is the largest at this location.

At location 1, because negative charged plate is closer, [math]\displaystyle{ \vec{E}_{net} }[/math] is to the right.

At location 3, because positive charged plate is closer, [math]\displaystyle{ \vec{E}_{net} }[/math] is to the left.

Step 2. Find the electric field of each plate

Assumption: Ignore the electric field due to the small charges on the outer surface of the capacitor since it's very small; Assume that separation between capacitor is very small compared to radius or a disk; consider that location 1 and 3 are just near the disk

To make equation valid at all locations, choose origin at the inner face of the left disk so Electric field of the negative charged plate is [math]\displaystyle{ \ E_{-}=\frac{Q/A}{2\epsilon_0 }[1-\frac{s}{R}] }[/math] to the left and Electric field of the positive charged plate is [math]\displaystyle{ \ E_{+}=\frac{Q/A}{2\epsilon_0 }[1-\frac{s-z}{R}] }[/math] to the left

Step 3. Add up

Location 2: Therefore, the location 2, middle of the capacitor, is located z from the negative charged plate and s-z from the positive plate. Since they are in same direction, we simply add them to find [math]\displaystyle{ \vec{E}_{net} }[/math]

[math]\displaystyle{ \vec{E}_{net} = \ E_{-} + \ E_{+} =\frac{Q/A}{2\epsilon_0 }[1-\frac{z}{R}] + \frac{Q/A}{2\epsilon_0 }[1-\frac{s-z}{R}] = \frac{Q/A}{\epsilon_0 }[1-\frac{s/2}{R}] = \frac{Q/A}{\epsilon_0 } }[/math]

Location 1: At location 1, we can see that [math]\displaystyle{ \vec{E}_{net} = \ E_{-} - \ E_{+} = \frac{Q/A}{2\epsilon_0 }[1-\frac{z}{R}] - \frac{Q/A}{2\epsilon_0 }[1-\frac{z+s}{R}] = \frac{Q/A}{2\epsilon_0 }(s/R) }[/math]

Location 3: At location 1, we can see that [math]\displaystyle{ \vec{E}_{net} = \ E_{+} - \ E_{-} = \frac{Q/A}{2\epsilon_0 }[1-\frac{z-s}{R}] - \frac{Q/A}{2\epsilon_0 }[1-\frac{z}{R}] = \frac{Q/A}{2\epsilon_0 }(s/R) }[/math]

Step 4. Check Unit

[math]\displaystyle{ \frac{C/m^2}{N*m^2/C^2} = \frac{N}{C} }[/math]

[math]\displaystyle{ (\frac{C/m^2}{N*m^2/C^2})(m/m) = \frac{N}{C} }[/math]

References

Matter and Interactions, 4th Edition