Ch. 24
Electric Potential Energy
Ch. 24
Warmup 05
•
Electric fields produce forces; forces do work
•
Since the electric fields are doing work,
they must have potential energy
•
The amount of work done is the change in
the potential energy
•
The force can be calculated from the charge
and the electric field
q
E
s
Electric Potential
Energy
•
If the path or the electric field are
not
straight lines,
we can get the change in energy by integration
•
Divide it into little steps of size
d
s
•
Add up all the little steps
d
s
J
JIT – charge free
to move in field
Warmup 05
•
Just like for electric forces, the electric potential
energy is always proportional to the charge
•
Just like for electric field, it makes sense to divide
by the charge and get the
electric potential V
:
The Electric Potential
•
Using the latter formula is a little tricky
•
It looks like it depends on which path you take
•
It doesn’t, because of conservation of energy
•
Electric potential is a scalar; it doesn’t have a direction
•
Electric potential is so important, it has its own unit, the volt (V)
•
A volt is a moderate amount of electric potential
•
Electric field is normally given as volts/meter
[V] = [E][s]=N
●
m/C=J/C=volt=V
Calculating the Electric Potential
To find the potential at a general point
B:
•
Pick a point
A
which we will assign potential 0
•
Pick a path from
A
to
B
•
It doesn’t matter which path, so pick the simplest possible one
•
Perform the integration
Example: Potential from a uniform electric field
E:
•
Choose
r
= 0 to have potential zero
E
V
high
V
low
•
Equipotential lines are perpendicular to
E
-field
•
E
-field lines point from high potential to low
•
Positive charges have the most energy at high
potential
•
Negative charges have the most energy at low potential
+
-
JIT
Ans (i) b (ii) a
Quick Quiz 24.2
The labeled points in the figure are on a series of
equipotential surfaces associated with an electric field.
Rank (from greatest to least) the work done by the
electric field on a positively charged particle that moves
from A to B, from B to C, from
C to D, and from D to E.
Ans
B to C,
C to D,
A to B,
D to E
Solve on Board
Solve on Board
JIT – like example 24.2
1.
It is a scalar quantity – that makes it easier to calculate and work with
2.
It is useful for problems involving conservation of energy
Why Electric Potential is useful
A proton initially at rest moves from an initial point with
V
= 0 to a
point where
V
= - 1.5 V. How fast is the proton moving at the end?
E
V
=0
V
= -1.5 V
+
•
Find the change in potential energy
•
Since energy is conserved, this must be counter-
balanced by a corresponding increase in kinetic energy
MCAT Practice Problem
T
Two parallel conducting plates are separated by a distance d.
One plate carries a charge +Q and the other charge carries a
charge of –Q. The voltage difference between the plates in 12 V.
If a +2
C charge is released from rest at the positive plate, how
much kinetic energy does it have when it reaches the negative
plate?
Warmup 05
W
a
r
m
u
p
0
6
The Zero of the Potential
We can only calculate the
difference
between the electric potential
between two places
•
This is because the zero of potential energy is arbitrary
•
Compare
U
=
mgh
from gravity
•
There are two arbitrary conventions used to set the zero point:
•
Physicists: Set
V
= 0 at
•
Electrical Engineers: Set
V
= 0 on the Earth
•
In circuit diagrams, we have a specific symbol
to designate something has
V
= 0.
V
= 0
Anything attached
here has
V
= 0
Potential From a Point Charge
q
•
Integrate from infinity to an arbitrary distance
r
•
For a point charge, the equipotential
surfaces are spheres centered on the charge
•
For multiple charges, or for continuous
charges, add or integrate
Calculating Potentials is Straight-Forward
q
q
q
q
Four charges
q
are each arranged symmetrically
around a central point, each a distance
a
from that
point. What is the potential at that point?
A) 0
B)
2
k
e
q
/
a
C)
4
k
e
q
/
a
D) None of the above
JIT – Like example 24.3B.
What is the change in potential energy
if now bring in a 5
th
charge Q from infinity to the central point.
JIT Quick Quiz 24.3
Ans (i) c, (ii) a
Ans B
Ans A
Equipotential Lines Are Like
Topographical Maps
•
Regions of high potential are like
“mountains”
•
For positive charges, they have a
lot of energy there
•
Regions of low potential are like
“valleys”
•
For positive charges, they have
minimum energy there
•
Electric fields point down the slope
•
Closely spaced equipotential
lines means big electric field
Understanding Equipotential Lines
In the graph below, what type of charge is at
X
, and what at
Y
?
A)
Positive, both places
B) Positive at X, negative at Y
C)
Negative at
X
, positive at
Y
D) Negative, both places
0
+1
+2
+3
+4
-1
-2
-3
-4
potentials in kV
X
Y
•
Positive charges don’t want to climb the high
mountain at
Y
•
Must be positive charge repelling
them!
•
Positive charges want to flow into
low valley at
X
•
Must be negative charge attracting them!
•
Electric fields are perpendicular to equipotential surfaces
Conductors and Gauss’s Law
•
Conductors are materials where charges are free to flow in response to
electric forces
•
The charges flow until the electric field is neutralized in the conductor
Inside a conductor,
E
=
0
•
Draw any Gaussian surface inside the
conductor
In the
interior
of a conductor,
there is no charge
The charge all flows to the surface
Electric Field at Surface of a Conductor
•
Because charge accumulates on the
surface of a conductor, there can be
electric field just outside the conductor
•
Will be perpendicular to surface
•
We can calculate it from Gauss’s Law
•
Draw a small box that slightly penetrates
the surface
•
The lateral sides are small and have no flux through them
•
The bottom side is inside the conductor and has no electric field
•
The top side has area
A
and has flux through it
•
The charge inside the box is due to the surface charge
•
We can use Gauss’s Law to relate these
Where does the charge go?
A hollow conducting sphere of outer radius 2 cm and inner radius 1 cm
has
q
= +80
nC of charge put on it. What is the surface charge
density on the
inner
surface? On the
outer
surface?
A) 20
nC/cm
2
B)
5
nC/cm
2
C)
4
nC/cm
2
D) 0
E) None of the above
cutaway
view
1 cm
2 cm
80
nC
•
The Gaussian surface is entirely contained in the
conductor; therefore
E
= 0 and electric flux = 0
•
Therefore, there can’t be any charge on the inner surface
•
From the symmetry of the problem, the charge will be
uniformly spread over the outer surface
The electric field:
•
The electric field in the cavity and in the conductor is zero
•
The electric field outside the conductor can be found from Gauss’s Law
Ans B
Solve on
Board
Serway 24-34
Solve on
Board
Conductors and Batteries
A
battery
or
cell
is a device that creates a fixed
potential difference
•
The circuit symbol for a battery looks like this:
•
The long side is at higher potential
•
It is labeled by the potential difference
1.5 V
•
A conductor has zero electric field inside it
•
Therefore, conductors
always
have constant potential
•
A wire is a thin, flexible conductor: circuit diagram looks like this:
•
A switch is a wire that can be connected or disconnected
What is the potential at point
X
?
A) 11 V
B) -11 V
C) +10 V
D) – 10 V
E) +8 V
F) -8 V
0 V
– 1 V
+ 8 V
open switch
closed switch
Conducting Spheres
•
Given the charge
q
on a conducting sphere of radius
R
, what is the potential everywhere?
•
Outside the sphere, the electric field is the same as
for a point charge
•
Therefore, so is the potential
•
Inside, the potential is constant
•
It must be continuous at the boundary
q
R
JIT – Figure 24.21
Sample Problem
q
2
Two widely separated conducting spheres, of radii
R
1
=
1.00 cm and
R
2
= 2.00 cm, each have 6.00
nC of charge
put on them
. What is their potential? They are then
joined by an electrical wire. How much charge do they
each end up with, and what is the final potential?
q
1
•
After connections, their potentials must be equal
Warmup 07
JIT
Ans A
Electric Fields near conductors
q
2
q
1
•
The potential for the two spheres ended up the same
•
The electric fields at the surface are
not
the same
•
The more curved the surface is, the higher the electric field is there
•
A sharp point can cause charged particles to spontaneously be shed into
air, even though we normally think of air as an insulator [ionize air]
•
Called “Corona discharge”
The Lightning Rod
•
Rain drops “rubbing” against the air can cause a separation of charge
•
This produces an enormous electric field
•
If electric field gets strong enough, it can cause breakdown of
atmosphere
•
Put a pointy rod on top of
the building you want to
protect
•
Coronal discharge drains
away the charge near the
protected object
•
Lightning hits somewhere
else
The Van de Graff Generator
•
Hollow conducting sphere, insulating belt, source of electric charge
•
Source causes charge to move to the belt
•
Belt rotates up inside sphere
•
Charge jumps to conductor inside sphere
•
Charge moves to outside of sphere
•
Since all the charge is on the outside of the sphere,
process can be repeated indefinitely.
-
slide 10 (warmup 6). slide 23 (line charge white board problem, "A long straight wire..."),
The relationship between electric potential, energy, and fields. Learn how forces and work relate to potential energy, and how to calculate electric potential. Discover the importance of electric potential in physics and its practical applications.
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Presentation Transcript
Electric Potential Ch. 24 Electric Potential Energy Warmup 05
Electric Potential Energy JJIT charge free to move in field = Electric fields produce forces; forces do work Since the electric fields are doing work, they must have potential energy The amount of work done is the change in the potential energy The force can be calculated from the charge and the electric field U If the path or the electric field are not straight lines, we can get the change in energy by integration Divide it into little steps of size ds Add up all the little steps F s F s = W W = U E q s = = F E q ds E s q = E s dU q d = E s U q d
The Electric Potential U = E s q d Just like for electric forces, the electric potential energy is always proportional to the charge Just like for electric field, it makes sense to divide by the charge and get the electric potential V: V U q U = = E s qV V d Using the latter formula is a little tricky It looks like it depends on which path you take It doesn t, because of conservation of energy Electric potential is a scalar; it doesn t have a direction Electric potential is so important, it has its own unit, the volt (V) A volt is a moderate amount of electric potential Electric field is normally given as volts/meter [V] = [E][s]=N m/C=J/C=volt=V N C V m =
Calculating the Electric Potential V d E s = B = E s V V d B A A To find the potential at a general point B: Pick a point A which we will assign potential 0 Pick a path from A to B It doesn t matter which path, so pick the simplest possible one Perform the integration Example: Potential from a uniform electric field E: Choose r= 0 to have potential zero ( ) 0 Equipotential lines are perpendicular to E-field E-field lines point from high potential to low Positive charges have the most energy at high potential Negative charges have the most energy at low potential V low V high r r = = E r E s = d r E s 0 V d E + 0 -
JIT Ans (i) b (ii) a
Quick Quiz 24.2 The labeled points in the figure are on a series of equipotential surfaces associated with an electric field. Rank (from greatest to least) the work done by the electric field on a positively charged particle that moves from A to B, from B to C, from C to D, and from D to E. Ans B to C, C to D, A to B, D to E
JIT like example 24.2 Example -. (a) Calculate the speed of a proton that is accelerated from rest through a potential difference of 120V. (b) Calculate the speed of an electron that is accelerated through the same potential difference. Solve on Board Ex-. An electron moving parallel to the x-axis has an initial speed of 3.7 x 106 m/s at the origin. Its speed is reduced to 1.4 x 105 m/s at the point x = 2.0 cm. Calculate the potential difference between the origin and this point. Which point is at a higher potential? Solve on Board
Why Electric Potential is useful 1. It is a scalar quantity that makes it easier to calculate and work with 2. It is useful for problems involving conservation of energy A proton initially at rest moves from an initial point with V = 0 to a point where V = - 1.5 V. How fast is the proton moving at the end? Find the change in potential energy V =0 V = -1.5 V ( ) )( = 1.5 V 19 q V = ( 1.602 10 = 2.4 10 = e U E ) + 1.5 V C 19 J Since energy is conserved, this must be counter- balanced by a corresponding increase in kinetic energy 1 2 K U mv = = 2 U v m 1.67 10 2 1.5 V ( ) 19 2 2.4 10 J = 2.87 10 m /s = 8 2 2 = 2 27 kg v = 17 km/s
MCAT Practice Problem TTwo parallel conducting plates are separated by a distance d. One plate carries a charge +Q and the other charge carries a charge of Q. The voltage difference between the plates in 12 V. If a +2 C charge is released from rest at the positive plate, how much kinetic energy does it have when it reaches the negative plate?
The Zero of the Potential V d E = s We can only calculate the difference between the electric potential between two places This is because the zero of potential energy is arbitrary Compare U = mgh from gravity There are two arbitrary conventions used to set the zero point: Physicists: Set V = 0 at Electrical Engineers: Set V = 0 on the Earth In circuit diagrams, we have a specific symbol to designate something has V = 0. Anything attached here has V = 0 V = 0
Potential From a Point Charge e k q r Integrate from infinity to an arbitrary distance r = 2 E r q ( ) U r = = 0 r r r r dr r e k q r e k q r ( ) r = = = = = | | | | E s E dr k q V d e 2 e k q r = V For a point charge, the equipotential surfaces are spheres centered on the charge For multiple charges, or for continuous charges, add or integrate k q V r e k dl e k dA = = = e i V V e k dV r = r r V i i
Calculating Potentials is Straight-Forward q q q q Four charges q are each arranged symmetrically around a central point, each a distance a from that point. What is the potential at that point? A) 0 B) 2keq/a C) 4keq/a D) None of the above JIT Like example 24.3B. What is the change in potential energy if now bring in a 5th charge Q from infinity to the central point.
JIT Quick Quiz 24.3 Ans (i) c, (ii) a
CT-1 -Two test charges are brought separately into the vicinity of a charge +Q. First, test charge +q is brought to point A, a distance r from +Q. Next, +q is removed and a test charge +2q is brought to point B a distance 2r from +Q. Compared with the electrostatic potential of the charge at A due to Q, that of the charge at B is A. greater. B. smaller. C. the same. Ans B
CT - 2 - Two test charges are brought separately into the vicinity of a charge +Q. First, test charge +q is brought to a point a distance r from +Q. Then this charge is removed and test charge q is brought to the same point. The electrostatic potential energy of which configuration is greater: A. +q B. q C. It is the same for both. Ans A
Equipotential Lines Are Like Topographical Maps Regions of high potential are like mountains For positive charges, they have a lot of energy there Regions of low potential are like valleys For positive charges, they have minimum energy there Electric fields point down the slope Closely spaced equipotential lines means big electric field
Understanding Equipotential Lines In the graph below, what type of charge is at X, and what at Y? A) Positive, both places C) Negative at X, positive at Y Positive charges don t want to climb the high mountain at Y Must be positive charge repelling them! Positive charges want to flow into low valley at X Must be negative charge attracting them! Electric fields are perpendicular to equipotential surfaces B) Positive at X, negative at Y D) Negative, both places potentials in kV -1 +1 0 -2 -3 +2 X Y -4 +4 +3
Conductors and Gausss Law Conductors are materials where charges are free to flow in response to electric forces The charges flow until the electric field is neutralized in the conductor Inside a conductor, E = 0 Draw any Gaussian surface inside the conductor = = E n = dA 0 q in 0 0 E In the interior of a conductor, there is no charge The charge all flows to the surface
Electric Field at Surface of a Conductor Because charge accumulates on the surface of a conductor, there can be electric field just outside the conductor Will be perpendicular to surface n We can calculate it from Gauss s Law Draw a small box that slightly penetrates the surface The lateral sides are small and have no flux through them The bottom side is inside the conductor and has no electric field The top side has area A and has flux through it The charge inside the box is due to the surface charge We can use Gauss s Law to relate these = = E n EA A E = q A in = n E = q = A EA in 0 E 0 0
Where does the charge go? A hollow conducting sphere of outer radius 2 cm and inner radius 1 cm has q = +80 nC of charge put on it. What is the surface charge density on the inner surface? On the outer surface? A) 20 nC/cm2 B) 5 nC/cm2 C) 4 nC/cm2 D) 0 E) None of the above The Gaussian surface is entirely contained in the conductor; therefore E = 0 and electric flux = 0 Therefore, there can t be any charge on the inner surface From the symmetry of the problem, the charge will be uniformly spread over the outer surface q A ( ) 4 2 cm The electric field: The electric field in the cavity and in the conductor is zero The electric field outside the conductor can be found from Gauss s Law 80 nC 1 cm 2 cm 80 nC 20 nC r cutaway view = = E r = = 2 5 nC/cm 2 2 0
CT - 5. Which of the following is true? A.The electric field inside a charged insulating sphere must be zero. B.The electric field inside a charged conducting sphere must be zero. C.The charge on a conducting spherical shell will always be equally distributed on the inner and outer surface regardless of the presence of other charges in the vicinity of the shell. D.Two of the above E. Three of the above Ans B
Serway 24-34 Solve on Board
Ex. Serway 24-36. A long, straight wire is surrounded by a hollow metal cylinder, the axis of which coincides with the wire. The wire has a charge per unit length of , and the cylinder has a charge per unit length of 2 . From this information, use Gauss's law to find (a) the charge per unit length on the inner and outer surfaces of the cylinder and (b) the electric field outside the cylinder a distance r from the axis. Solve on Board
Conductors and Batteries A conductor has zero electric field inside it Therefore, conductors always have constant potential A wire is a thin, flexible conductor: circuit diagram looks like this: A switch is a wire that can be connected or disconnected = = E s 0 V d closed switch open switch A battery or cell is a device that creates a fixed potential difference The circuit symbol for a battery looks like this: The long side is at higher potential It is labeled by the potential difference 1.5 V 0 V What is the potential at point X? A) 11 V B) -11 V D) 10 V E) +8 V 1 V 3 V C) +10 V F) -8 V 1 V 9 V + 8 V X
Conducting Spheres Given the charge q on a conducting sphere of radius R, what is the potential everywhere? Outside the sphere, the electric field is the same as for a point charge Therefore, so is the potential Inside, the potential is constant It must be continuous at the boundary q e k q r e k q r = 2 E r = V out R R R for for k q r k q R r r e = V e k q R e = Sphere V JIT Figure 24.21
Sample Problem q1 q2 Two widely separated conducting spheres, of radii R1 = 1.00 cm and R2 = 2.00 cm, each have 6.00 nC of charge put on them. What is their potential? They are then joined by an electrical wire. How much charge do they each end up with, and what is the final potential? ( ( 10 m After connections, their potentials must be equal k q k q = 1 q q + k q = e R i V i i V = 2700 V 2 )( ) 8.988 10 N m /C 6 10 C 9 2 2 9 = = V 5390 V ) 1 2 = 2q q = = = 3600 V V V 1 2 e e 1 2 = 1 2 12 nC 1 cm 2 cm 2 q = q = 4 nC 8 nC 1 2
JIT Ans A
Electric Fields near conductors q1 q2 The potential for the two spheres ended up the same The electric fields at the surface are not the same k q k q R R ( 1 E R ( ) ( ) 2 2 1 2 e E R R k q The more curved the surface is, the higher the electric field is there e k q R e k q R = 1 2 e e ) ( ) = = 1 2 E R 1 2 2 2 2 2 1 2 1 E R 2 2 k q R R R = 1 1 = 1 e 2 2 1 Very strong electric field here A sharp point can cause charged particles to spontaneously be shed into air, even though we normally think of air as an insulator [ionize air] Called Corona discharge
The Lightning Rod Rain drops rubbing against the air can cause a separation of charge This produces an enormous electric field If electric field gets strong enough, it can cause breakdown of atmosphere + Put a pointy rod on top of the building you want to protect Coronal discharge drains away the charge near the protected object Lightning hits somewhere else + + + + + + + + + + + + + + + + + + + +
The Van de Graff Generator Hollow conducting sphere, insulating belt, source of electric charge Source causes charge to move to the belt Belt rotates up inside sphere Charge jumps to conductor inside sphere Charge moves to outside of sphere Since all the charge is on the outside of the sphere, process can be repeated indefinitely. -
