Download work done by the electric force

Electric Potential and Electric Energy
Chapter 17
Potential Energy
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Let's go back to junior physics for a second
:)
What is gravitational potential energy?

Energy that depends on an object's mass
and its position relative to some point

i.e. To calculate someone's potential
energy relative to the surface of the Earth
you'd need mass, g and height above the
surface
Electric Potential Energy

The idea of electric potential energy is
similar to that of gravitational potential
energy

Electric potential energy for a charge is
calculated based on the magnitude of the
charge and its position relative to some
point
Gravitational vs Electric Potential
Energy (p. 504)
Caption:
(a) Two rocks are at the same height. The larger rock
has more potential energy. (b) Two charges have the same electric
potential. The 2Q charge has more potential energy.
Recall point charges in electric fields
(ch 16)

Let's say you have an electric field of
magnitude 4500 N/C pointing toward the
right
E
+

If you place a proton in that field, what is the
magnitude and direction of the force acting
on that proton?

F=qE=7.2 x 10-16 N

Right
The proton in the electric field


So since the force acting on the proton is
toward the right, it will accelerate toward
the right
What will happen to the proton's kinetic
energy and electric potential energy?

Kinetic Energy will increase

EPE will decrease (conservation of
energy)
Two charged plates (capacitor)
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-
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Let's say we've got two
charged plates that are
separated by a small
distance (this is a
capacitor)
The E-field points from left
to right
Two charged plates (capacitor)
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E-field between
Two charged plates
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A proton between these
two plates would move
towards the negative
plate (right)
An electron between
these two plates would
move towards the
positive plate (left)
Two charged plates (capacitor)
p. 503
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E-field between
Two charged plates
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A proton has the highest
potential energy when it's
near the positive plate
An electron has the
highest potential energy
when it's near the negative
plate
Potential for pos/neg charges

By convention, the positive plate is at a
higher potential than the negative plate

Positively charged objects move from
higher potential to lower potential (i.e.
towards negative plate)

Negatively charged objects move from
lower potential to higher potential (i.e.
towards positive plate)
Potential for positive/negative
charges
Electric Potential (V)

Electric Potential, V, is the potential energy
per unit charge


Unit is Volts (1 V= 1J/1 C)
If a point charge, q, has an electric potential
energy at some point a, then the electric
potential is
¿q> ¿
¿PEa>
V= ¿

V= PE/q
¿
¿
Electric potential and Potential
Energy

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The change in potential energy of a charge,
q, when moved between two points a and b
ΔPE = PEb-PEa=qVba
Sample Problem p. 505

An electron in a television set is accelerated
from rest through a potential difference
Vba=+5000 V

What is the change in PE of the electron?

What is the speed of the electron as a
result of the acceleration?

Repeat for a proton that accelerates
through a potential difference of -5000 V
Change in PE of electron

ΔPE = Peb-PEa=qVba

ΔPE = qVba=(-1.6 x 10-19 C)(5000 V)

ΔPE = -8 x 10-16 J

Potential Energy was lost!
What is the speed of the electron as
a result of the acceleration?

Conservation of Energy!

The amount of PE lost, must be equal to
the amount of KE gained!

KE= 8 x 10-16 J=0.5mv2

V=4.2 x 107 m/s
For the proton

ΔPE = qVba=(1.6 x 10-19 C)(-5000 V)
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ΔPE = -8 x 10-16 J (Same as electron)
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Velocity is less because speed is greater
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V=9.8 x 105 m/s
Potential Difference

Since potential energy is always measured
relative to some other point, only
differences in potential energy are
measurable

Potential Difference is also known as
voltage
Potential Difference

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In order to move a charge between two
points a and b, the electric force must do
work on the charge
Vab=Va-Vb= -Wba/q

The potential difference between two
points a and b is equal to the negative of
the work done by the electric force to
move the charge from point b to point a,
divided by the charge
Sample Problem p. 522 #2

How much work is needed to move a proton
from a point with a potential of +100 V to a
point where it is -50 V?
Break it down

We're moving the proton from +100 V to -50
V

Therefore point A is +100 V, point B is -50
V

We're looking for the work done by the field

-Wba= qVab=q(Va-Vb)
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
-Wba= (1.6 x 10-19 C)(100V -(-50V))
Wba= -2.4 x 10-17 J
Back to the parallel plates!
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E-field between
Two charged plates
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For two parallel plates, the
relationship between
electric field and electric
potential is below
E=Vba/d

d is the distance
between the plates
The electron volt

The electron volt is another unit for energy

1 ev= 1.6 x 10-19 J

Problem: A proton has 2 MeV of kinetic
energy, how fast is it moving?
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2 x106 eV= 3.2 x 10-13 J= 0.5mv2
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V= 1.96 x 107 m/s
Section 17-Equipotential Lines
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Equipotential lines are used to represent
electric potential
Equipotential lines are always perpendicular
to electric field lines
Equipotential Lines (p. 507)

Equipotential lines
(green) are perpendicular
to the electric field lines
(red)
17-5 Electric Potential due to Point
Charges (p. 509)
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The electric potential at a distance r from a
single point charge q is : V=kQ/r

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Potential is zero at infinity
The potential near a positive charge is large
and decreases toward zero at large
distances
Electric potential p. 509

The potential near a negative charge is
negative and increases toward zero at large
distances
Bringing charges together
Ex 17-3 p. 509

What minimum work is required by an
external force to bring a charge q = 3.00
microC from a great distance away to a
point 0.500 m from a charge Q= 20.0
microC?
Analyze the problem

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Basically, we're taking the charge q from a
place of zero potential, to a place of nonzero
potential
Use our trusty equation:Vab=Va-Vb= -Wba/q
Figure out the work done
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The charge is coming from infinity, so Va=0
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What is Vb?
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
Vb=KQ/r=(9x109 Nm2/C2)(20x106C)/0.500m

Vb= 360,000 V
Wba= -q(Va-Vb)=-(3.00x10-6C)(0-360000V)
W= 1.08 J
Electric potential of multiple charges
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Electric fields are vectors, but electric
potential is a scalar!
When determining the electric potential at a
point you can just add the electric potential
from each charge, just be sure to include the
correct sign of the charge when calculating
potential
Example
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Calculate the electric field at a point midway
between a -0.5 microC charge and a -0.8
microC charge that are separated by 0.50
m.
For the -0.5 microC charge, E= 72000 N/C
left
For the -0.8 microC charge, E= 115,200 N/C
right
Therefore E is 43200 N/C right
Electric Potential

Calculate the electric potential at a point
midway between a -0.5 microC charge and
a -0.8 microC charge that are separated by
0.50 m.
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For the -0.5 microC charge,
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V=kQ/r= (9x109 Nm2/C2)(-0.5 x 10-6 C)/0.25m
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V= -18000 N/C
Electric Potential
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For the -0.8 microC charge,
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V=kQ/r= (9x109 Nm2/C2)(-0.8 x 10-6 C)/0.25m
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V= -28800 V
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Total V= -46800 V

This is much easier! No directions...just
make sure you include the sign!
Section 17-7- Capacitance
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E-field between
Two charged plates
A capacitor stores electric
charge and consists of two
conducting objects that are
placed next to each other
but not touching
Capacitance p. 513
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• If a voltage is applied to a
capacitor (i.e. connected to a
battery), then it becomes
charged
• Amount of charge for each
plate: Q  CV
• C= Capacitance of the capacitor
(different for each capacitor)
• Unit for C is farad (F)
Capacitance of the Capacitor
A
C 0
d
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d
•A= Area of plates
•If A increases, C
increases
•d= distance between the
plates
•If d increases, C
decreases
•ε0 = 8.85 x 10-12 C2/Nm2
(This is the permitivity of
free space)
Storage of Electrical Energy
A charged capacitor stores electric energy
2
1
1
1Q
2
U  Stored Energy  QV  CV 
2
2
2 C
Sample Problem p. 524 #41
A 7.7 µF capacitor is charged by a 125 V
battery and then is disconnected from the
battery. When this capacitor (C1) is
connected to a second, uncharged capacitor
(C2), the voltage on the first drops to 15 V.
What is the value of C2? (Charge is
conserved)
Solve the Problem
• For the first capacitor:
6
4
Q  CV  (7.7 x10 F )(125V )  9.625x10 C
• When the capacitors are connected, the
voltage on the first one is 15 V. That means
the new charge on C1 is:
6
4
Q  CV  (7.7 x10 F )(15V )  1.155x10 C
Solving the problem
• What happens to the rest of the charge?
• It must be on capacitor 2 because charge is
conserved
Q2  9.625x104 C  1.155x104 C  8.47 x104 C
• Since the two capacitors are connected, the
voltage for the second one must also be 15 V
Q2 8.47 x10 4 C
C

 5.6 x10 5 F
V
15V
Connected Capacitors
• Capacitors can be
connected in series or
parallel
• When capacitors are
connected in parallel, the
equivalent capacitance is
the sum
• The voltage across each
capacitor is the same
Capacitors in Parallel
V1  V2  V3
Ceq  C1  C2  C3  .....
Q  Q1  Q2  Q3  ...
Capacitors in Series
If the capacitors are connected in series, the
equivalent capacitance is given by the
following expression
1
1
1
1



 ...
C C1 C2 C3
Capacitors in Series
For capacitors in series, the total voltage must
equal the sum of the voltages across each
capacitor
Vtotal  V1  V2  V3
The charge on each capacitor is the same as
the charge on the equivalent capacitor for
capacitors in series
Q1  Q2  Q3
Sample Problem
•What is the equivalent
capacitance for this
combination of capacitors?
•C2 and C3 are connected in
parallel
•Combine them into one
capacitor
C1 = 12 µF
C2 = 25 µF
C3 = 10 µF
•C23=C2 + C3 = 35 µF
Simplify the Combination
C23 and C1 are connected
in series
1
1
1


C123 C1 C23
1
1
1


C123 12µF 35µF
C1 = 12 µF
C23 = 35 µF
C123  8.94F
Sample Problem Continued
•How much charge is
stored on each
capacitor?
•Q=CV
•V1= 50 V (this is the
voltage across C1)


Q1  C1V1  12 x106 F 50V   6.0 x104 C
Sample Problem Continued
C1 and C23 are connected
in series, therefore the
charge on C23 is the same
as the charge on C1
4
Q1  Q23  6.0 x10 C
Sample Problem Continued
C2 and C3 are connected
in parallel, therefore:
V23  V2  V3
Q23  C23V23
Q23 6.0 x104 C
V23 

 17.14 V
6
C23 35 x10 F
Sample Problem Finished!
The charge on C2 is:


Q2  C2V2  25x106 F 17.14V   4.29 x104 C
The charge on C3 is:


Q3  C3V3  10 x10 F 17.14V   1.71x10 C
6
4