Transcript Document

How do most objects interact with each other?
•Electric Charge (q or Q)
–An intrinsic property of matter (as is mass)
–S.I. unit: coulombs (C), (also: μC)
–Two types (+) or (-): add algebraically to give a net
charge.
–Like charges repel; unlike charges attract.
–Coulomb’s Law: The force between two charges (q1)
and q2) separated by a distance r is: Fe=k q1 q2 / r2.
– k is Coulomb’s constant: k = 9 × 109 Nm2/C2
–Force by q1 on q2 is equal and opposite to the force by
q2 on q1.
More about Charge.
• Quantization of charge (e): charge comes in
discrete amounts of e = 1.6×10-19C
• e is the “elementary charge”; we always treat e
as positive.
• Electrons have a charge of –e
• Protons have a charge of +e and about 2000
times the mass of electrons.
• Conservation of charge: In any reaction, net
charge remains the same. (Is this different
from conservation of mass?)
Examples
• Book sitting on a table
• Pushing off
– A hard wall
– A soft wall
– Electric Force is a measure of hardness in repulsion.
• Sodium Metal vs. Aluminum Metal.
– Electric Force is a measure of hardness in attraction.
• What happens if we hold a
– Negatively charged rod held near a neutral atom?
– Positively charged rod held near a neutral atom?
How can charge be transferred
between objects?
1. Friction
•
•
•
Electrons transferred by rubbing.
Rubbing plastic with fur; fur loses electrons.
Rubbing glass with silk; silk gains electrons.
2. Induction
•
•
•
A charged object causes electrons to move in a
neutral object.
Atoms become polarized; they turn into dipoles.
Charged rod (+) or (-) near a neutral object; what
happens?
How can charge be transferred
between objects? (cont’d)
3. Conduction
• Electrons can move freely through some
materials, called conductors.
• Conductors can be charged directly through
contact with charged objects (not friction).
• Conductors can be charged through induction
 Electrons move freely to one side or another;
object as a whole becomes polarized.
Van de Graf Generator
• Device to concentrate positive
charge on a conducting sphere.
• Operation:
– Lower roller (G) rubs e- off belt.
– Belt becomes (+)
– (+) belt attracts e- from brush (B),
bleeding them off the conducting
sphere.
– Belt is neutralized as it travels back
to lower roller.
– Sphere (A) becomes (+) more and
more.
Electrical Properties of Materials
• Conductors: Outer electrons can
move nearly freely through the
material. (What is it about metallic
bonding that allows this?)
• Insulators: Outer electrons are
bound tightly to the atoms. Cannot
move freely.
Electrical Properties of Materials (cont’d)
• Semiconductors: Outer electrons are bound,
but not so tightly. (e.g. silicon, germanium)
– Can be made to conduct electricity in very
controlled ways.
– Doping: impurities to increase or decrease
electrons in crystal structure (e.g. transistors)
– Photoelectric Effect: electrons can be
energized by absorbing light ot leave their
atoms and travel through the material. (e.g.
photocells)
Using Coulomb’s Law
1. A charge (Q) of +2µC is fixed in the ground.
A charge (q) of +3µC having a mass of 0.24 kg
is held 0.15m directly above the fixed charge
and released.
How does the charge q move?
Give the magnitude and direction of its
acceleration.
Use (g = 10 m/s2 and k = 9 × 109 Nm2/C2)
Using Coulomb’s Law (cont’d)
2. Two protons are separated by 5×10-15m.
a. Calculate the force on one proton due to the other.
(qp= + e = 1.6×10-19 C). Does the force seem
small or large as felt by:
a person?
a fly? a proton?
b. Calculate the acceleration of the proton. Is
this a large number? (mp= 1.67×10-27 kg)
Using Coulomb’s Law (cont’d)
3. Three charges (q1 and q3 are fixed):
q1 = -86 µC at (52 cm, 0)
q2 = +50 µC at (0, 0)
q3 = +65 µC at (0,30 cm)
What is the net force on q2? (magnitude and
direction)
What is its acceleration if it has a mass
of 0.5 kg?
Electric Field
• How does one charge know that another charge is
there? (“Action at a distance”).
• A charge actually changes the space around it; a
force field is caused.
• Other charges interact with this “Electric Field”.
• The Electric Field due to a “source” charge (qs):
E = k qs /r2 ř (a vector)
The direction of E at a point is in the same direction
that a positive point charge would move if placed at
that point. (BY DEFINITION)
Electric Field (cont’d)
• Suppose we have many charged objects
creating an E-field in space.
• At some point (r) (position vector from
origin) in space:
– Put a small positive test charge (qT) at r.
– The magnitude of E at r equals the Force on qT
divided by qT. E = Fe/qT (By definition)
– The direction of E at r is in the direction of the
force on qT. (By definition)
Advantages of the E-Field
• The Electric fields caused by several charges add up
as vectors.
• For a system of charges:
– Find the resultant (net) E-Field
– Now you can easily find the force on any charge (q) you
place in the system: Fe= q E
– If q is (+), then Fe and E are in the same direction,
otherwise they are opposite to each other.
• If you change q, you don’t need to recalculate the
E-field.
• The E-field is physical; energy is actually stored in
space.
Common Types of E-Field
Problems
• Given a collection of point charges:
– Find the resulting E-field at a specified location
– Find the location where the E-field has a
specified magnitude and direction (such as zero.)
• Given the E-field:
– find the forces and/or accelerations produced on
a charge or charged system.
Common Mistake to Avoid
• E = k q /r2 ONLY when it is CAUSED by a
POINT CHARGE we call “q”.
• If you have are given an E-field due to some
other source or sources, the above equation is
FALSE.
• Example: A uniform E-field has the same
value everywhere in space. For such a field, the
above equation is FALSE.
Using the E-field
1. For charges: q1 = -3 μC placed at (0,0)
q2 = +6 μC placed at (2 m, 0)
a. Find the E-field at (-1m, 0)
b. Find the force on an electron placed
at (-1 m , 0)
c. Where on the x-axis is E = 0?
Using the E-field (cont’d)
2. A pendulum with: charge (+Q)
and mass (m) is in equilibrium
in a uniform E-field applied in
the +x-direction.
In terms of Q, m, and θ,
a.
b.
Find the magnitude of the Efield.
Find the tension in the string.
Using the E-field (cont’d)
3. A charge of (+q) and mass (m) is at the
origin and moving in the +x-direction with
a speed v. A uniform E-field (with
magnitude: E) acts in the –x-direction.
a. Find the acceleration of the charge.
b. Find the maximum value of +x reached by the
charge.
c. Find the time for the charge to return to its start
point.
Electric Field Lines
• Used to visualize an
electric field’s
– Direction: Directed
path along which a
positive test charge
would move.
– Magnitude:
Proportional to the
number of field lines
coming through an
area (line density).
Drawing Electric Field Lines
To draw the electric field lines for a single charge
1. Draw lines directed away from (+) charges.
2. Draw lines directed towards (-) charges.
3. The number of lines entering or leaving the
charge is proportional to the amount of charge.
4. Examples to do: Using 4 lines for a (+1 C)
charge, draw the field lines for:
a. +1C charge, +2C charge, -1C charge, -2C
charge.
Drawing Electric Field Lines (cont’d)
To draw electric field lines for a system of
charges:
1. Draw field lines near each charge in the
system.
2. Connect field lines smoothly between
(+) and (-) charges.
3. Field lines between like charges can go to
infinity.
4. Draw field lines perpendicular (normal)
to conducting surfaces (why?).
Examples
1. Dipole (+ -)
2. Two (+) charges (+ +)
3. Quadrupole
+
- +
4. Oppositely charged plates that are close
together.(*Important*)
Check answers at Electric Field Applet
How does a Conductor behave in an E-Field?
• In electrostatics, charges don’t move. (Fields due
to stationary charges are called electrostatic
fields.)
• Recall what happens when we put a charged rod
near a conductor:
– Transient current
– Polarization of entire object
– Electrons moved so as to cancel the E-field inside the
conductor. (Why must this be so?)
• Excess charges move to surface and distribute
themselves out of mutual repulsion.
• And so ....
How does an E-field behave in a Conductor?
• No electrostatic field can exist in the material of
a conductor.
• At the surface of a charged conductor:
– E-field is perpendicular to the surface.
– Component tangential to the surface is zero.
– Higher concentrations of charge occur at surfaces
of higher curvature.
• In an empty cavity within a conductor
– E-field is zero
– Region is shielded from any external charges.
Charge densities at curved surfaces
 On flat surfaces of low curvature,
repulsive forces are directed mostly
parallel to surface, keeping charges
further apart.
 On highly curved surfaces,
parallel components of the repulsive
forces are smaller, allowing charges
to be closer together.
Result: Strong E-fields near highly
curved surfaces. (Ex: Lightning
Rods)
Electric Field Lines near a
Conductor
Conductor in an E-field
(summary)
Field is normal to surface
Field is ZERO in
an empty cavity.
Charge concentrates
at high curvature.
Field is NOT zero in a
cavity containing charge.
How do we describe the ability of an
Electric Field to do Work?
• We want to discuss the ability of a field to do
work without regard to the material being
worked on.
• Approach:
– Develop an intuitive gravitational analog.
– Extend the concept to the electrical case.
– Apply the concept to different combinations of
positive/negative source/test charges.
What is the Gravitational Equivalent to an
E-field?
• For uniform fields:
Fg = mg
Fe = q E
• For fields due to a point mass or charge:
Fg = m GM/r2 ř
Fe = q kqs/r2 ř
So E-field is to charge as ____________ is to mass.
Gravitational Potential Energy
b _______________
a
•
•
•
_______________
Mass falling from b to a loses P.E.: ΔUg = -mgh
Work done by the gravitational field: Wfield = mgh
It takes work to lift mass from a to b:
Wexternal = mgh
• More mass will lose more P.E.
• More mass will require more work to lift.
Gravitational Potential
• Define Gravitational Potential: Vg = Ug/m
(gravitational potential energy per unit mass).
• Gravitational Potential Difference:
ΔVg = -gh in going from b to a.
• ΔVg is a measure of
How fast and how far material will fall from b to a.
How difficult it is to lift material from a to b.
• ΔVg describes the terrain through which we can
climb or fall without regard to our mass.
• For a uniform field (such as near Earth’s Surface)
Vg = gh
or
Vg = -GM/r
Topographical Maps
• Topo map: maps contours or lines of equal
gravitational potential.
• No work is done to move objects along these
lines.
• Moving across lines requires work
– Closely spaced lines indicates steepness.
– Steepness indicates the strength of the field in the
direction of your movement.
(e.g. “field strength” here would be analogous to an an
inclined plane: g sinθ; direction of fall would be down
the plane.)
Topo map
Electrical Potential Energy
Let’s extend the concept to the E-field:
• A (+q) charge “falls” from high (P.E.)b
to low (P.E.)a : ΔUe = Uea - Ueb < 0
• Work done by the field in this falling:
Wfield = F d = qE d
• Change in P.E. :
ΔUe = -qEd
• More charged material means
– More work done by the field in falling
– More P.E. lost in falling
– More external work to push the charge
“uphill” against the field.
+
|
|
(b)
|
|
(a)
E 
Electric Potential
• Define Electric Potential: V= Ue/q
(Electric potential energy per unit charge).
(Units: joule/coulomb or “volt”; J/C or V)
• Potential Difference or (“Voltage”): (going from b to a)
ΔV = Va-Vb=ΔUe/q = -qEd/q = -Ed
• ΔV is a measure of
 How fast and how far charged material will fall from b to a.
 How difficult it is to lift charged material from a to b.
 ΔUe = q ΔV (the signs of q and ΔV are important)
• ΔV describes the electrical terrain through which we
can climb or fall without regard to our charge.
• BUT: The direction of your fall depends on the sign
of your charge.
Electric Potential and Charged Material
• E-field direction specifies
region of high V to region
of low V.
• (+) charges fall to:
___________ potential
_________ potential energy
• (-) charges fall to:
___________ potential
_________ potential energy
+
|
|
(b)
High V
E 
|
|
(a)
Low V
ΔUe = q ΔV
Some Examples
1. An electron is accelerated across a
voltage of 5000 volts between two plates.
a. What is the change in potential energy of the
electron?
b. What is the work done on the electron by the
E-field between the two plates?
c. What speed does the electron attain?
d. Who works with these kinds of problems?
e. What is an “electron-volt (eV)”?
The Electron Volt
• The amount of work needed to push an electron
across a potential difference of 1 volt.
• Allows MUCH easier calculations of energies.
• Allows MUCH easier grading of the calculations.
• Use them when charges are given in units of “e”.
• 1 eV = (1.60×10-19C)(1 J/C) = 1.60×10-19J
• Notice: instead of multiplying voltage by the electronic
charge in coulombs, you just put an “e” in front of it.
• If you’re asked for speeds; you’re back to S.I. units.
Examples (cont’d)
2. Two plates are charged to a voltage 50V. If
their separation is 5.0 cm, what is the E-field
between them?
The E-field is uniform
+
between the plates.
|
|
|
|
+50 V
|
|
|
|
0V
| d = 5 cm |
Visualizing Electric Potential
• Draw Equipotentials: Lines along which
the electric potential is constant.
• Similar to contours on a topo map.
• Questions
– How would you calculate the work done in
moving a charge along an equipotential?
– How are equipotentials related to electric
field lines?
They are perpendicular to each
other at every point.
Mapping Equipotentials
• What do equipotentials look like for
– A point charge?
– A dipole?
– Two like charges?
– Applet as before: Electric Field Applet
• For a charge distribution, how could you
experimentally map
– Equipotentials?
– Electric field lines?
Finding the Electric Potential due to a Point
Charge
• Suppose we have a positive charge: Q
– How can we calculate (estimate) the work done
by Q’s E-field as a (+) charge q falls away from
rA to rB?
• WTOT = Σ Wi = Σ Fi Δr = Σ q Ei Δr = q Σ kQ/ri2 Δr
= q (kQ/ rA - kQ/ rB) (By Calculus)
– What is the resulting change in potential energy
of q?
• ΔUe = - WTOT = q (kQ/rB - kQ/rA)
– What is the potential difference (voltage) between
rA and rB?
• ΔV = ΔUe /q = kQ/rB - kQ/rA
Electric Potential due to a Point Charge
• Due to a point charge (Q)





V = kQ/r at a point a distance r from Q
V  0 as r  ∞
V increases as you get closer to a (+) charge.
V decreases as you get closer to a (-) charge.
Be able to plot V vs. r for a point charge.
• Potential Difference or Voltage between two points
(rA and rB) (due to a point charge):
 ΔV = kQ/rB - kQ/rA
 Work done in moving a charge from one point to another is
independent of the path taken between the two points.
 ΔV between two points is independent of any path between the
two points.
Electric Potential Energy
(of two point charges)
• If we have a (+q1) charge fixed in space
 If we push another (+q2) charge towards it from ∞,
how does the P.E. of the system of two charges
change?
 If a (-q2) charge falls towards it from ∞, how does the
P.E. of the system change?
• Change in P.E. of two point charges (q1 and q2)
(The q’s include the signs)
 ΔUe = q2 ΔV = q2 (V1 - V∞)
= q2 (k q1 /r12 - 0) = k q1 q2/r12 = U12
 How is this different from gravitational P.E. ?
Electric Potential Energy
(for various charge configurations)
• For several charges, the potential energies add
(same as in the gravitational case):




For three charges:
Utotal = U12 + U13 + U23
For four charges
Utotal = U12 + U13 + U23 + U14 + U24 + U34
• The potentials at a point due to a collection of
charges (q1, q2, q3, …) also add:
 Vtotal = V1 + V2 + V3 + ….
Examples
Ex 1: Looking at some charge configurations
(Two charges at different spacing as shown)
a) +q -q
b) +q
-q
c) +q
+q
1. Which set has a positive P.E.?
2. Which set has the most negative P.E.?
3. Which set requires the most work to separate
the charges to ∞?
Examples (cont’d)
2. Potential above two charges:
Q1 is at (0, 26 cm)
40 cm
B
A
Q2 is at (0, -26 cm)
40 cm
30 cm
60 cm
Q2=+50μC
Q1=-50μC
Find the potential at
points A and B due to
charges Q1 and Q2.
At A: VA = VA1 + VA2 = k(Q1/rA1 + Q2/rA2) =
At B: VB = VB1 + VB2 = k(Q1/rB1 + Q2/rB2) =
Examples (cont’d)
3. For the configuration of charges shown
a. Calculate the energy necessary to build the
configuration.
b. Calculate the potential at the center of the triangle.
q1=+4µC
S=.2m
q2=+4µC
q3= -4µC
How can we store Electric Energy?
(How can we store Electric Charge?)
A Comprehensive Capacitor Problem
I.
Start with a charged capacitor of known
dimensions. Calculate: C0, V0, E0-field, UC0.
(The “0” subscript refer to values for an air filled
or vacuum capacitor.)
II. Insert a dielectric of known dielectric constant.
Calculate: Q, C, V, E-field, UC.
III. Apply the voltage (V0) calculated in (I) to the
new capacitor. Calculate: Q, C, V, E-field, UC.
IV. How much more charge and energy are we
able to store for the same applied voltage?
Capacitor Example
I.
A parallel plate capacitor has area: A= .05 m2
separation distance: d = 2.00 mm and is
charged to Q0 = 6 μC. The capacitor is
disconnected from any voltage source and the
gap is filled with air which can be treated as a
vacuum.
C0 =
V0 =
E0-Field =
UC0=
Capacitor Example (cont’d)
II.
The gap is now filled with an insulator
having a dielectric constant of K=3.
Q=
C=
V=
E-Field =
UC=
Capacitor Example (cont’d)
III. The capacitor is now hooked up to a battery
maintaining the same voltage as calculated in
part (I).
Q=
C=
V=
E-Field =
UC =
Capacitor Example (cont’d)
IV. How much more charge and energy can we
store at the same applied voltage of parts
(I) and (III)?
Charge:
Energy:
Dielectric Strength
• Applied E-field beyond which the dielectric
fails as an insulator.
• High E-field liberates bound electrons.
• Freed electrons can collide with other electrons
and free them (“avalanche effect”)
• Dielectric material becomes partially
conductive.
• We now have a “leaky capacitor”; electrons are
lost from (-) plate to (+) plate.
• Ex: Glass (11.5 MV/m), Water (67.5 MV/m),
Neoprene Rubber(16-28 MV/m)