Lightning Review
General Physics (PHY 2140)
Last lecture:
Lecture 3
Q
1. Flux. Gauss’
Gauss’s law.
Φ net = ∑ EA cos θ =
εo
9 simplifies computation of electric fields
¾ Electrostatics
ΔPE
2. Potential and potential energy
ΔV = VB − VA =
q
9 electrostatic force is conservative
9 potential (a scalar) can be introduced as potential
energy of electrostatic field per unit charge
9 Electrical energy
9 potential difference and
electric potential
9 potential energy of charged
conductors
9 Capacitance and capacitors
Review Problem: Perhaps you have noticed sudden
gushes of rain or hail moments after lightning strokes in
thunderstorms. Is there any connection between the
gush and the stroke or thunder? Or is this just a
coincidence?
http://www.physics.wayne.edu/~alan/
Vcloud
JG
E
JJG
JG
FC = qE
JG
mg
Chapter 16
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2
16.2 Electric potential and potential energy
due to point charges
Electrostatic Precipitation
Gushes of Rain and Hail After Lightning
Issn:
Issn: 15201520-0469 Journal: Journal of the Atmospheric Sciences Volume: 21 Issue: 6
Pages: 646646-665
Authors: Moore, C.B., Vonnegut, B., Vrablik,
Vrablik, E.A., McCaig,
McCaig, D.A.
Article ID:10.1175/1520
ID:10.1175/1520--0469(1964)021<0646:GORAMA>2.0.CO;2
ABSTRACT
Observations of thunderstorms in New Mexico were made with a vertically
vertically--scanning,
3-cm radar on a mountain top. Prior to a cloudcloud-toto-ground lightning discharge nearby,
the radar echo overhead was usually quite weak, indicating low intensities
intensities of
precipitation there. Following the lightning it was observed sometimes
sometimes that in the
region of the cloud where the discharge occurred the radar echo intensity rapidly
increased, and shortly thereafter a gush of rain or hail fell nearby.
nearby.
These studies confirm earlier radar observations, made by the authors
authors at Grand
Bahama Island, B.W.I., in which it was found that lightning is often followed
followed in the
cloud by a rapidly intensifying echo and then by a gush of rain at the ground. The
increases in radar reflectivity in small volumes of the cloud following
following lightning suggest
that the electric discharge is influencing the size of particles in the cloud.
An analysis indicates that within 30 seconds after a lightning discharge,
discharge, the
mass of some droplets may increase as much as 100100-fold as the result of an
electrostatic precipitation effect.
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Electric circuits: point of zero potential is defined by
grounding some point in the circuit
Electric potential due to a point charge at a point in
space: point of zero potential is taken at an infinite
distance from the charge
With this choice, a potential can be found as
V = ke
q
r
Note: the potential depends only on charge of an object,
q, and a distance from this object to a point in space, r.
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1
Potential energy of a system of point
charges
Superposition principle for potentials
If more than one point charge is present, their electric
potential can be found by applying superposition
principle
The total electric potential at some point P due to several
point charges is the algebraic sum of the electric
potentials due to the individual charges.
Consider a system of two particles
If V1 is the electric potential due to charge q1 at a point P,
then work required to bring the charge q2 from infinity to P
without acceleration is q2V1. If a distance between P and
q1 is r, then by definition
q2
P
q1
r
PE = q2V1 = ke
A
q1q2
r
Remember that potentials are scalar quantities!
Potential energy is positive if charges are of the same
sign and vice versa.
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Recall that work is opposite of the change in potential
energy,
Three ions, Na+, Na+, and Cl-, located such, that they
form corners of an equilateral triangle of side 2 nm in
water. What is the electric potential energy of one of the
Na+ ions?
?
W = − PE = −q [VB − VA ]
q q
q q
q
PE = ke Na Cl + ke Na Na = ke Na [ qCl + qNa ]
r
r
r
No work is required to move a charge between two points
that are at the same potential. That is, W=0 if VB=VA
Recall:
1. all charge of the charged conductor is located on its surface
2. electric field, E, is always perpendicular to its surface, i.e. no work is
done if charges are moved along the surface
but : qCl = −qNa !
Na+
Na+
PE = ke
6
16.3 Potentials and charged conductors
MiniMini-quiz: potential energy of an ion
Cl-
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qNa
[ −qNa + qNa ] = 0
r
Thus: potential is constant everywhere on the surface of a
charged conductor in equilibrium
… but that’s not all!
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2
Because the electric field is zero inside the conductor, no
work is required to move charges between any two
points, i.e.
W = −q [VB − VA ] = 0
If work is zero, any two points inside the conductor have
the same potential, i.e. potential is constant everywhere
inside a conductor
Finally, since one of the points can be arbitrarily close to
the surface of the conductor, the electric potential is
constant everywhere inside a conductor and equal to its
value at the surface!
The electron volt
A unit of energy commonly used in atomic, nuclear and
particle physics is electron volt (eV
(eV))
The electron volt is defined as the energy that electron
(or proton) gains when accelerating through a potential
difference of 1 V
Relation to SI:
1 eV = 1.60×10-19 C·V = 1.60×10-19 J
Note that the potential inside a conductor is not necessarily zero,
even though the interior electric field is always zero!
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Example : ionization energy of the electron
in a hydrogen atom
ProblemProblem-solving strategy
Remember that potential is a scalar quantity
Superposition principle is an algebraic sum of potentials due
to a system of charges
Signs are important
Just as in mechanics, only changes in electric potential
are significant, hence, the point you choose for zero
electric potential is arbitrary.
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Vab=1 V
In the Bohr model of a hydrogen atom, the electron, if it is in the
ground state, orbits the proton at a distance of r = 5.29×10-11 m. Find
the ionization energy of the atom, i.e. the energy required to remove
the electron from the atom.
Note that the Bohr model, the idea of electrons as tiny balls orbiting the nucleus, is not
a very good model of the atom. A better picture is one in which the electron is spread
out around the nucleus in a cloud of varying density; however, the Bohr model does
give the right answer for the ionization energy
11
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3
16.4 Equipotential surfaces
In the Bohr model of a hydrogen atom, the electron, if it is in the ground state,
orbits the proton at a distance of r = 5.29 x 10-11 m. Find the ionization energy,
i.e. the energy required to remove the electron from the atom.
They are defined as a surface in space on which the potential
is the same for every point (surfaces of constant voltage)
The electric field at every point of an equipotential surface is
perpendicular to the surface
The ionization energy equals to the total energy of the
electron-proton system,
Given:
r = 5.292 x 10-11 m
me = 9.11×10-31 kg
mp = 1.67×10-27 kg
|e| = 1.60×10-19 C
E = PE + KE
with
PE = − ke
e2
v2
, KE = me
2
r
Find:
The velocity of e can be found by analyzing the force
on the electron. This force is the Coulomb force;
because the electron travels in a circular orbit, the
acceleration will be the centripetal acceleration:
E=?
me ac = Fc
or
me
v2
e2
e2
= ke 2 , or v 2 = ke
,
r
r
me r
Thus, total energy is
E = − ke
convenient to represent by drawing
equipotential lines
e 2 me ⎛ ke e 2 ⎞
e2
+
= −2.18 ×10−18 J ≈ -13.6 eV
⎜
⎟ = − ke
2 ⎝ me r ⎠
2r
r
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Examples of Equipotential Surfaces
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16.6 The definition of capacitance
Capacitor: two conductors (separated by an insulator)
„
A Positive and Negative Charge
( A Dipole)
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usually oppositely charged
a
b
+Q
-Q
The capacitance, C, of a capacitor is defined as a ratio of
the magnitude of a charge on either conductor to the
magnitude of the potential difference between the
conductors
Q
Two Positive Charges
C=
15
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ΔV
16
4
1. A capacitor is basically two parallel
conducting plates with insulating
material in between. The capacitor
doesn’t have to look like metal
plates.
2. When a capacitor is connected to an
external potential, charges flow
onto the plates and create a
potential difference between the
plates.
3. Capacitors in circuits
symbols
analysis follows from
conservation of energy
conservation of charge
Units of capacitance
The unit of C is the farad (F), but
most capacitors have values of C
ranging from picofarads to
microfarads (pF to μF).
1 F =1C V
Capacitor for use in
high-performance
audio systems.
-
+-
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16.7 The parallelparallel-plate capacitor
+Q
d
A
-Q
ke =
18
A parallel plate capacitor has plates 2.00 m2 in area, separated by a
distance of 5.00 mm. A potential difference of 10,000 V is applied
across the capacitor. Determine
the capacitance
the charge on each plate
The electric field between the plates
A
where A is the area of one of
the plates, d is the separation,
ε0 is a constant called the
permittivity of free space,
space,
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Problem: parallelparallel-plate capacitor
The capacitance of a device
depends on the geometric
arrangement of the conductors
A
C = ε0
d
Recall, micro ⇒10-6, nano ⇒10-9,
pico ⇒10-12
If the external potential is
disconnected, charges remain on
the plates, so capacitors are good
for storing charge (and energy).
1
4πε 0
E = ΔV / d = 10000V / 5.0 ×10−3 m = 2.0 ×106 V / m
ε0= 8.85×10-12 C2/N·
/N·m2
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5
A parallel plate capacitor has plates 2.00 m2 in area, separated by a distance of 5.00 mm. A
potential difference of 10,000 V is applied across the capacitor. Determine
the capacitance
the charge on each plate
Solution:
It is very often that more than one capacitor is used in an
electric circuit
We would have to learn how to compute the equivalent
capacitance of certain combinations of capacitors
Since we are dealing with the parallel-plate capacitor,
the capacitance can be found as
Given:
ΔV=10,000 V
A = 2.00 m2
d = 5.00 mm
A
2.00 m2
= 8.85 ×10−12 C 2 N ⋅ m2
d
5.00 ×10−3 m
−9
= 3.54 ×10 F = 3.54 nF
C = ε0
Find:
(
)
(
C2
C1
C3
C2
C5
C3
C1
Once the capacitance is known, the charge can be
found from the definition of a capacitance via charge
and potential difference:
C=?
Q=?
C4
)
Q = C ΔV = 3.54 ×10−9 F (10000V ) = 3.54 ×10−5 C
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a. Parallel combination
C1
V=Vab
+Q1 C2
+Q2
−Q1
−Q2
b
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Analogous formula is true for any number of capacitors,
Ceq = C1 + C2 + C3 + ...
V1 = V2 = V
The total charge transferred to the system
from the battery is the sum of charges of
the two capacitors,
Q1 + Q2 = Q
By definition,
Q1 = C1V1
Thus, Ceq would be
Q Q + Q2 Q1 Q2 Q1 Q2
Ceq ≡ = 1
=
+
=
+
V
V
V
V
V1 V2
Q2 = C 2V2
CCeqeq ==CC11++CC22
22
Parallel combination: notes
Connecting a battery to the parallel
combination of capacitors is equivalent to
introducing the same potential difference for
both capacitors,
a
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16.8 Combinations of capacitors
Q1 + Q2 = Q
V1 = V2 = V
C1
(parallel combination)
It follows that the equivalent capacitance of a parallel
combination of capacitors is greater than any of the
individual capacitors
C2
23
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6
A 3 μF capacitor and a 6 μF capacitor are connected in parallel across an 18 V
battery. Determine the equivalent capacitance and total charge deposited.
Problem: parallel combination of capacitors
A 3 μF capacitor and a 6 μF capacitor are connected in parallel
across an 18 V battery. Determine the equivalent capacitance
and total charge deposited.
a
Given:
C1
V=Vab
V = 18 V
C1= 3 μF
C2= 6 μF
Find:
Ceq=?
Q=?
+Q1 C2
+Q2
−Q1
−Q2
b
First determine equivalent capacitance of C1 and C2:
C12 = C1 + C2 = 9 μ F
Next, determine the charge
(
)
Q = C ΔV = 9 ×10−6 F (18V ) = 1.6 ×10−4 C
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b. Series combination
V=Vab
c
C2
b
1
C1
−Q1
+Q2
−Q2
A voltage induced in the system from the
battery is the sum of potential differences
across the individual capacitors,
V = V1 + V2
By definition,
Q1 = CV
1 1 Q2 = C2V2
Thus, Ceq would be
1 V V1 +V2 V1 V2 V1 V2
≡ =
= + = +
Ceq Q
Q
Q Q Q1 Q2
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Analogous formula is true for any number of capacitors,
1
1
1
1
= +
+ + ...
Ceq C1 C2 C3
Q1 = Q2 = Q
+Q1
C1
11 11 11
== ++
CCeqeq CC1 1 CC
2 2
26
Series combination: notes
Connecting a battery to the serial
combination of capacitors is equivalent to
introducing the same charge for both
capacitors,
a
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Q1 = Q2 = Q
V1 +V2 = V
1
C1
1
C2
27
(series combination)
It follows that the equivalent capacitance of a series
combination of capacitors is always less than any of the
individual capacitance in the combination
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7
A 3 μF capacitor and a 6 μF capacitor are connected in series across an 18 V
battery. Determine the equivalent capacitance and total charge deposited.
Problem: series combination of capacitors
A 3 μF capacitor and a 6 μF capacitor are connected in series
across an 18 V battery. Determine the equivalent capacitance.
a
+Q1
C1
Given:
V=Vab
c
C2
V = 18 V
C1= 3 μF
C2= 6 μF
−Q1
+Q2
−Q2
b
First determine equivalent capacitance of C1 and C2:
Find:
Ceq =
Ceq=?
Q=?
C1C2
= 2 μF
C1 + C2
Next, determine the charge
(
)
Q = C ΔV = 2 ×10−6 F (18V ) = 3.6 ×10−5 C
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Find electric field energy density (energy per unit volume) in a
parallel-plate capacitor
+
+
+
+
+
+
−
−
−
−
−
−
−
−
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31
−
Q
+
1 ε0 A
( Ed )2 /( Ad )
2 d
and so, the energy density is
q
−
=
+
Thus,
−
V
1
U = CV 2
2
ε0 A
volume = Ad V = Ed
C=
d
u ≡ U / volume = energy density
+
Recall
+
V
1
Q2 1
U = QV =
= CV 2
2
2C 2
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Example: electric field energy in parallelparallel-plate
capacitor
16.9 Energy stored in a charged capacitor
Consider a battery connected to a capacitor
A battery must do work to move electrons
from one plate to the other. The work done
to move a small charge Δq across a voltage
V is ΔW = V Δq.
As the charge increases, V increases so the
work to bring Δq increases. Using calculus
we find that the energy (U) stored on a
capacitor is given by:
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1
u = ε0E2
2
32
8
In the circuit shown V = 48V, C1 = 9μF, C2 = 4μF and C3 = 8μF.
(a) determine the equivalent capacitance of the circuit,
(b) determine the energy stored in the combination by
V
calculating the energy stored in the equivalent capacitance,
Example: stored energy
In the circuit shown V = 48V, C1 = 9μF, C2 = 4μF and C3 = 8μF.
(a) determine the equivalent capacitance of the circuit,
(b) determine the energy stored in the combination by calculating
the energy stored in the equivalent capacitance.
C1
Given:
V = 48 V
C1= 9 μF
C2= 4 μF
C3= 8 μF
Find:
Ceq=?
U=?
C3
C2
C3
First determine equivalent capacitance of C2 and C3:
C23 = C2 + C3 = 12 μ F
Next, determine equivalent capacitance of the circuit
by noting that C1 and C23 are connected in series
Ceq ≡ C123 =
V
C2
C1
C1C23
= 5.14 μ F
C1 + C23
The energy stored in the capacitor C123 is then
(
)
1
1
2
U = CV 2 = 5.14 ×10−6 F ( 48V ) = 5.9 ×10−3 J
2
2
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16.10 Capacitors with dielectrics
A dielectrics is an insulating material (rubber, glass, etc.)
Consider an insolated, charged capacitor
−Q
+Q
V0
+Q
−Q
Insert a dielectric
V
Notice that the potential difference decreases (k = V0/V)
Since charge stayed the same (Q=Q0) → capacitance increases
C=
„
Q0
Q
κQ
= 0 = 0 = κ C0
V V0 κ V0
dielectric constant: k = C/C0
Dielectric constant is a material property
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9