Download Capacitors

Capacitors
We will use a parallel plate capacitor
for our discussions here. We already know that the
electric field between the plates is a uniform field
with equipotentials parallel to the plates and equally
spaced. The lines of force are perpendicular to
these, running straight from the positive plate to the
negative one.
When there is a potential difference of
ΔV between the plates, the potential and electric
field can be mathematically represented as
V = ΔV x/d where x is the distance from the negative plate.
E = ΔV/d anywhere between the plates and directed from the positive to the
negative plate.
We need to relate these to the charge on the plates. The charge density on the
plates, σ = q/A, increases with increasing potential, ΔV, and with decreasing
plate separation, d. Both these factors strengthen the field and provide the force
necessary to pack the charges together closer to increase the charge density.
So, we have
σ = q/A  ΔV/d
Rearranging gives
q = constant(AΔV/d)
It turns out that the amount of charge on the plate is also affected by the material
between the plates. Materials respond to electric fields in different ways.
A conductor placed in the field becomes a dipole
with charges accumulating at each end.
Insulating materials respond differently. To the
extent the material allows charges to move, they
respond similarly to conductors, but to a lesser
degree. Some of these materials, also called
dielectrics, have molecules with positive and
negative ends, called dipoles. Water is a good
example of this. These molecules can align with
the field to modify the effective field strength in
the material.
When the dielectric material has electric dipoles which can align, they strengthen
the field between the plates, increasing the
charge stored on the plates. The
parameter used to model this effect is the
dielectric constant, . This factor
multiplies the permittivity of free space, o,
which is 4 divided by the constant in the
Coulomb force law:
o = 4/K = 4/9E9
Putting this factor in as the constant in the capacitor equation,
q = o (AΔV/d)
Rearranging,
q = (o A/d) ΔV = C ΔV
The collection of variables called C above all involve characteristics of the
parallel plate capacitor.
(o A/d) = C
o
A
d
: property of the material between the plates.
: Area of a plate of the capacitor.
: the distance between the plates.
A physical capacitor, once constructed, will have a capacitance determined by
the geometry and choice of dielectric used to make the capacitor. Commercial
capacitors are labeled with the capacitance value.
The value of the dielectric constant, , of a material is often determined by
placing a measured amount of the material between the plates of a capacitor and
measuring its capacitance.
Energy:
Storing charge in a capacitor also stores energy. The charge on the plates has a
potential energy due to the repulsion between charges. This potential energy is
stored in the electric field. The amount of potential energy can be determined by
considering charge to have been moved from one plate to the other through the
potential difference ΔV.
As charge is separated, the potential difference grows from 0 to ΔV. To figure
the energy, use the average potential difference, ΔV/2, and the total charge, Q,
to give the total potential energy as
PE = Q ΔV/2 = ½ Q2/C = ½ C (ΔV)2 or ½ CV2
The second and third forms are obtained using the capacitor equation
Q = C ΔV to substitute for either Q or ΔV. The potential difference, ΔV, is often
written simply as V.
Adding capacitors
Two capacitors can be put in a circuit in two different ways; side-by-side (parallel)
and in line (series).
Capacitors wired in parallel effectively
increase the area of the plates, adding the
two areas. This gives a total capacitance
equal to the sum of the individual
capacitances.
Ctot = C1 + C2
Wired in series, the simplest approach is to
notice that the connection between the
capacitors, along with the plates connected
is completely insulated by the dielectrics, so
the total charge must remain 0. This
means that –Q on the plate of one capacitor
is the same amount of charge as +Q on the
plate of the other capacitor.
This means that the amount of charge on each
capacitor is the same.
We also know that the total potential
difference across both capacitors is equal to
the sum of the potential differences across
each capacitor.
Using the capacitor equation and putting it all together:
ΔV1 = Q/C1
ΔV2 = Q/C2
ΔV = Q/Ctot (same Q on outer plates),
Q/C1 + Q/C2 =
1/C1 + 1/C2 =
Q/Ctot
1/Ctot