
CAPACITORS
and DIELECTRICS (INSULATORS)  Chapter 24.
Following Chapter
24 we cover (below) Section 26.4 ResistanceCapacitance (RC) Circuits.
KEY TERMS: capacitor,
capacitance, Farad, parallelplate capacitor, series connection, parallel
connection, equivalent capacitance, energy density, dielectric, dielectric
constant, polarization, permittivity.

A capacitor
can be "charged" and can store charge. When a capacitor
is being charged, negative charge is removed from one side of the
capacitor and placed onto the other, leaving one side with a negative
charge (q) and the other side with a positive charge (+q). The
net charge of the capacitor as a whole remains equal to zero.

Fig.
24.1 Any two conductors
insulated from one another
form a CAPACITOR. 

CAPACITOR
Fig. 24.2 (a) A charged
parallel plate capacitor.
(b) When the plate separa
tion is small compared to
the plate area, the fringing
of the electric field
at the edges is negligible.

The amount of
charge that can be placed on a capacitor is proportional to the
voltage pushing the charge onto the positive plate. The larger the
potential difference (voltage) between the plates, the larger the
charge on the plates:
Q
= C V
The constant of proportionality is called the "capacitance"
and is proportional to the area (A) of one of the plates and inversely
proportional to the separation between the plates (d):
C = e A / d
for a parallel plate capacitor,
where e is the permittivity
of the insulating material (or DIELECTRIC) between the plates.
Recall that
we used Gauss's Law to calculate the magnitude of the electric field
(E) between the plates of a charged capacitor:
E = s / eo
where the space between the plates is a vacuum.
Vab = E d,
so
E = Vab / d





The unit of
capacitance is called the Farad (F).
One Farad is equal to one Coulomb per Volt. ( F = C / V )
Capacitors, like
resistors, can be connected in a circuit in two ways, series or parallel
(or combinations of series and parallel).
The equivalent capacitance
of a series connection is
1 / Ceq = 1 / C1 +
1/ C2
(Just the opposite from resistors!)


Fig.
24.8 (a) Two capacitors in series and
(b) The equivalent capacitor. 

The equivalent capacitor
of a parallel connection is
Ceq = C1 + C2
(Just the opposite from resistors!)


Fig.
24.9 (a) Two capacitors in parallel and
(b) the equivalent circuit. 

Omit spherical and
cylindrical capacitors. We will discuss only parallel plate
capacitors this semester.
For an interesting
website (HyperPhysics) showing some nice pictures of capacitors and their
properties go to:
http://230nsc1.phyastr.gsu.edu/hbase/hframe.html
and click on Electricity
and Magnetism,
then Capacitors,
and then Parallel
Plate Capacitors.
24.3 ENERGY
STORAGE in CAPACITORS and ELECTRIC FIELD ENERGY
Capacitors
can store charge and ENERGY. We can calculate the potential
energy (U) stored in a charged capacitor from the ideas we learned
in the last chapter (Electric Potential). Since the charging
of a capacitor can be thought of as moving charge from one plate
DIRECTLY onto the other plate through a potential difference
of V, the voltage between the capacitor plates. Recall: DU
= q DV, and the potential V increases
as the charge is placed on the plates (V = Q / C). Since the
V changes as the Q is increased, we have to integrate over all
the little dq being added to a plate: DU
= q DV leads to
U = ò
V dq = ò
q/c dq = 1/C ò
q dq = Q^2 / 2C.
And using Q = C V, we get
U = Q^2
/ 2C = 0.5 C V^2 = 0.5 Q V
So the energy
stored in a capacitor can be thought of as the potential energy
stored in the system of positive charges that are separated
from the negative charges, much like a stretched spring has
potential energy associated with it.
ELECTRIC
FIELD ENERGY
Here's another
way to think of the energy stored in a charged capacitor. If
we consider the space between the plates to contain the energy
(equal to 1/2 C V^2) we can calculate an energy DENSITY (Joules
per volume). The volume between the plates is Area x distance
between plates, or A d. Then the energy density (u) is
u = 1/2
C V^2 / A d = 1/2 eo E^2
where we have used C = Eo A / d and V =E d.
This is an important
result because it tells us that
empty space can contain energy,
i.e., if there is an electric field in the "empty"
space.
If we can
get an electric field to travel (or propagate) through empty space
we can send or transmit energy!!!





Fig.
24.14 Effect of a dielectric between the plates of a parallel plate
capacitor. The electrometer measures the potential difference. (a)
With a given charge, the potential difference is Vo.
(b) With the same charge but with a dielectric between the plates,
the potential difference V is smaller than Vo. 
Most
capacitors have an insulating (or DIELECTRIC) material between the
plates. The presence of this dielectric INCREASES the capacitance
of the capacitor compared to when the space between the plates was
empty (a vacuum). Here's why: From measurements like the one above
we know that insertion of a dielectric between the plates of a capacitor
causes the potential difference Vo between the plates to decrease
to V. The original capacitance, from Q = C V, is given by Co = Q /
Vo and since Q must stay constant (it has nowhere to go!) we can write
Q = Q, or Co Vo = C V, and if Vo decreases to V, Co must increase
to C to keep the equation balanced. We can now define
K = C / Co = "dielectric constant" = the ratio
of the capacitances. And from Co Vo = C V (above) we can write
V = Vo Co / C = Vo / K. The dielectric constant (K)
is a positive number equal to 1 for a vacuum and greater than 1 for
other dielectric materials (Teflon  2.1, glass = 7, water = 80, etc.),
so we can say that the potential of a capacitor decreases by a
factor of K when a dielectric material is added and the charge
on the capacitor stays constant (i.e., the capacitor is
not connected to a battery while the dielectric is inserted). 

And since
V = E d, we can conclude that the electric field between the
plates must be reduced as a dielectric is inserted Q = (constant).
See below for a physical picture of what is happening.



Fig.
24.15 (a) Electric field lines with vacuum between the plates.
(b) The induced charges on the faces of the dielectric decrease
the electric field. (See figure below for a physical picture of
how the (blue) induced charges on the faces of the dielectric
are established.) 
The
electric field between the capacitor plates is reduced by the
presence of the dielectric because the induced surface charges
on the dielectric (see figure below) cause an electric field in
the opposite direction of the original field in the charged capacitor.
These fields tend to cancel each other resulting in a reduction
of the original field. 




Molecules in
the dielectric material have their positive and negative charges
separated slightly, causing the molecules to be oriented slightly
in the electric field of the charged capacitor.

Fig.
24.20 Polarization of a dielectric in an electric field gives rise
to thin layers of bound charges on the surfaces, creating positive
and negative surface charge densities. The sizes of the molecules
are greatly exaggerated for clarity. 



Fig.
24.21 (a) Electric filed of magnitude Eo between two charged plates.
(b) Introduction of a dielectric of dielectric constant K. (c) the
induced surface charges and their field (thinner lines). (d) Resultant
filed of magnitude Eo/K when a dielectric is between charged
plates. 

RC CIRCUITS (from Section 4, Chapter 26)
It
takes time to charge a capacitor and it takes time to discharge one.
This time is dependent on the sizes of the capacitor and the resistor
in the circuit. The product of RC has units of seconds.
The case for charging a capacitor is described first, then discharging
a capacitor. 





Fig.
26.21 CHARGING a capacitor. Just before the switch is closed the charge
on the capacitor is zero. When the switch is closed (at time t= 0)
the current jumps from zero to: i = e
/ R. 


For a CHARGING
capacitor,
I(t) = dq/ dt =
e / R exp ( t/RC) =
Io exp ( t/RC)
q(t) = Ce
[1  exp ( t/RC)] = Qf [1  exp (
 t/RC)]
Both these functions
are plotted to the left.

Fig.
26.22 CHARGING capacitor 



Fig. 26.23 DISCHARGING
a capacitor. Just before the switch is closed (at t= 0) the charge
on the capacitor is Qo, and the current is zero. At the time t after
the switch is closed, the charge on the capacitor is q, the current
is i. 


For a DISCHARGING
capacitor,
I(t) = dq/ DT =
 Qo / RC exp ( t/RC)
q(t) = Qo
exp ( t/RC)
Both these functions
are plotted to the left.

Fig.
26.24 DISCHARGING capacitor 

© 2009 J. F. Becker


