Transcript AP C UNIT 7 - student handout

Electric Potential Energy
& Electric Potential
Unit 7
Recall ‘Work’ from earlier
• Work done by a force is given by:
– W = F d cos(q) or W   F ( x)dx
+W: Force is in direction moved
-W: Force is opposite direction moved
W=0: Force is 90o to direction moved
• Conservative Forces
DU = -Wfield
Electric Potential Energy
General Points:
ΔU = -Wfield
1) Potential Energy increases if the particle moves
in the direction opposite to the field force.
Work will have to be done by an
external agent for this to occur
2) Potential Energy decreases if the particle moves
in the same direction as the field force on it
Electrical Potential Energy
Graphical look at EPE
The potential
energy is taken to
be zero when the
two charges are
infinitely separated
Potential Energy of a System of Charges
Start by putting first charge in position
No work is necessary to do this
Next bring second charge into place
Now work is done by the electric field of the
first charge. This work goes into the potential
energy between these two charges.
Now the third & fourth charge are put into place
Work is done by the electric fields of the two previous charges.
The total EPE is
then given by (signs
matter)
Example
A test charge is brought separately to the vicinity of a positive
r
charge Q at pt B
Q
Charge +q is brought to pt A, a
distance r from Q
q
A
A
B
I) Compare the potential energy of q to that of Q.
(b) UA = UB
(c) UA > UB
(a) UA < UB
II) Suppose charge q has mass m and is released from rest from
the above position (a distance r from Q). What is its velocity vf as
it approaches r = ∞ ?
(a)
vf 
1 Qq
4 0 mr
(b)
vf 
1 Qq
2 0 mr
(c)
vf 0
Work Done by Uniform Electric Field
Force on charge is
Work is done on the
charge by field
ELECTRIC
POTENTIAL
Consider a ball of mass, m, placed at a point in space
(height, h, above Earth). It would possess a certain PE per
unit mass due to it being in the gravitational field of Earth.
If the ball was replaced by a bowling ball of mass, M, it too,
would possess the SAME potential energy per unit mass.
Similarly, 2 different sized
charges at the same
distance from the charged
sphere (green) will have the
same EPE/charge.
The electric potential energy per unit charge for
some location in an electrical field is called electric
potential.
Electric Potential
Electric potential is
defined as the potential
energy per unit charge
at a point in space
Comparison:
Electric Potential Energy vs. Electric Potential
• Electric Potential Energy (U) - the energy of a
charge at some location.
• Electric Potential (V) - tells what the EPE would
be if a charge were located there
Voltage and Potential Energy
• Positive charge will naturally move towards
lower electrical potential energies, lower
voltage.
General Points for either positive or negative
charges:
Positive potential is taken to be higher by
definition due to positive test charge.
Relation between
Potential and Field
The work done by the electric field force in moving
a charge from point a to point b is given by
Electric Potential for a Point Charge
• The direction of the electric field from a point charge is
always radial. We integrate from distance r (distance
from the point charge) along a radial line to infinity:
What is the electric potential difference between A and B?
Rank (a), (b) and (c) according to the net electric
potential V produced at point P by two protons.
Greatest first.
A: (b), (c), (a)
B: all equal
C: (c), (b), (a)
D: (a) and (c) tie, then (b)
Question…
The electric potential at point A is _______ at point B
1) greater than
2) equal to
3) less than
Points A, B, and C lie
in a uniform electric
field.
A
E
C
B
1) If a positive charge is moved from point A to point
B, its electric potential energy
a) Increases b) decreases c) doesn’t change
2) Compare the potential differences between
points A and C and points B and C.
a) VAC > VBC
b) VAC = VBC
c) VAC < VBC
Two Charges
• In region II (between the two charges) the electric
potential is
1) always positive
2) positive at some points, negative at others.
3) always negative
I
II
Q=+7.0mC
III
Q=-3.5 mC
Potential for two charges
Calculate electric potential at point A due to charges
4m
A
How much work do you have to
do to bring a 2 mC charge from
far away to point A?
Q=+7.0mC
6m
Q=-3.5 mC
Potential of a solid conducting sphere
(radius R) with charge +Q
Find V at the following locations:
+Q
R
i) At r > R
ii) at r = R
R
iii) at r < R
E vs V graph for conductor
V for uniformly charged,
nonconducting sphere (radius R)
R
a) Find ΔV for r > R moving from ∞ to a
distance r from center
b) Find ΔV moving from ∞ to a distance r where
r < R.
R
Equipotentials
4/5/2016
Equipotential (EP) Surfaces &
Their Relation to Electric Field
An equipotential surface is a surface on which
the electric potential is the same everywhere.
The EP surfaces that surround the point
charge +q are spherical. The electric
force does no work as a charge moves on
a path that lies on an EP surface, such as
the path ABC.
However, work is done by the electric
force when a charge moves along the
path AD.
Equipotential Surfaces &
Their Relation to Electric Field
The radially directed electric field of
a point charge is perpendicular to
the spherical equipotential surfaces
that surround the charge. The
electric field points in the direction
of decreasing potential.
Equipotential surfaces (in blue) of an
electric dipole. The surfaces are drawn so
that at every point they are
perpendicular to the electric field lines
(in red) of the dipole.
Consider two conducting spheres
with differing radii Ra and
Rb sitting on insulating stands far
apart. The sphere with radius
Ra has an electric charge +Q. If
we connect a thin, conducting
line between the spheres, then
disconnect it, what are the
charges on the spheres?
2 identical spheres question
•
+Q
+Q/2
higher V?
• Attach wire btw spheres…what happens?
• What is final charge of each?
Potential difference
between charged plates
Capacitance
Describes how much charge an
arrangement of conductors can
hold for a given voltage applied.
-
+
When the battery is connected to the pair of plates, charges
will flow until the top plate’s potential is the same as the +
side of the battery, and the bottom plate’s potential is the
same as the – side of the battery. No potential difference.
Q is the amount of charge on a
plate and ΔV is the voltage applied
to the plates
Units = Farads (F)
Work to charge conductor
• Consider a spherical uncharged conductor,
radius R. After a small amount of charge is
placed on conductor, its potential becomes V =
kQ/R (where V∞=0).
• To further charge conductor, work must be
done to bring additional increments of charge,
dQ, to place on surface. W = ΔV dQ…the
amount of work increases as each dQ is added
and sphere becomes more charged.
EPE of conductor
Capacitance for Parallel Plates
– The E field is constant
– The geometry is simple, only
the area and plate separation
are important.
• To calculate capacitance, we first
need to determine the E-field
between the plates. We did this
using Gauss’ Law:
separation
d
area A
V+
E and dA
parallel
VTotal charge q
on inside of plate
Need to find potential difference. Since E=constant
area A
V+
E and dA
parallel
VTotal charge q
on inside of plate
How large is 1 Farad?
If a parallel plate capacitor has plates that are
separated by only 1mm, in order to achieve 1F, the
area of each plate would be…
C
o A
A = 1.1x108m2
d
This corresponds to about 6 miles on each side of plate!
Obviously, this is impractical to achieve large
capacitance. Therefore, what do they do?
Energy per unit Volume
It is necessary at times
to relate energy per unit
volume to electric field
of capacitor (parallel
plate)
Example
Charge 8.0uF capacitor
(C1) by connecting it to a
120V potential difference.
a) Find charge on capacitor
b) Find energy stored
in capacitor
Now remove the power supply.
c) Now connect C1 to another capacitor, C2= 4.0uF, initially
uncharged.
C1
C2
What will be the potential difference
across each capacitor & charge on
each after equilibrium is reached?
Conservation of charge
++++++
++++++
-----
-----
Method for finding C for various
geometries of plates
1) We are trying to calculate C where C = Q / ΔV
2) In order to acquire ΔV, we must use
rb
DV  -  Edr
ra
3) Therefore we must find expression for E
first using Gauss’ Law.
4) Find E, then ΔV, and then C.
Find capacitance of concentric cylindrical
conductors with radius a (inside) & radius b.
Inside charge is +,
outside is -
Field is radially
outward
End view
Spherical Capacitor
(2 concentric spheres, inner radius a &
outer radius b as shown)
b
a
+Q
-Q
Combination of Capacitors
&
Equivalent Capacitance (CEQ)
1) Capacitors in Parallel & CEQ
Consider 3 identical capacitors in parallel
connected to battery of voltage, V. Find CEQ
V
C1 C
C2 C
C3C
All top plates are at same potential
and so are bottom plates, so…
Splitting capacitor into 3
separate capacitors in
parallel all with equal
potential difference btwn
them (same as battery)
Capacitors in Series
If you place
2 capacitors
in series, the
charge
remains the
same, but
the potential
difference is
less for each
capacitor
Charge on each
capacitor is the
same, Q.
• Consider the individual voltages
Since q
across each capacitor
is the
same for
each
The sum of these voltages is the
total voltage of the battery, V
Combo Example
A 12 battery is connected to the combination of
capacitors as shown.
8uF
C1
C2
12V
2uF
Find Ceq, Q1, Q2, Q3, V1, V2, V3, & Qtotal
C3
4uF
DIELECTRICS
E
• If the strength of the electric field between the plates of
an air filled capacitor becomes too strong, then the air can
no longer insulate the charges from sparking (discharging)
between the plates. For air, this breakdown occurs when
the electric field is greater than 3x106 V/m. (this is what
occurs during a lightning strike)…V/m is equivalent to N/C.
In order to keep this from happening, an insulator, or
dielectric, is often inserted between the plates to
reduce the strength of the electric field, which yields a
larger capacitance.
Why does dielectric reduce E?
Dielectric material is polar and molecules polarize as shown.
The charge alignment creates an E-field within the material
which OPPOSES the original E-field between the plates.
Electrical forces create a torque to
rotate and align molecules
The dielectric is measured in terms of a
dimensionless constant,
κ (greek kappa) ≥ 1. (see table)
Assuming a capacitor is charged with no power
source present, dielectric reduces E which reduces V
(according to V = Ed) while d remains constant. If V
reduces, then C increases (according to C = Q / V)
example
A parallel-plate air capacitor is charged by placing a 90-V
battery across it. The battery is then removed. An insulating,
dielectric fluid is inserted between the plates. The voltage
across the capacitor is now 28V. What is the dielectric constant
of the fluid?
Example 2
A parallel-plate air capacitor holds a charge of 30nC when a
voltage of V is placed across its plates. If the battery is not
removed and a dielectric fluid is inserted between the plates,
the charge on the plates increases to 87nC. Find the
dielectric value.
The charged plates of an air-filled capacitor are 10
cm by 20 cm and the gap between the plates is 6
mm. What is the capacitance when its gap is only
half-filled with a dielectric having κ = 3.0?
We treat this problem as 2 dielectrics in SERIES
Dielectric application - computer key
on keyboard
When the key is pressed, the
plate separation is decreased
and the capacitance
increases. Each key
corresponds to a different
capacitance.
Dielectric application: Stud-finder
(how you doin!)
The dielectric constant of
wood (and of all other
insulating materials, for that
matter) is greater than 1;
therefore, the capacitance
increases. This increase is
sensed by the stud-finder's
special circuitry, which causes
an indicator on the device to
light up.