Download q 2 - UMN Physics home

Summary of some properties of electrical forces.
They can be attractive or repulsive.
charges can be of two signs, + and -
charges of the same sign repel
charges of opposite sign attract.
The forces are long range.
All materials contain both positive charges(protons in
the nuclei) and negative charges which exactly
balance in charge neutral material.
In conductors, electrons move easily (metals)
In insulators they don't.
Quantitative characterization of the electrical forces.
We measure charge in coulombs. The charge of an
electron is
-1.6 x 10-19 Coulombs and is called -e
The charge of a proton is + 1.6 x 10-19 Coulombs
Consider two point charges (very small, like a nucleus).
Let them have charges q1 and q2 and suppose that
the distance between them is r12 . Then the force which
1 exerts on 2 has magnitude
K|q1||q2|/r12
2
and is along the line between the points. The direction
is determined by the signs of the charges as we have
discussed. The value of K is 9.0 x 109 Nm2/C2 .
Electric field.
By making calculations with a test , (that is probe) charge
qt at various points around a collection of charges
(such as a dipole or a quadrupole) one can find the
forces which a test charge WOULD EXPERIENCE if
it were at each point. All those forces are
proportional to the test charge qt .
The ELECTRIC FIELD E at each point in space around
the collection of charges is DEFINED to be the
force which a test charge at that point would
experience, divided by the value qt of the test charge.
The value of the electric field
is independent of qt . The electric field is
regarded as a property of the space surrounding the
collection of charges.
The electric field due to a plane of
charge with surface charge density Q/A is
(Q/A) 2πK independent of the distance from the plane.
The electric field between 2 oppositely charged
capacitor plates has magnitude (Q/A) 4πK=Q/Aε0
In metals, when no charges are moving,
(we say in equilibrium) the electric field
is zero everywhere inside the metal.
If the metal in equilibrium has a net charge,
it is all at its surface.
Just outside the surface of a charged metal,
the electric field is perpendicular to the
surface.
A charge qt in an electric field E has electrical
potential energy. The potential energy
of such a charge can change as the charge
moves between points in the field. When the
charge is moved a small distance between two
points 1 and 2, the amount ΔU21 of its potential energy
change is the work required to move the charge slowly
between the two points 1 and 2 and is given by
ΔU21 =-qt E Δh21
where Δh21 is measured along the direction of
the field. The electrostatic potential difference ΔV21
between the two points is the electrical
potential energy change ΔU21 of the test charge divided by qt
ΔV21 = ΔU21/qt and is measured in volts,
which are joules/coulomb.
The electric field is perpendicular everywhere
to surfaces of constant electrostatic potential.
The electrostatic potential of the field
around a point charge q is given by
Kq/r where r is the distance from the
point charge.
Electrostatic potential is scalar quantity:
It does not have a direction.
Energy stored in a charged
capacitor:
We return to a charged capacitor.
+Q/A
d
.q
-Q/A
We are going to think about the process of charging
it up and how much energy it takes,
In fact, capacitors are usually charged up using a
battery or other voltage source which we will discuss
shortly. However we will imagine doing it another way
which costs an equivalent amount of energy, in order
to compute the energy required.
Start with uncharged capacitor, move one tiny charge at a time
+0/A
from bottom to top.
d. q
-0/A
.q
+q/A
work required =0
d
-q/A
+q/A
d. q
.q.q
d
-q/A
+2qA
-2q/A
work required
=q2d/ͼ0A
(Note that the bottom plate was originally charge
neutral so removing a + charge from it leaves an
equal minus charge behind.)
+2q/A
d
.q
-2q/A
How much work is required to move a third tiny charge q
from bottom to top?
a. q2d/ͼ0A
b. 2q2d/ͼ0A
c. 3q2d/ͼ0A
d. (1/2) q2d/ͼ0A
Now suppose we want to charge the capacitor up
to a total charge Q. We do it one tiny charge q at a
time where q=Q/N and N is large (>>1, say 10,000).
The total work required is
0+q2d/ͼ0A + 2q2d/ͼ0A +3q2d/ͼ0A + ...+(N-1)q2d/ͼ0A=
q2d/ͼ0A(0+1+2+...+(N-1))
The sum
0+1+2+...+(N-1)= N(N-1)/2 (You can check to by hand
for a few values of N, check it
with Excel or try to prove it.)
and with N>>1 we have the total work =q2(N2/2) d/ͼ0A
=(1/2) Q2d/ͼ0A
Summary:
The work required to charge up the capacitor
is
(1/2) Q2d/ͼ0A which can also be written (1/2)QEd=(1/2)VQ
The resulting increase in the energy of the capacitor
is in the form of electrostatic potential energy U.
It is convenient to write the above expression
using the (final) electric field E= -Q/ͼ0A :
U = (1/2) E2 ͼ0 dA
dA is the volume inside the capacitor. We regard
the potential energy as STORED IN THE FIELD
and see that the energy stored per unit volume
in the field is (1/2) E2 ͼ0 . This turns out to be
true quite generally for any field distribution.
Capacitance.
The magnitude of the
electric field in a parallel plate capacitor is
E= Q/Aε0
and the electrostatic potential drop across the
plates separated by a distance d has magnitude
V=Ed= Q(d/Aε0 )
so Q = (Aε0 /d)V
which is often written Q= CV with C = Aε0 /d
called the capacitance.
Capacitance for electrodes which are
not parallel plates.
If two pieces of metal near each other but
not touching are attached to the two
poles of a voltage source, then in most
cases, the charge Q which flows onto the
electrodes (>0 on one piece and <0 on
the other) will be proportional to V, the
voltage drop supplied by the source to
the two pieces of metal. If so, then the
proportionality constant is also called C
so the relationship
Q=CV
applies to such cases as well. C can be measured
or, in many cases calculated by more complicated
methods than we will use.
Dielectrics.
The preceding exercises were worked out for
a capacitor with a vacuum between the plates.
That is usually not the case. There is some material
between the plates. Let's think about what
what kind of material to put there.
If there is metal, then the negative charges immediately
move from the negative to the positive side and
both plates become charge neutral giving zero field
and zero potential drop in the capacitor.
in the insulator, the charges can't move (very far)
BUT THE DIPOLES OF THE MOLECULES IN THE
MATERIAL CAN BECOME ALIGNED IN THE FIELD,
BECAUSE OF THE TORQUE WE DISCUSSED
EARLIER.
The resulting picture at the molecular level looks like
this:
The charges on
the dipoles in the
middle cancel each
other out, but the
charges on the
dipoles next to the
capacitor plate leave
an uncompensated
charge which produces
a field opposite to
the applied one.
If the surface charge on the plates is fixed at Q/A
and -Q/A then what will happen to the electric field
and the voltage drop across the capacitor
if the material between the plates is a
a.
b.
c.
d.
metal or an
field voltage
big
big
0
0
little
little
both less than
vacuum
insulator
field voltage
little little
both less than vacuum
big big
0 0
Answer: b.
Capacitance.
With a vacuum between the plates we had the relation
Q =(Aε0/d) V
which says that the charge on the plates increases
with the voltage applied, The proportionality constant
Aε0/d is called the capacitance C of the capacitor
Q = CV for a capacitor and, for a parallel plate
capacitor in a vacuum
C=Aε0/d
Consider a capacitor with no material in it (ie in
vacuum) and suppose that it is charged to give
a voltage difference of V between the plates.
since V=Ed =Qd/Aε0 the resulting charge is
Q =(Aε0/d) V . Now consider two experiments:
In the first you detach the leads to the battery
charging the capacitor and slip a dielectric
material between the plates. In the second,
you leave the battery attached and slip the
same material between the plates. What happens to
the charge Q, the field E inside the capacitor and
the voltage V between the two plates in the two cases?
exp 1, detached
expt 2 attached
Q
V
E
Q
V
E
a.
same down up
up same down
b.
up down down
down up up
c.
same down down
up
same same
d.
0
0
0
same same same
e.
down same same
same down down
We saw in the exercise that when a dielectric material
(insulator) is inserted between the plates,
then either the charge Q goes up if the voltage
remains same or the voltage goes down if the
charge remains the same. In either case the
capacitance C= Q/V goes up.
We characterise the amount by which the capacitance
goes up by a parameter characteristic of the
material and called the dielectric constant κ
giving, for a parallel plate capacitor
C= κAε0/d
The energy stored in the capacitor is
U=(1/2) V Q=1/2 C V2 and increases at fixed voltage
when you put in a dielectric.
At fixed charge Q on the capacitor plates,
what is the electric field between the
capacitor plates if the material in between
the plates is a dielectric with dielectric
constant κ? Plate area A, plate separation d.
a. κ Q/Aε0
b. Q/κAε0
c. 0
d. very large
Answer: b.
C= κAε0/d =Q/V
so E=V/d = (Q/Aε0κ)
What is the capacitance of a
parallel plate capacitor into which
a piece of metal (close to but not touching
the plates) has been inserted?
a.0
b. Aε0//d
c. very large
d. -Aε0//d
Answer c.
C=Q/V
but V must be 0.
What is the dielectric constant of an ideal
metal?
a. 0
b. infinite
c. depends on which metal.
d. much less than that of an insulator.
Answer:b.
The field inside a metal placed between
the plates of a capacitor (but not touching
the plates) must be Q/κAε0 but it must also
be zero. All the other quantities in the expression
for the field are finite. So κ must be infinite.
+
V
Q/A
-Q'/A
-
+Q'/A
Q/A
In this picture, we said that
there is a layer of charge induced
on the surfaces of the dielectric when a
voltage is applied to the capacitor and it reduces the
electric field inside.
Here I denote that induced charge by Q'. What is
the relationship between Q' , Q and κ ?
a. Q' =Q/κ b. Q' =κQ c. Q'=Q(1-κ) d. Q' =Q(1-1/κ)
Answer: d.
With E being the magnitude of the field between
the capacitor plates we have
E =Q/Aε0κ
But using the charges to calculate the field
E= (1/Aε0)(Q-Q')
Equating the two expressions:
(1/Aε0)(Q-Q')=Q/Aε0κ Solve for Q'
Supercapacitors to store a lot of energy.
Tricks for making capacitance big: Use a solution
(liquid) containing + and i ions in it (like salt in water).
The - ions are attracted to the + terminal and the + ions
are attracted to the - terminal, but they don't get all the
way to the plates because of thermal agitation.
The result is TWO capacitors as shown, each with very
small d. C= κAε0/d for each. κ is also big (water).
Circuits:
Most electrical applications involve closed circuits
which are engineered paths which charges move
through. The reason that circuits are closed is
that, if they are not, and one starts to move charge
from one part of the path to another, then charges
of one sign build up at one end of the path
and charges of the other sign build up at the
other end. The whole path then acts like a capacitor
and the field builds, opposing the motion of the
charges until the energy source cannot
supply enough force to the charges to move them.
That's actually what happens in an experiment you
are doing in Laboratory 3.
Elements in circuits:
The objects in circuits can be characterized as
of three types:
Power sources: batteries, power plants, etic
As charge passes through these elements
the electrostatic energy of the charge increases.
Dissipative elements (light bulbs, appliances,
...) As charge passes through these elements
the electrostatic energy of the charge decreases.
Capacitive elements: Charges cannot pass through.
As charge flows onto the capacitor, energy is
stored in the capacitor (as we have seen).
Current
To quantitatively characterise the motion of
charges through a circuit we use the
concept of current, usually denoted I,
which is the number of coulombs of charge
passing through a point in the circuit per
second. Current has units of Coulombs/second
which are called amperes or amps.
In many, but not all, cases dissipative elements
have the property that the voltage V between the
ends of the dissipative element is proportional
to the current passing through it. The proportionality
constant is called the resistance R. Units of
R are volts/amp or joules/(coulomb2 sec) usually
called ohms. The relation V=IR is called ohm's law.
Power:
In all these circuit elements you can
get the change in the energy of
the charges per unit time by dividing
the energy change by the time.
Power = qVBA/t = (q/t) VBA =IVBA
which will be negative for a dissipative
element (wire or appliance) and positive
for a power source (power plant or battery)
I=q/t is the current. If I is in amps and
V is in volts, then P comes out in watts.
q
A
B
qVB > qVA for a power plant or battery between
A and B but
qVB <qVA for a wire or appliance between
A and B
Power sources in circuits. One could transfer
energy to charges in circuits by rubbing things
together as we did in some of the demonstrations
to transfer excess charge to an object. However
that is very seldom the way that it is done in
practise. As you know, we use batteries and
, as we will discuss later, electrical generators.
Batteries add energy to the charges that pass
through them. The energy comes from the electrical
potential energy stored in chemical constituents of
the battery.
e-
e-
Cu
Zn
H+
2H++2eH2
2Zn +H2SO4
Zn2SO4+
2H++2e-
H2SO4
Illustration of the operation of the first battery invented
by Allesandro Volta in the 18th century. The potential energy
of the Zn and sulfuric acid is higher than the potential energy
of the hydrogen gas and zinc sulfate. Part of the difference
is transferred to the electrons which pass through the
circuit when the battery is connected.
A more common, somewhat more complicated
example is a car battery....
Pb + H2SO4- ->
PbSO4 +2H+
++2e-+H SO
PbO
+2H
2
2
4
+2e
PbSO4+2H2O
The current inside the car battery is being carried by
the H+ ions which have positive charge equal in magnitude
(but not in sign) to the electron charge.
p. 327
Batteries using the energy in lithium ions are
used in cell phones and laptops. The principle
is simpler and the material is less heavy so more
energy is stored per kilogram. However lithium
metal is dangerous (can burn in water) and the
lithium is usually embedded in graphite (lithium
ion battery) which makes it safer but less powerful.
Circuit elements: Ok we have an energy source
(source of emf) which is likely to be a battery. What
is going on in the rest of the circuit, indicated by an
arrow?
It could be a capacitor. Consider that case
1st. Introduce the traditional symbols
+
-
C
V
Before I close the switch what charge is on the
capacitor? (Suppose the capacitance associated
with the switch is very small.)
a. nearly zero b. almost infinite
c. +CV on right and -CV on left
d. -CV on right and +CV on left.
Answer: a.
+
-
C
V
Now I close the switch and wait until
charges stop moving. What charge is on the
capacitor?
a. zero b. infinite c. +CV on right and -CV on left
d. -CV on right and +CV on left.
Answer: c.
Another thing we can insert between the poles
of our source of emf (battery) is a light
bulb. That is characteristic of the class of
circuit elements which DISSIPATE energy,
which means that they convert the electrical
potential energy of the charges that enter them
into another form of energy (thermal,
gravitational potential energy, light carrying
energy in electromagnetic waves,..) which
leaves the circuit. The charges stay in the
circuit, but the energy leaves the circuit.
By contrast, capacitors STORE electrical
potential energy which stays in the circuit
and can be recovered.
When attached to a battery, charges can keep moving
through such dissipative elements indefinitely, whereas
in capacitors, one quickly builds up so much charge that
motion stops.
Resistance and Ohm's law:
For dissipative elements (like
wires and applicances) it is often
the case that the voltage drop is
proportional to the current. If
that is the case, then the
proportionality constant is called
the electrical resistance and is
denoted R. The units of R, which
are volts/amp are called ohms.
When this is true we call the
circuit elements 'ohmic'
V=IR for ohmic circuit
elements, including wires.
A dissipative circuit element obeys
ohm's law and has resistance R. If
it has a current I running through, at what
rate is it dissipating energy into the
environment?
a. IR b. 0 c. I2/R d. I2 R e. I/R
Answer: d.
P=IV=I(IR)= I2R
R
-
+
V
When I close the switch the current will
flow and current and the voltage at the ends of the
resistor will be
current
voltage
a.
left to right
bigger on left
b.
right to left
bigger on right
c.
left to right
bigger on right
d.
right to left
bigger on left
Answer: c.
power plant
I
II
wire
wire
III
IV
appliance
If the arrows show the direction of the current,
then the electrons (negative charge) are flowing
in the opposite direction. At what points in the
circuit is the energy of the electrons highest
and lowest?
highest
lowest
A.
IV
III
B.
I
II
C.
III
IV
D.
II
III
Answer: B
power plant
I
II
wire
wire
III
IV
appliance
Which of the following statements is true concerning the application of
the law of conservation of charge to this circuit?
A. Charge is conserved because are electrons lost
in the appliance and wires and then are put back
in the power plant so nothing has changed.
B. Charge is conserved because no charge is wasted.
C. Charge is conserved because the current is alternating.
D. Charge is conserved because
no electrons leave the circuit at any point.
Answer: d
In the series circuit containing 3 light
bulbs , I will close the switch which
goes across the second light bulb
as sketched.
What will happen?
A. The middle light bulb will get dimmer
and the other two will say the same.
B. The middle light bulb will go out and
the other two will stay the same.
C.. The middle light bulb will go out and
the other two will get brighter
D. The middle light bulb will go out
and the other two will get dimmer.
E. They will all go out.
Answer: C
The voltage drop across the other two
goes up so that the total voltage drop
will be the same as the total applied
voltage supplied by the battery. Therefore
the current in those two lights goes up
and they are brighter.
What is the ratio of the power dissipated
in each lighted bulb before the switch is
closed to the power dissipated in each
bulb after the switch is closed?
A. 3/2
B. 4/9
C. 1
D. 2/3
Answer: B.
Before: I=V/3R, Power=(V/3R)(V/3)=V2/9R
After: I=V/2R, Power=(V/2R)(V/2)=V2/4R
What is the ratio of the TOTAL power dissipated
before the switch is closed to the TOTAL power
after it’s closed?
A. 2/3
B. 3/2
C. 1
D. 9/4
Answer: A.
Before: 3 x V2/9R=V2/3R
After: 2xV2/4R=V2/2R
If I close the switch do the other two lights get
A Brighter
B. Dimmer
C. Stay the same
D. Go out.
Answer: C
Same current in each of the remaining bulbs.
Similar question for the parallel circuit:
What is the ratio of the power dissipated per
lighted bulb before the switch is opened to the power
dissipated after the switch is opened?
A. 2/3
B. 3/2
C. 9/4
D. 1
Answer: D.
Before: I=V/R, Power=V2/R
After: Same
What about the ratio of the TOTAL power
dissipated before the switch is opened to
the TOTAL power dissipated after it’s
opened.
A. 1
B. 3/2
C. 2/3
D. 4/9
Answer: B
Parallel
Series
V
R1
R2
V
R1
R2
In the two cases in the two resistors:
a.
b.
c.
d.
Current is
series Parallel
same
different
different same
same
same
same
different
voltage is
series parallel
same different
same different
different different
different same
Answer: D.
Series
V
R1
What is the voltage drop across
R2
R1
a. VR1/( R1+R2)
VR2/( R1+R2)
b
VR2/(R1+R2)
c.
d.
VR2/R1
VR1/R2
VR1/(R1+R2)
VR1/R2
VR2/R1
R2
Current is the same in both. The voltage
between a point to the left of the first
resistor and a point to the right of the
second resistor is
V = IR1+IR2 . solve for I:
I=V/(R1+R2 )
and the voltage across R1 is
IR1= VR1 /(R1+R2 )
Parallel
V
R1
R2
What is the voltage drop across
R2
R1
a. VR1/( R1+R2)
VR2/( R1+R2)
b
VR1/(R1+R2)
V
VR2/R1
VR2/(R1+R2)
c. V
d. VR1/R2
Answer: C.
Parallel
V
R1
R2
What is the current through
R2
R1
a. V (R1+R2)/R1R2
V(R1+R2)/R1R2
b
V/(R1+R2)
c.
d.
V/R1
V/R2
V/(R1+R2)
V/R2
V/R1
Answer C.
Parallel
Series
V
R1
R2
Equivalent Resistance Req
Req =V/I=V/(V/R1 +V/R2 )=
1/(1/R1 +1/R2 )
V
R1
R
2
Req =V/I=(IR1 +IR2 )/I=
(R1 +R2 )
Review principles:
I=q/t is charge passing through circuit element per
unit time (current)
qV is energy gained or lost by charge q as it
passes through element, where V is
change in voltage from one end to the other.
Power dissipated or gained is qV/t= IV
Charge is always conserved.
For ohmic dissipative elements the magnitude of
the voltage across the circuit element is IR .
R can be different for each element.