part 1: energy in an electric field
Consider the 2 plates of a capacitor, one charged positively and one charged negatively. If you move the upper plate a distance d you are doing work against the
electric field E between the plates. By moving the plate you are creating a new field . Your work is stored as electric energy inside this new field.
V 1/2 εo E2
work = force x distance = F d with F = Q E'. The charge Q moves between 2 region : 1 region with the electric field E
and 1 with no field (0) so the effective field is E' = E /2
so work = Q E/2 d with Q = σ /A and E =σ /εo
work = 1/2 E2εo A d with A d = volume where the energy is stored
so Energy per volume = 1/2 εo E2 units = J/m2
So we have a new way to compute energy. Before electrical potential energy = work to produce to bring charges together.
Now, if the electric field os known. you can find the integral over the volume of space.
U = integral ( 1/2 εo E2 dV )
For a capacitor, the field is constant between the plates so E = 1/2 εo E2 A d = 1/2 Q |VAB| with V voltage between plate A and plate B
part2 capacitance
By definition the ratio Q/V is the capacitance C . this physical quantity measures the capacity to store charges. As Cincrease, more charge
can be stored. C depends on the material and the structure of the capacitor. the unit is the farad/
example : A sphere of radius R has a potential of V = Q / (4pi εo R) so its capacitance is 4pi εo R
Let's compute the capacitance of spheres of different radius. (1/ (4pi εo) about 9 109 ). fill table. use 1 pF = 10-12 F
| capacitance (farad) | radius of sphere (m) |
| _________ | 9109 m (huge) |
| __________ | 6000 km (earth) |
| _________ | 0.3 m (van graaf) |
| _________ | 1cm (tiny) |
The capacitance of a capacitor depends on the difference of potential between the plate. It has to be smaller than
the capacitance of 1 plate alone or 1 sphere alone. (it is easier to charge 1 plate if there is another one, next to it, holding opposite charges).
the capacitance of a capacitor is C = |Q/VAB| where VAB is the difference of potential between the plates.
C=σ A / ( E d ) = σA/dE = εo A / d
Using this formula you can compute the capacity of a capacitor with plates 25m x 5m area and d=0.01mm (realsitic measures, the surface
area A is rolled up in a capacitor). C = _______ F = ______ microfarads.
To multiply the capacity by 1000, the distance between the plates can be divided by 1000.
A lot of energy can be stored in a capacitor U = 1/2 Q V = 1/2 C V2
Compute the energy stored in a capacitor C=1000 micro farad in serie with a generator of 100 V ?
U = ______ J.
This energy can be used to light a bulb in a a microsecond. We can use this set up for photography (flash of light)
or for stroboscope (to analyze motion)and to take picture of fast moving object. (bullet)
see stroboscope