Electric Fields
• Review of gravitational fields
• Electric field vector
• Electric fields for various charge configurations
• Field strengths for point charges and uniform fields
• Work done by fields & change in potential energy
• Potential & equipotential surfaces
• Capacitors, capacitance, & voltage drops across capacitors
• Millikan oil drop experiment
• Excess Charge Distribution on a Conductor
Gravitational Fields: Review
Recall that surrounding any object with mass, or collection of objects with mass, is a
gravitational field. Any mass placed in a gravitational field will experience a
gravitational force. We defined the field strength as the gravitational force per unit
mass on any “test mass” placed in the field: g = F / m. g is a vector that points in
the direction of the net gravitational force; its units are N / kg. F is the vector force
on the test mass, and m is the test mass, a scalar. g and F are always parallel. The
strength of the field is independent of the test mass. For example, near Earth’s
surface mg / m = g = 9.8 N / kg for any mass. Some fields are uniform (parallel,
equally spaced fields lines). Nonuniform fields are stronger where the field lines are
closer together.
uniform field
10 kg
98 N
Earth’s surface
nonuniform
field
Earth
F
m
Electric Fields: Intro
Surrounding any object with charge, or collection of objects with charge, is a electric
field. Any charge placed in an electric field will experience a electrical force. We
defined the field strength as the electric force per unit charge on any “test charge”
placed in the field: E = F / q. E is a vector that points, by definition, in the
direction of the net electric force on a positive charge; its units are N / C. F is the
vector force on the test charge, and q is the test charge, a scalar. E and F are only
parallel if the test charge is positive. Some fields are uniform (parallel, equally
spaced fields lines) such as the field on the left formed by a sheet of negative charge.
Nonuniform fields are stronger where the field lines are closer together, such as the
field on the right produced by a sphere of negative charge.
uniform field
+q
F
nonuniform
field
F
+
--------------
q
Overview of Fields
Charge, like mass, is an intrinsic property of an object. Charges produce electric
fields that affect other charges; masses produce gravitational fields that affect other
masses. Gravitational fields lines always point toward an isolated mass. Unlike
mass, though, charges can be positive or negative. Electric field lines emanate from
positive charges and penetrate into negative charge.
We refer to the charge producing a field as a field charge. A group of field charges
can produce very nonuniform fields. To determine the strength of the field at a
particular point, we place a small, positive test charge in the field. We then measure
the electric force on it and divide by the test charge: E = F / q.
For an isolated positive field charge, the field lines point away from the field
charge (since the force on a positive charge would be away from the field charge).
The opposite is true for an isolated negative field charge. No matter how complex
the field, the electric force on a test charge is always tangent to the field line at that
point.
The coming slides will reiterate these ideas and provide examples.
Electric & Gravitational Fields Compared
Field
strength
Force
Intrinsic
Property
SI units
Gravity:
g
=
W
/
m
N / kg
Electric
Force:
E
=
FE
/
q
N/C
Field strength is given by per unit mass or force per unit charge,
depending on the type of field. Field strength means the magnitude of a
field vector. Ex #1: If a +10 C charge is placed in an electric field and
experiences a 50 N force, the field strength at the location of the charge is
5 N/C. The electric field vector is given by: E = 5 N/C, where the
direction of this vector is parallel to the force vector (and the field lines).
Ex #2: If a -10 C charge experiences a 50 N force, E = 5 N/C in a
direction opposite the force vector (opposite the direction of the field
lines).
Electric Field Example Problem
A sphere of mass 1.3 grams is charged via friction, and in the
process excess electrons are rubbed onto it, giving the sphere a
charge of - 4.8 μC. The sphere is then placed into an external
uniform electric field of 6 N/C directed to the right. The sphere is
released from rest. What is its displacement after 15 s? (Hints on
next slide.)
E
-
E
Sample Problem Hints
1. Draw a vector as shown. Note that
FE = q E, by definition of E, and
that FE is to the left (opposite E)
since the charge is negative.
2. Instead of finding the net force
(which would work), compute the
acceleration due to each force
separately.
qE
mg
3. Find the displacement due to each force using the time given and
kinematics.
4. Add the displacement vectors to find the net displacement
vector.
Drawing an E Field for a Point Charge
Let’s use the idea of a test charge to produce the E field for an isolated positive field
charge. We place small, positive test charges in the vicinity of the field and draw the
force vector on each. Note that the closer the test charge is to the field charge, the
greater the force, but all force vectors are directed radially outward from the field
charge. At any point near the field charge, the force vector points in the direction of
the electric field. Thus we have a field that looks like a sea urchin, with field lines
radiating outward from the field charge to infinity in all direction, not just in a plane.
The number of field lines drawn in arbitrary, but they should be evenly spaced around
the field charge. What if the field charge were negative?
+
Test charges and force vectors
surrounding a field charge
+
Isolated, positive point charge
and its electric field
Single Positive Field Charge
This is a 2D picture
of the field lines that
surround a positive
field charge that is
either point-like or
spherically
symmetric. Not
shown are field lines
going out of and into
the page. Keep in
mind that the field
lines radiate
outwards because,
by definition, an
electric field vector
points in the
direction of the force
on a positive test
charge.
+
The nearer you
get to the
charge, the
more uniform
and stronger
the field.
Farther away
the field
strength gets
weaker, as
indicated by
the field lines
becoming
more spread
out.
Single Negative Field Charge
The field surrounding an
isolated, negative point (or
spherically symmetric)
charge looks just like that of
an isolated positive charge
except the field lines are
directed toward the field
charge. This is because, by
definition, an electric field
vector points in the direction
of the force on a positive test
charge, which, in this case is
toward the field charge. As
before, the field is stronger
where the field lines are
closer together, and the force
vector on a test charge is
parallel to the field.
-
Point Charges of Different Magnitudes
Let’s compare the fields on two separate isolated point charges, one with a
charge of +1 unit, the other with a charge of +2 units. It doesn’t matter how
many field lines we draw emanating from the +1 charge so long as we draw
twice as many line coming from the +2 charge. This means, at a given
distance, the strength of the E field for the +2 charge is twice that for the +1
charge.
+1
+2
Equal but Opposite Field Charges
Pictured is the electric field produced by two equal but opposite
charges. Because the charges are of the same magnitude, the field is
symmetric. Note that all the lines that emanate from the positive
charge land on the negative charge. Also pictured is a small positive
charge placed in the field and the force vector on it at that position.
This is the vector sum of the forces exerted on the test charge by each
field charge. Note that the net force vector is tangent to the field line.
This is always the case. In fact, the field is defined by the direction of
net force vectors on test charges at
various places. The net force on a
negative test charge is tangent to the
+
field as well, but it points in the
opposite direction of the field.
(Continued on next slide.)
Link #1
Link #2
Link #3
Equal but Opposite Field Charges (cont.)
D
C
-
+
A
B
Here is another view of the field. Since the net force on a charge can only be
in one direction, field lines never intersect. Draw the electric force on a
positive charge at A, the electric field vector and B, and the electric force on a
negative charge at C. The net force on a + charge at D charge is directly to the
left. Show why this is the case by drawing force vectors from each field
charge and then summing these vectors.
Multiple Charges: How to Determine the Field
To determine the field surrounding two field charges, Q1 and Q2, we pick
some points in the vicinity and place test charges there (red dots). Q1
exerts a force on each, directly away from itself (blue vectors), as does
Q2 (purple vectors). The resultant vectors (black) show the direction of
the net electric force and define the direction of the electric field.
The net force vector on each test
charge is tangent to the E field
there. If we place little a tangent
segment parallel with the net
force at each test charge and do
this at many different points, we
will build a picture of the electric
field. The same procedure can be
used regardless of the number of
field charges.
Q1 +
Q2
+
Two Identical Charges
+
+
With two identical field charges, the field is symmetric but all field
lines go to infinity (if the charges are positive) or come from infinity (if
the charges are negative). As with any field the net force on a test
charge is tangent to the field. Here, each field charge repels a positive
test charge. The forces are shown as well as the resultant vectors, which
are tangent to the field lines.
Coulomb’s Law Review
The force that two point charges, Q and q, separated by a distance r,
exert on one another is given by:
K Qq
F= 2
r
where K = 9  109 Nm2/C2 (constant).
This formula only applies to point charges or spherically
symmetric charges.
Suppose that the force two point charges are exerting on one
another is F. What is the force when one charge is tripled, the
other is doubled, and the distance is cut in half ?
Answer: 24 F
Field Strengths: Point Charge; Point Mass
Suppose a test charge q is placed in the electric field produced by a
point-like field charge Q. From the definition of electric field and
Coulomb’s law
K Qq / r 2
KQ
F
E= =
=
q
q
r2
Note that the field strength is independent of the charge placed in it.
Suppose a test mass m is placed in the gravitational field produced
by a point-like field mass M. From the definition of gravitational
field and Newton’s law of universal gravitation
GM
F G Mm / r 2
g=
=
= 2
m
m
r
Again, the field strength is independent of the mass place in it.
Uniform Field
Just as near Earth’s surface the gravitational field is approximately
uniform, the electric field near the surface of a charged sphere is
approximately uniform. A common way to produce a uniform E field is
with a parallel plate capacitor: two flat, metal, parallel plates, one
negative, one positive. Aside from some fringing on the edges, the field
is nearly uniform inside. This means everywhere inside the capacitor the
field has about the same magnitude and direction. Two positive test
charges are depicted along with force vectors.
-
-
-
-
-
-
+
+
+
+
+
+
-
+
-
+
Two + Field Charges of Different Magnitude
• More field lines emanate from the greater charge; none of the
field lines cross and they all go to infinity.
• The field lines of the greater charge looks more like that of an
isolated charge, since it dominates the smaller charge.
• If you “zoomed out” on this picture, i.e., if you looked at the
field from a great distance, it would look like that of an
isolated point charge due to one combined charge.
+
+
Although in this pic the
greater charge is depicted as
physically bigger, this need
not be the case.
Opposite Signs, Unequal Charges
The positive charge has a greater magnitude than the negative charge.
Explain why the field is as shown. (Answer on next slide.)
+
-
Opposite Signs, Unequal Charges
+
(cont.)
-
More field lines come from the positive charge than land on the negative.
Those that don’t land on the negative charge go to infinity. As always, net
force on a test charge is the vector sum of the two forces and it’s tangent to the
field. Since the positive charge has greater magnitude, it dominates the
negative charge, forcing the “turning points” of the point to be closer to the
negative charge. If you were to “zoom out” (observe the field from a distance)
it would look like that of an isolated, positive point with a charge equal to the
net charge of the system.
Summary of Fields due to Unequal Charges
You should be able to explain each case in some detail.
Review of Induction
Valence electrons of a conductor
are mobile. Thus they can
respond to an electric force from
a charged object. This is called
charging by induction. Note: not
all of the valence electrons will
move from the bottom to the top.
The greater the positive charge
brought near it, and the nearer it
is brought, the more electrons that
will migrate toward it. (See
animation on next slide.)
+
+ - + - + - +- +
conductor
+ - + - + - +-+
Review of Induction
+
(cont.)
Because of the
displaced electrons, a
charge separation is
induced in the
conductor.
+ - + - + - + -+ + + + + +
Positive Charge Near a Neutral Conductor
• The + charge induces a
charge separation on the
neutral conductor.
+
+
• Since it is neutral, as many
lines land on the conductor as
leave it.
• The number of field lines that
go off to infinity is the same
as if the + charge were
isolated.
• Viewed from afar, the field
would look like that of an
isolated, + field charge.
Overview of Field Types
For the following scenarios, you should be able to draw the
associated electric fields correctly:
1.
2.
3.
4.
5.
6.
7.
8.
A uniform field
An isolated + point charge
An isolated – charge
Two identical + point charges
Two identical – point charges
Point charge (either sign) near neutral conductor
Unequal point charges of the same sign
Unequal point charges of the opposite sign
Note that a field drawn without a direction indicated (without arrows)
is incorrect. You should be able to draw vector forces on positive or
negative charges placed in any field. Also, for complex fields you
should be able to describe them as the appear from a distance.
Work done by Fields & Applied Forces
To lift an object of mass m a height h in a uniform gravitational field
g without acceleration, you must apply a force m g. The work you do
is +m g h, while the work done by the field is - m g h. When you lower
the object, you do negative work and the field does positive work.
Near the surface of a negatively charged object, the electric field is
nearly uniform. To lift without acceleration a positive charge q in a
downward field E requires a force q E. You do positive work in lifting
the charge, and the field does negative work. The signs reverse when
you lower the charge.
m
mg
Earth’s surface
g
E
+q
qE
Negatively charged surface
Fields: Work & Potential Energy
The work your applied force does on the mass or on the charge can go into
kinetic energy, waste heat, or potential energy. If there is no friction and no
acceleration, then the work you do goes into a change of potential energy:
U = m g h for a mass in a gravitational field and U = q E h for a
charge in a uniform electric field. The sign of h determines the sign of
U. (If a charged object is moved in a vicinity where both types of fields
are present, we’d have to use both formulae.) Whether or not there is
friction or acceleration, it is always the case that the work done by the field
is the opposite of the change in potential energy: Wfield = - U.
m
mg
Earth’s surface
g
E
+q
qE
Negatively charged surface
Work-Energy Example
Here the E field is to the right and approximately uniform. The applied
force is FA to the left, as is the displacement.
The work done by FA is + FA d.
The work done by the field is WF = - q E d.
The change in electric potential energy is U = - WF = + q E d.
Since FA > q E, the applied force does more positive work than the field
does negative work. The difference goes into kinetic energy and heat.
The work done by friction is Wfric < 0. So, Wnet = FA d - q E d - |Wfric|
= K by the work-energy theorem.
d
+
+
+ FA
+
+
q
+
qE
-
Work-Energy Practice
For each situation a charge is displaced by some applied force while
in a uniform electric field. Determine the sign of: the work done by
the applied force; the work done by the field; and U.
1. q is positive and displaced to the right.
2. q is negative and displaced to the right.
3. q is positive and displaced to the left.
4. q is negative and displaced to the left.
+
+
+
+
+
q
-
Potential
Gravitational potential is defined to be gravitational potential energy per unit
mass. At any given height above Earth’s surface, the gravitational potential is a
constant since U / m = m g h / m = g h.Thus potential is independent of mass.
If M > m and they’re at the same height, M has more potential energy than
m, but they are at the same potential.
Similarly, electric potential, V, is defined to be electric potential energy per
unit charge. At any given distance from a charged surface in a uniform field,
the electric potential is a constant since U / q = q E d / q = E d. Thus potential
is independent of charge. If Q > q and they’re the same distance from the
surface, Q has more potential energy than q, but they are at the same
potential. In a uniform field V = E d.
g
m
E
M
q
h
Earth’s surface
Q
d
Negatively charged surface
SI Units for Potential
By definition, electric potential is potential energy per unit charge. So,
U
V= q
The SI unit for electric potential is the volts. Both potential and its
unit are notated by the capital letter “V.” Based on the definition
above, a volt is defined as joule per coulomb:
1J
1V=
C
Ex: If an object with a 10 C charge is placed at a certain point in an
electric field so that its potential energy is 50 J, every coulomb of
charge in the object contributes to 5 J of its energy, and its potential is
5 J / C, that is, 5 V.
Equipotential Surfaces
As with gravitational potential energy, the reference point for electric potential energy,
and hence potential, is arbitrary. Usually what matters is a change in potential, so we
just pick a convenient place to call potential energy zero. The dotted lines on the left
represent equipotential surfaces--planes in which masses all have the same potential,
regardless of the mass. On the 30 J/kg surface, for example, every kilogram of every
mass has 30 J of potential energy. Note that equipotentials are always perpendicular to
field lines.
The equipotentials on the right are labeled in volts. Potential decreases with distance
from a positively charged surface since a positive charge loses potential energy as it
recedes from the surface. Here again the equipotentials are perpendicular to the field
lines. On the -45 V surface, every coulomb of charges has -45 J of potential energy.
A -2 C charge there has a potential energy of +90 J.
Earth’s surface
40 J/kg
-60 V
30 J/kg
20 J/kg
10 J/kg
0 J/kg
-45 V
-30 V
-15 V
0V
Positively charged surface
Contour Map Analogy
Earth’s gravitational field doesn’t diminish much over the height of a mountain, so
the field is nearly uniform and the equipotentials are evenly spaced, parallel planes.
Thus the dotted lines are equally spaced (side view). As seen from above, though, the
corresponding contour lines are not equally spaced. They are closer together where
the potential energy changes rapidly (steep part of the mountain), and they’re far
apart where the energy changes gradually (gentle sloping part of mountain). Contour
lines connect points of equal elevation, so walking along one mean your potential
energy remains constant. They are analogous to equipotentials.
top view
side view
steep
not steep
Equipotential Surfaces: Positive Point Charge
Imagine a positive test charge, q, approaching an isolated, positive, pointlike field charge, Q. The closer q approaches, the more potential energy it
has. So, potential increases as distance decreases. Next year we’ll derive this
formula for potential due to a point charge: V = KQ / r. This shows that V is
proportional to Q, that V → 0 as r → ∞, and that V → ∞ as r → 0.
Equipotential surfaces are always perpendicular to the field lines, for any
charge configuration. For a point charge the
surfaces are spheres centered at Q.
Here the surfaces could be labeled
from the inside out: 100V, 90 V,
80 V, and 70 V. Every 10 V step is
bigger than the previous, since the field
+
is getting weaker with distance. The
gap between the 50 V and 40 V
surfaces would be very large, and the
gap between 10 V and 0 V would be
infinite.
Equipotential Surfaces: Negative Point Charge
The field and the equipotentials look just like that of the isolated,
positive point charge. However, the field lines point in the opposite
direction and the potential decreases with distance. Imagine a positive
test charge, q, approaching an isolated, negative, point-like field
charge, -Q. The closer q approaches, the more negative its potential
energy becomes. So, V → 0 as r → ∞ (as with the positive field
charge), but V → - ∞ as r → 0.
Here the surfaces could be labeled
from the inside out: -100V, -90 V,
-80 V, and -70 V. Every 10 V step is
bigger than the previous, since
V = zero at infinity. (The step size
to be drawn is a matter of choice.) A
+3 C charge placed on the -70 V
surface has a potential energy of
-210 J.
Equipotentials Surfaces for Multiple-charge Configurations
In class practice: First experiment with the link, then draw equipotentials
on the board on top of this picture. Here are the rules:
Link
• Equipotentials are always
perpendicular to the field lines.
• Equipotentials never intersect
one another.
• The potential is large &
positive near a positive charge,
large & negative near a
negative charge, and near
zero far from all the charges.
• Equipotentials are close
together where potential
energy changes quickly
(close to charges).
+
-
Moving in an Electric Field
Electric and gravitation fields are called conservative fields because, when
a mass/charge moves about one, any change in potential energy is
independent of path. A charge taking a straight-line path from A to B
undergoes a change in potential of 10 V (V = +10 V). If a charge takes the
long, curvy path, its energy increases as
it approaches the field charge, and
D
decreases as it recedes, but the change
is the same as the straight-line path.
In either case each coulomb of
C
charge gains 10 J of potential energy.
A
B
No matter what path is taken:
VC→A = -20 V, and VD→A = 0.
+
V is independent of path!
From A to D along the equipotential the
field can do no work, since the
displacement if always to E, which is
|| to F. Recall: W = F · x = F x cos.
Capacitors - Overview
• A capacitor is a device that stores electrical charge.
• A charged capacitor is actually neutral overall, but it maintains
a charge separation.
• The charge storing capacity of a capacitor is called its
capacitance.
• An electric field exists inside a charged capacitor, between the
positive and negative charge separation.
• A charged capacitor store electrical potential energy.
• Capacitors are ubiquitous in electrical devices. They’re used in
power transmission, computer memory, photoflash units in
cameras, tuners for radios and TV’s, defibrillators, etc.
Parallel Plate Capacitor
capacitor
The simplest type of capacitor is a
-Q
parallel plate capacitor, which consists
+Q
of two parallel metal plates, each of
Area, A
area A, separated by a distance d. When
one plate is attached via a wire to the +
d
terminal of a battery, and the other plate
is connected to the - terminal, the
V
battery pulls e-’s from the plate
wire
connected to the - terminal and
battery
deposits them on the other. As a whole
the capacitor remains neutral, but we say it now has a
C
charge Q, the amount of charge moved from one plate to the
other. Without a resistor in the circuit, the capacitor charges
very quickly. Thus the current, i, which by definition is in
+Q -Q
the opposite direction of the flow of e-’s, lasts but a short
time. As soon as the voltage drop across the capacitor (the
i
potential difference between its plates) is the same as that of
V
the battery, V, the charging ceases. The capacitor can
remain charged even when disconnected from the battery.
Note the symbols used in the circuit diagram to the right.
Parallel Plate Capacitor: E & U
Because of the charge separation, an electric field exists between the plates of a
charged capacitor. If it is a parallel plate variety, the field is very nearly uniform inside,
with some fringing on the edges, as we’ve seen before. Outside the plates the field is
very weak. The strength of the E field inside is proportional to how much charge is on
the capacitor and inversely proportional to how the capacitors area. (Less area means
the charge is more concentrated and the field is stronger.) A charged capacitor also
stores potential energy (in an amount proportional to the square of the charge) since
energy is required to separate the charges in the first place. Touching a charged
capacitor will allow it to discharge quickly and will result in a shock. Once discharged,
the electric field vanishes and the potential energy is converted to some other form.
-
-
-
-
-
-
+
+
+
+
+
+
-
+
-
+
Capacitance
Capacitance, C, is the capacity to store charge. The amount of charge,
Q, stored on given capacitor depends on the potential difference
between its plates, V, and its capacitance C. In other words, Q is
directly proportional to V, and the constant of proportionality is C:
Q = CV
Ex: A 12 V battery will cause a capacitor to store
twice as much charge as a 6 V battery. Also, if
capacitor #2 has twice the capacitance of capacitor
#1, then #2 will store twice as much charge as #1,
provided they are charged by the same battery.
C depends on the type of the capacitor. For a
parallel plate capacitor, C is proportional to the
area, A and inversely proportional to the plate
separation, d.
C
Q
V
Capacitance: SI Units
The SI unit for capacitance is the farad, named for the famous
19th century scientist Michael Faraday. Its symbol is F. From the
defining equation for capacitance, Q = CV, we define a farad:
implies
Q = CV
1 C = (1 F)(1 V)
So, a farad is a coulomb per volt. This means a capacitor with a
capacitance of 3 F could store 30 C of charge if connected to a 10 V
battery. This is a tremendous amount of charge for a reasonable
potential difference. Thus a farad is a large amount of capacitance.
Many capacitors have capacitances measured in pF or fF (pico or
femtofarads).
m: milli = 10-3, μ: micro = 10-6, n: nano = 10-9,
p: pico = 10-12,
f: femto = 10-15
Capacitance Problem
A parallel plate capacitor is fully charged by a 20 V battery, acquiring
a charge of 1.62 nC. The area of each plate is 3.5 cm2 and the gap
between them is 1.3 mm. What is the capacitance of the capacitor?
From Q = C V, C = Q / V = (1.62  10-9 C) / (20 V)
= 8.1  10-11 F = 81  10-12 F = 81 pF.
The gap and area are extraneous.
- 1.62 μC
+ 1.62 μC
3.5 cm2
1.3 mm
20 V
V = Ed
As argued on the slide entitled “Potential,” in a uniform field,
V = E d. This argument was based on an analogy with gravity and
applies only to uniform fields:
gravitational: U = m g h
electric: U = q E d  U / q = E d  V = E d
Since E is uniform inside a parallel plate capacitor, the voltage drop
across it is equal to the magnitude of the electric field times the
distance between the plates.
d
V = E d (formal derivation)
U
V=
q
(from the definition of potential)
|Wfield |
V=
q
(since Wfield = - U )
Fd
V=
q
(since W = F d )
V= E d
F
)
(since E =
q
Millikan’s Oil Drop Experiment
In 1909, Robert Millikan performed an experiment to determine the
charge of an electron. The charge to mass ratio of the electron had
already been calculated by J. J. Thomson (discoverer of the electron) in
1897. But until Millikan’s experiment, neither the mass nor the charge
was known, only the ratio. By examining the motion of the oil droplets
falling between two highly charged plates, he found the charge to be
-1.6 10-19 C. The charged plates were similar to that of a parallel plate
capacitor.
Millikan Apparatus and
Experiment
A battery connected to the plates
kept the top (positive) plate at a higher
potential than the lower (negative) plate. So, a nearly uniform, downward E
field existed between the plates. An atomizer sprayed tiny oil droplets (of radii
about 1 μm) from above the plates, some of which fell through a hole in the
positive plate into the E field. Due to friction during the spraying, some of
the drops were charged, either positively or negatively. A negatively charged
drop that makes it into the hole will not undergo free fall, since it experiences
an upward electric force between the plates. The radius, mass, and charge of
the drops varied, but by adjusting the potential difference across the plates,
Millikan could make a drop hover. Continued…
switch
Oil Drop Experiment
(cont.)
A drop suspended in midair has no net force on it. This means the downward
weight, m g, was negated by an upward electric force, q E. Millikan could
vary E by adjusting the potential difference across the plates (V = E d). So,
the excess charge on the drop is: q = m g / E = m g d / V.
But m needed to be calculated in order to determine q.
To find the drop’s mass, he turned off the electric field by opening the switch
and disconnecting the battery. The drop then began to fall, but it quickly
reached terminal velocity in the air. The greater the falling speed, the greater
the drag force, and by measuring terminal speeds, Millikan could calculate the
mass. Continued…
qE
mg
switch
d
Oil Drop Experiment
(cont.)
At this point Millikan could calculate the charge on a drop. But without
knowing the number of excess electrons on the drop, he couldn’t determine
the charge of an electron. So, he altered the charge on the drop with X-rays
(not shown). The X-rays ionized the surrounding air, which, in turn, altered
the charge on the drops. The drops were now no longer in equilibrium, so
Millikan adjusted the E field until equilibrium was reestablished.
Since at equilibrium q = mgd / V, and all quantities on the right side of the
equation were known, Millikan could repeat this X-ray procedure numerous
times and calculate the many different charges that a drop could attain. He
found that the charge on a drop was always a multiple of 1.6  10-19 C. As
you know, we now call this amount of charge e, the elementary charge.
His experiment showed that charge is quantized, existing in discrete bundles
(in this case, electrons) and that the charge on an electron is -1.6  10-19 C.
Excess Charge on a Conductor
Any excess charge placed on a conductor will immediately distribute itself
over the surface of the conductor. No excess charge will remain inside. On
a spherical conductor the excess charge will be distributed evenly. If
electrons are added, they themselves will spread out. If electrons are
removed, electrons in the conductor will replace them, leaving all excess
positive charge on the surface. Excess charge placed on an insulator pretty
much stays put.
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Now lets add some extra
charges.
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The new charges repel
themselves and reside only on
the surface.
Excess Charge on a Pointy Conductor
Excess charge, which always resides on the surface of a conductor, will collect in
high concentrations at points. In general, the smaller the radius of curvature, R, the
greater the charge density (charge per unit area). The reason for this is that when R
is large, neighboring charges push a charge nearly tangent to the surface (left pic).
But where R is small (as near a point), neighboring charges are mostly pushing a
charge outward, away from the surface instead of away from each other (right pic).
This allows the charges be reside closer together.
vector forces due to neighboring charges
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uniform R, uniform charge density
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small R, high
charge density
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large R, low
charge density
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Electric Fields In & Around Charged Conductors
E is always zero inside any conductor, even a charged one. If this were not
the case, mobile valence electrons inside the conductor would be
accelerated by the E field, leaving them in a state of perpetual motion.
Outside a charged conductor E is greater where the charge density is
greater. Near points, E can be extremely high. Surrounding a sphere the
field is radially symmetric, just the field due to a point charge.
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E=0
inside
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small R,
strong E
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E is radially symmetric outside.
large R,
weak E
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E=0
inside
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Shielding Electric Fields
A box or room made of metal or with a metal liner can shield its
interior from external electric fields. Valence e-’s in the metal will
respond to the field and reorient themselves until the field inside
the box no longer exists. The external field (black) points right.
This causes a charge separation in the box (e-’s migrating left),
which produces its own field (red), negating the external field.
Thus, the net field inside is zero. Outside, the field persists.
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Credits
http://images.google.com/imgres?imgurl=http://buphy.bu.edu/~duffy/PY106/2e.GIF&imgrefurl=http://physics.bu.edu
/~duffy/PY106/Electricfield.html&h=221&w=370&sz=4&tbnid=y0qny4b133kJ:&tbnh=70&tbnw=117&start=3&pre
v=/images%3Fq%3Delectric%2Bfield%26hl%3Den%26lr%3D
Spark Picture: http://cdcollura.tripod.com/tcspark2.htm
electric field lines: http://www.gel.ulaval.ca/~mbusque/elec/main_e.html
java, placing and moving test charges and regular charges:
http://www.physicslessons.com/exp21b.htm
java animation, placing test charges:
http://www.colorado.edu/physics/2000/waves_particles/wavpart3.html
http://www.slcc.edu/schools/hum_sci/physics/tutor/2220/e_fields/
lesson, pictures, units: http://www.pa.msu.edu/courses/1997spring/PHY232/lectures/efields/
java electric field: http://www.msu.edu/user/brechtjo/physics/eField/eField.html
lesson with animations, explanations: http://www.cyberclassrooms.net/~pschweiger/field.html
http://library.thinkquest.org/10796/ch12/ch12.htm
Robert Millikan: http://www.nobel.se/physics/laureates/1923/millikan-bio.html
Millikan Oil Drop: http://www.mdclearhills.ab.ca/millikan/experiment.html
http://www.glenbrook.k12.il.us/gbssci/phys/Class/estatics/u8l4c16.gif
http://www.physchem.co.za/Static%20Electricity/Graphics/GRDA0008.gif
http://www.eng.uct.ac.za/~victor/electric/charge_opposite_particles.gif