Download Part 2 Set 1

Electrical Energies
Just as Newton’s Laws worked completely,
but were difficult, so to, working with
Electric Forces will be difficult.
Just as with gravitation, in electricity we can
solve many problems using the
Conservation of Energy, a scalar
equation that does not involve time or
direction. This requires that we find an
expression for the electric energy.
Electric Potential Energy
Since Coulomb’s Law has the same form as
Newton’s Law of Gravity, we will get a
very similar formula for electric potential
energy: DPE = -  (kq1q2/r122) dr and
choosing rstandard = infinity, gives:
PEel = k q1 q2 / r12
Recall for gravity, PEgr = - G m1 m2 / r12 .
Note that the PEelectric does NOT have a
minus sign. This is because two like
charges repel instead of attract as in gravity.
Voltage
Just like we did with forces on particles to get
fields in space,
(Edue to 1 = Fon 2/ q2)
we can define an electric voltage in space (a
scalar):
Vdue to 1 = PEof 2 / q2 .
We often use this definition this way:
PEof 2 = q2 * Vdue to 1 .
Units
The unit for voltage is, from the definition:
Vdue to 1 = PEof 2 / q2
volt = Joule / Coulomb .
Note that voltage, like field, exists in space,
while energy, like force, is associated with
a particle!
Voltage due to a point charge
Since the potential energy of one charge due
to another charge is:
PEel = k q1 q2 / r12
and since voltage is defined to be:
Vdue to 1 = PEof 2 / q2
we can find a nice formula for the voltage in
space due to a single charge:
Vdue to 1 = k q1 / r12 .
Voltages due to several point
charges
Since voltage, like energy, is a scalar, we can
simply add the voltages created by
individual point charges at any point in
space to find the total voltage at that point
in space:
Vtotal = S k qi / r1i .
For continuous distribution of charges, we can
integrate: Vtotal =  k dq / r .
Voltages and Electric Fields
Just like force and work, and hence potential
energy, are related:
D PE = - W = - F Ds,
so too are electric field and voltage:
D V = - E Ds .
Note that voltage changes only in the
direction of electric field. This also means
that there is no electric field in directions
in which the voltage is constant.
Voltage and Field
D V = - E Ds , or Ex = -DV / Dx .
Note also the minus sign means that electric
field goes from high voltage towards low
voltage.
Note also that this means that
positive charges will tend to “fall” from high
voltage to low voltage, but that
negative charges will tend to “rise” from low
voltage to high voltage!
Voltage and Field
D V = - E Ds , or Ex = -DV / Dx .
Note that the units of electric field are (from its
definition: E = F/q) Nt/Coul.
But from the above relation, they are
equivalently Volts/m.
Hence: Nt/Coul = Volt/m.
Voltage, Field and Energy
The Computer Homework, Vol 3, #3, has an
Introduction and problems concerning these
ideas that relate voltage to field
D V = - E Ds
and voltage to energy
PEof 2 = q2 * Vat 1
for use with the Conservation of Energy Law.
Review
F1on2 = k q1 q2 / r122 PE12 = k q1 q2 / r12
Fon2 = q2 Eat2
PEof2 = q2 Vat2
Eat2 = k q1 / r122
Vat2 = k q1 / r12
use in
use in
S F = ma
KEi + PEi = KEf +PEf +Elost
VECTOR
scalar
Ex = -DV / Dx
Energy example
Through how many volts will a proton have to
be accelerated if it is to reach a million
miles per hour?
DV = ?
qproton = 1.6 x 10-19 Coul
mproton = 1.67 x 10-27 kg
vi = 0 m/s
vf = 1 x 106 mph * (1 m/s / 2.24 mph) =
4.46x105 m/s .
Example, cont.
We can use the Conservation of Energy
including the formulas for kinetic energy
and potential energy:
KEi + PEi = KEf + PEf + Elost , where
KE = (1/2)mv2 and PE = qV:
(1/2)mpvi2 + qpVi = (1/2)mpvf2 + qpVf + Elost
Since vi=0 and Elost=0, and bringing qpVf to the left
side, we have: qp(Vi-Vf) = (1/2)mpvf2 .
Example, cont.
qp(Vi-Vf) = (1/2)mpvf2
We note that (Vi-Vf) = -DV since the change
is normally final minus initial. Thus,
DV = -(1/2)mpvf2 / qp =
-(1/2)(1.67x10-27kg)(4.46x105m/s)2 / 1.6x10-19 C
= -1040 volts.
We see that the proton must fall down (DV is
negative) through 1040 volts to reach a
million miles/hour.
Voltages and Fields
We’ve already seen that one way of
calculating a voltage at a location is to add
up all the voltages at that point due to all the
point charges.
However, another way is to use what we’ve
already done with fields, and use the
relation between voltage and field:
Ex = -DV / Dx .
Voltages from Fields
Inside a hollow sphere of charge, we saw
using Gauss’s Law that the Electric Field
was zero. From Ex = -DV / Dx , with E=0
we see that DV = 0, and so the voltage is
constant.
Voltages and Fields
Outside a hollow sphere of charge, we saw
using Gauss’s Law that the Electric Field
was the same as that due to a point charge.
Thus, the voltage outside of a point charge
is simply: V = kq/r (assuming V=0 at
r=infinity).
Voltages and Fields
For a line of charge, or outside a cylinder of
charge, the Electric Field was (from
Gauss’s Law): Eline = 2 k l / r .
Using Ex = -DV / Dx (or DV = -  E ds )
Vat 2 - Vat 1 = 2 k l ln(r1/r2) .
Voltages and Fields
Between parallel plates, E = s/e = constant,
using Ex = -DV / Dx (or DV = -  E ds )
gives: E = -DV / Dx = -V / d .
Change in Voltage
sphere, cylinder and plate
Change in Voltage
0.00
0
0.02
0.04
0.06
0.08
0.1
0.12
0.14
-2.00
change in voltage
-4.00
Sphere
Cylinder
-6.00
Plates
-8.00
-10.00
-12.00
radius or distance
Voltage, Force and Energy
The Computer Homework, Vol. 3, #4, on Electric
Deflection, provides problems involving energy (PE
= qV) and force (F = qE, where Ey = -DV/Dy ).
This situation is what occurs in an old-style TV or
computer monitor (CRT).
There is a supplementary ppt slide set that helps set up
this problem that is accessed from the course web
page.
Electric Circuits
Now that we have the concept of voltage, we
can use this concept to understand electric
circuits.
Just like we can use pipes to carry water, we
can use wires to carry electricity. The flow
of water through pipes is caused by pressure
differences, and the flow is measured by
volume of water per time.
Electric Circuits
In electricity, the concept of voltage will be
like pressure. Water flows from high
pressure to low pressure; electricity flows
from high voltage to low voltage.
But what flows in electricity? Charges!
How do we measure this flow? By Current:
current = I = Dq / Dt
UNITS: Amp(ere) = Coulomb / second
Circuit Elements
In this second part of the course we will
consider two of the common circuit
elements:
capacitor
resistor
The capacitor is an element that stores charge
for use later (somewhat like a water tower).
The resistor is an element that “resists” the
flow of electricity.
Capacitance
A water tower holds water. A capacitor holds
charge.
The pressure at the base of the water tower
depends on the height (and hence the
amount) of the water. The voltage across a
capacitor depends on the amount of charge
held by the capacitor.
Capacitance
We define capacitance as the amount of charge
stored per volt: C = Qstored / DV.
UNITS: Farad = Coulomb / Volt
Just as the capacity of a water tower depends on
the size and shape, so the capacitance of a
capacitor depends on its size and shape. Just
as a big water tower can contain more water
per foot (or per unit pressure), so a big
capacitor can store more charge per volt.
Parallel Plate Capacitor
For a parallel plate capacitor, we can pull
charge from one plate (leaving a -Q on that
plate) and deposit it on the other plate
(leaving a +Q on that plate). Because of the
charge separation, we have a voltage
difference between the plates, DV. The
harder we pull (the more voltage across the two
plates), the more charge we pull: C = Q /DV.
Note that C is NOT CHANGED by either Q
or DV; instead, C RELATES Q and DV!
Parallel Plate Capacitor
C = Q/V. For a parallel plate capacitor, the
charge is related to the electric field, and the
electric field is related to the voltage. We
can then relate the charge to the voltage:
E = s/eo where s=Q/A; and E = V/d. Recall
also that k = 1/[4peo]. This leads to:
V = Ed = Qd/eoA. Thus C = Q/V becomes
C = Q/V = Q / (Qd/eoA) = eoA/d .
Parallel Plate Capacitor
For this parallel plate capacitor, the
capacitance is related to charge and voltage
(C = Q/V), but the actual capacitance
depends on the size and shape:
Cparallel plate = A / (4 p k d)
where A is the area of each plate, d is the
distance between the plates, and k is
Coulomb’s constant (9 x 109 Nt-m2 / Coul2).
Example
What is the capacitance of a parallel plate
capacitor where each plate is 5 cm by 5 cm
in size, the two plates are separated by 2
mm with vacuum between the plates, and
there is 20 volts difference between the
plates?
Examples: Parallel Plate
Capacitor
The capacitance depends on A, k and d:
Cparallel plate = A / (4 p k d)
where A = 5 cm x 5 cm = 25 cm2 = 25 x 10-4
m2, d = 2 mm = 2 x 10-3 m, and k = 9 x 109
Nt-m2/Coul2 , so C =
(1) * (25 x 10-4 m2) / [4 * 3.14 * 9 x 109 Nt-m2/Coul2
* 2 x 10-3 m] = 1.10 x 10-11 F = 11 pF .
Examples: Parallel Plate
Capacitor
Thus,
one Farad would be a HUGE capacitance!
If we have a DV = 20 volts, then to calculate
the charge, Q, we can use: C = Q/V to get:
Q = C*V = 11 x 10-12 F * 20 volts =
2.2 x 10-10 Coul = 0.22 nCoul = 220 pCoul.
Note that we often drop the D in front of the V
since we often are concerned by the change in
voltage rather than the absolute value of the
voltage - just as we do when we talk about height!
Capacitance
Note that if we doubled the voltage, we
would not do anything to the capacitance.
Instead, we would double the charge
stored on the capacitor.
However, if we try to overfill the capacitor by
placing too much voltage across it, the
positive and negative plates will attract each
other so strongly, they will spark across the
gap and destroy the capacitor. Thus
capacitors have a maximum voltage!
Another Example
Consider a coaxial cable used as a capacitor.
Charges are removed from the inner wire
and deposited on the outer cylinder. What
is the capacitance of this cable if the cable is
2 meters long, the wire has a radius of 2 mm
and the cylinder has a larger radius of 1 cm
(10 mm) ?
Example #2 (cont.)
Start with the definition of capacitance:
C = Q/V . Then relate the voltage to the
charge that has been moved. In the coaxial
case, we know that the electric field
between the wire and the cylinder is that
due to the wire itself: E = 2kl/r, where
l=Q/L . From this we can get the voltage,
V = -Edr = - (2kl/r)dr = 2k(Q/L) ln(rcyl/rwire).
Example #2 (cont.)
Therefore, C = Q/V =
C = Q / [2k(Q/L) ln(rcyl/rwire)] , or
Ccoax = L / [2k ln(rcyl/rwire)].
Note that C does NOT depend on the Q or the V
(it does relate them), but rather depends on
how the capacitor is made (L, rcyl, and rwire).
Example 2 (cont.)
Ccoax = L / [2k ln(rcyl/rwire)].
In this case, we are given: L = 2 meters;
rcyl = 1 cm = 10 mm; rwire = 2 mm.
Thus, C = 2 m / [2 x 9 x 109 Nt-m2/Coul2 x
ln(10 mm / 2 mm)] =
6.9 x 10-11 Coul2/Nt-m
(UNITS: Nt-m = Joule, and a Joule/Coul = Volt, so
units are Coul/Volt = Farad. C = 69 pF .
Energy Storage
If a capacitor stores charge and carries
voltage, it also stores the energy it took to
separate the charge.
Note that we move the charge piece by piece
from one plate to the next, and as each
charge is moved, the voltage across the
plates increases. Note that the whole charge
is not moved across the whole voltage!
Gravitational analogy
This is like digging a hole: the first shovel of
dirt (charge) is lifted (raised in voltage) only
a slight amount. The next shovel of dirt is
deeper and it must be lifted higher. Only
the final shovel of dirt goes the whole
distance between the bottom of the hole and
the top of the hill.
Energy Stored
Starting from DPE = q DV, and adding each
contribution of each charge, we have for the
total stored energy:
Estored = S qi Vi =  V dq =  (q/C) dq =
(1/2)Q2/C (or using C = Q/V)
Estored = (1/2)QV = (1/2)CV2 .
Energy Storage
Note that previously we had:
PE = q*V ,
and now for a capacitor we have:
E = (1/2)*Q*V .
Why the 1/2 factor for a capacitor?
Energy Storage
The reason is that in charging a capacitor, the first bit
of charge is transferred while there is very little
voltage on the capacitor (recall that the charge
separation creates the voltage!). Only the last bit of
charge is moved across the full voltage. Thus, on
average, the full charge moves across only half the
voltage!
But the battery moves the full charge with the full
voltage, so where does that other half of the PE go?
It goes into heat in the resistance in the circuit!
Effects of Materials
Our derivation for the capacitance of parallel
plates assumed we had vacuum between the
plates. Would placing some material
between the plates affect the capacitance?
To answer that question, we need to consider
what material is made of, and how it reacts
to the presence of electric fields.
Effects of Materials
Materials are composed of atoms and molecules.
Atoms and molecules are composed of charged
parts (positive nucleus and negative electrons,
positive and negative ions). When placed in an
electric field, the charged particles are going to be
pulled apart - the negative charges pulled towards
the positive plate, the positive charges pulled
towards the negative plate.
+Q
-q +q
-Q
Effects of Materials
The net effect of this pulling is to create a separation
of charges. This separation of charges will create
an opposing electric field. This opposing electric
field will tend to reduce the total field between the
plates, and hence reduce the voltage between the
plates (since E = -DV/Ds). This reduced voltage
(for the same charge on the plates) will increase
the capacitance (C=Q/V) !
+Q
-Q
-q
-q
-q
-q
+q
+q
+q
+q
Dielectric Constant
Some materials are more “stretchable” than
others, and so will cause the capacitance to
increase more than others. Vacuum can not
stretch (no charges to stretch!), so it will
provide the smallest possible capacitance.
To describe this “stretchiness”, we define a
dielectric constant, K, such that
K = Cwith material / Cwith vacuum , or
Cwith material = K Cwith vacuum .
(Be sure to distinguish between k and K!)
Capacitor terminals
For a parallel plate capacitor with vacuum
between the plates, it does not matter which
plate is connected to the positive and which
to the negative voltage.
However, for large capacitors, a dielectric
material is used that “stretches” better one
way than another. This means that some
capacitors will specify which terminal
should be connected to the positive and
which to the negative.
Capacitance and Max V
If we apply too much voltage across the
plates, the charges will stretch to the point of
breaking – which produces lightning in the
capacitor which will destroy the capacitor.
Hence capacitors will have a maximum
voltage listed in their properties along with
their capacitance.