Download PH504lec0809-6

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project

Document related concepts

Conservation of energy wikipedia , lookup

Internal energy wikipedia , lookup

Gibbs free energy wikipedia , lookup

Woodward effect wikipedia , lookup

Casimir effect wikipedia , lookup

Potential energy wikipedia , lookup

Electric charge wikipedia , lookup

Electrostatics wikipedia , lookup

Transcript
PH504
Capacitance; definition
In a conductor, electrons are free to roam. Any net charge resides
on the surface – the electric field is zero inside.
When an isolated, finite size conductor is given a charge Q, its
potential (with respect to a zero at infinity) is V. It can be shown
that for any body that Q is proportional to V and the constant of
proportionality is known as the capacitance (C) of the conductor.
C = Q/V
The capacitance can be thought of expressing the amount of
charge the conductor can carry for a given potential V.
Farad: The units of capacitance are the farad (symbol F).
The Farad, F, is the SI unit for capacitance, and from the definition of
capacitance is seen to be equal to a Coulomb/Volt.
The capacitance of a body is a property of its shape and size.
Example: It can be shown (see below) that a conducting sphere of
radius a and carrying a charge Q has a potential
.
Hence from C = Q / V 
, just depending on its size!
Note: for a = 1cm, C = .01/(9x109), this is about 1 pF.
Practical capacitors generally consist of two conductors, in
operation one carries a charge +Q the other a charge –Q. The
definition of capacitance is still C = Q / V but now V is the potential
difference between the bodies.
Calculating Capacitance
Place a charge +Q on one conductor and –Q on the other.
Find the potential difference between the conductors by
(a)
using a suitable equation for the potential appropriate to the
symmetry of the problem or
(b)
find the form of the E-field in the region between the two
conductors and then integrate E with respect to a suitable
spatial co-ordinate to find the potential difference.
The result will be an equation for V in terms of Q and the spatial
dimensions of the conductors. Finally use the definition C=Q/V to
find an expression for C.
Example 1: 2 Spheres (HyperPhysics website)
Spherical Capacitor
The capacitance for spherical or cylindrical conductors
can be obtained by evaluating the voltage difference
between the conductors for a given charge on each. By
applying Gauss' law to a charged conducting sphere,
the electric field outside it is found to be
The voltage between the two spheres can be found by integrating the electric
field along a radial line:
From the definition of capacitance, the capacitance is
Does an isolated charged sphere have capacitance?
Isolated Sphere
Capacitor?
An isolated charged conducting sphere has
capacitance. Applications for such a capacitor may not
be immediately evident, but it does illustrate that a
charged sphere has stored some energy as a result of
being charged. Taking the concentric sphere
capacitance expression:
and taking the limits
gives
Further confirmation of this comes from examining the potential of a charged
conducting sphere:
Example 2: Coaxial Cable
Consider a cylindrical surface of radius R.
Take a charge per unit length 
Then Gauss's Law yields
E 2r = o E = 2ro) for r > R.
E=0
for r < R
Suppose the permittivity within a coaxial cable is ko
k = relative permittivity of the dielectric material between the
plates.
k=1 for free space, k>1 for all media, approximately =1 for air.
Cylindrical
Capacitor
For a cylindrical geometry like a
coaxial cable, the capacitance is
usually stated as a capacitance per
unit length. The charge resides on
the outer surface of the inner
conductor and the inner wall of the
outer conductor. By applying
Gauss' law to an infinite cylinder,
the electric field outside a charged
cylinder is found to be
The voltage between the cylinders (in a vacuum)can be found by integrating
the electric field along a radial line:
The capacitance per unit length ( with rel. permittivity k) is:
Example 3 : 2 parallel plates, charge Q & -Q, area A, separation
d:
Apply Gauss’s law: E = Q/(oA)
- constant
Potential diffence: V = d Q/(oA)
and so
C = Ao/d
Parallel Plate Capacitor
The capacitance of flat, parallel metallic plates of area A and separation d is
given by the expression above where:
= permittivity of space and
Capacitance of Parallel Plates
The electric field between two large
parallel plates is given by
Show
The voltage difference between the two plates can be expressed in terms of the
work done on a positive test charge q when it moves from the positive to the
negative plate.
It then follows from the definition of capacitance that
Energy stored by a capacitor
In charging a capacitor from zero to a finite charge Q work must
be done (i.e. to remove electrons from the positive plate and
overcome the repulsion from the negative plate).
If at some point the charge is q and the potential is v (hence
q=Cv) then to add an additional charge dq work dW=vdq must be
done.
Hence total work in charging from 0 to Q is given by an integral:
or
This is the work done in charging the capacitor and hence also
equals the potential energy stored by a charged capacitor
assuming that zero potential energy corresponds to zero charge.
Energy Stored on a Capacitor
The energy stored on a capacitor
can be calculated from the
equivalent expressions:
This energy is stored in the
electric field.
From the definition of voltage as the energy per unit charge, one might expect that
the energy stored on this ideal capacitor would be just QV. That is, all the work
done on the charge in moving it from one plate to the other would appear as
energy stored. But in fact, the expression above shows that just half of that work
appears as energy stored in the capacitor.
For a finite resistance, one can show that half of the energy supplied by the
battery for the charging of the capacitor is dissipated as heat in the resistor,
regardless of the size of the resistor.
Stored energy in terms of the E-field
For a parallel plate capacitor we can show that
and
where C is the capacitance, A is the area of the plates, d their
separation and E is the E-field between the plates (the only
region where E is non-zero).
From above, the potential energy U is given by
where E has been used to eliminate Q.
The final result is simply (1/2)0E2x(volume between plates)
This result suggests a general one that the potential energy is
given by (1/2)0E2 multiplied by the volume over which E is nonzero or if E is not constant
Electric Field Energy in Capacitor
The energy stored on a
capacitor is in the form of
energy density in an electric
field is given by
This can be shown to be
consistent with the energy
stored in a charged parallel
plate capacitor
Capacitors
Capacitance is typified by a
parallel plate arrangement
and is defined in terms of
charge storage:
where


Q = magnitude of
charge stored on
each plate.
V = voltage applied
to the plates.
Capacitor Combinations
Capacitors in parallel add ... simply increasing Area A
C = C1 + C2 + …..
Capacitors in series combine as reciprocals ... since they share
the voltage V: Q/C = V = V1 + V2 + V3 + …..
Charge on Series Capacitors
Since charge cannot be added or taken away from the
conductor between series capacitors, the net charge
there remains zero.
You store less charge on series
capacitors than you would on either
one of them alone with the same
voltage!
Does it ever make sense to put capacitors in series? You get less
capacitance and less charge storage than with either alone. It is sometimes
done in electronics practice because capacitors have maximum working
voltages, and with two "600 volt maximum" capacitors in series, you can
increase the working voltage to 1200 volts.
Conclusions


Capacitance – definition and determination
Potential energy stored in a capacitor