Transcript Electric Potential, Energy, and Capacitance

Electric Potential, Energy, and
Capacitance
Objective: TSW understand, transfer
and apply energy concepts to electric
fields and charges by solving problems
involving electric fields and forces.
Let’s revisit energy concepts for a gravitational field. A 250
gram baseball is thrown upward with an initial velocity of
25m/s. What is the maximum height reached by the ball?
v = 0m/s
E0  W  E f
K Ug
1 2
mv  mgh
2
2
v
h
2g
v = 25m/s
EARTH
(25) 2
h
2(10)
h  31.3m
Let’s make a parallel comparison of Gravitational Potential
Energy (Ug) to Electric Potential Energy (UE).
U g  mgh
Units:
U E  qEd
Units:
N
kg   m  Nm  J
kg
N
C   m  Nm  J
C
UE = Energy that a charge has due to its position in an
electric field.
Now we will do an example:
An electron is released from rest in an electric field of
2000N/C. How fast will the electron be moving after traveling
30cm?
E0  W  E f
v=?
_
v = 0m/s
_
30cm
UE  K
1 2
qEd  mv
2
2qEd
v
m
2(1.6 1019 )( 2000)(.3)
v
9.111031
m
v  1.45 10
s
7
A new quantity is defined called the Electric Potential
Difference. It is the electric potential energy difference per
unit charge between two points in an electric field.
Electric Potential Energy
Electric Potential Difference 
Charge
UE
V 
q
qEd
V 
q
V  Ed
The Electric Potential Difference can also
called the Electric Potential, the Potential,
or the voltage.
Units:
J
 volt  V
C
Remember Energy and voltage are scalars, so
you don’t have to deal with vectors (direction)
The electric potential energy can now be written in
terms of the electric potential.
U E  qEd
V  Ed
U E  qV
Check the units qV = (C)(J/C) = J
Let’s do an example using this new concept:
The potential difference between two charge plates is 500V.
Find the velocity of a proton if it is accelerated from rest from
one plate to the other.
E W  E
0
High
Low
Potential
+
Potential
-
+
-
+
+
-
+
-
+
-
500V
Positive charges move from high to low potential
Negative charges move from low to high potential
f
UE  K
1 2
qV  mv
2
2qV
v
m
2(1.6 10 19 )(500)
v
1.67 10  27
5 m
v  3.110
s
Let’s revisit a gravitational field again:
The gravitational potential energy if the field is constant is
given by:
U g  mgh
But, what if the field is not constant. In other words the
gravitational potential energy of a mass located far away
from the earth.
m
r
The gravitational potential energy of mass m is given by:
M Em
U g  G
r
Notice r is not squared
The gravitational potential energy at
infinity is zero
Let’s use this new equation to do an example:
With what velocity must a rocket be fired in order to escape
the earth’s gravitational field? (Neglect air resistance)
E0  W  E f
K Ug  0
1 2
M Em
mv  G
0
2
r
1 2
M Em
11
24
mv  G
2
(
6
.
67

10
)(
5
.
98

10
)
2
r
v
6.38 106
2GM E
v
m
v  11000
r
s
Let’s make a parallel comparison of Gravitational Potential
Energy (Ug) to Electric Potential Energy (UE) when dealing
with individual masses or charges.
The gravitational potential energy between two masses
is given by:
M Em
U g  G
r
The electric potential energy between two masses is
given by:
q1q2
UE  k
r
Example:
E0  W  E f
U E  K1  K 2
q1q2 1 2 1 2
k
 mv  mv
r
2
2
q2
k
 mv 2
r
2
kq
v2 
rm
Two 40 gram masses each
with a charge of -6µC are
20cm apart. If the two
charges are released, how
fast will they be moving when
they are a very, very long way
apart. (infinity)
kq2
(9  109 )(6  10 6 ) 2
m
v

 6.36
rm
(.2)(. 04)
s
Let’s use the electric potential energy between two charges
to derive an equation for the electric potential (voltage) due
to a single point charge.
The electric potential energy is given by:
q2P
r
q1
q1q2
UE  k
r
Let’s remove charge q2 and consider the electric
potential (voltage) at point P.
kq1q2
UE
q1
r
V

k
q
q2
r
To find the potential due to more than one point charge simply
add up all the individual potentials:
i
q
V  k
r
i
Example 1: The electron in the Bohr model of the atom can exist at only certain
orbits. The smallest has a radius of .0529nm, and the next level has a
radius of .212nm.
a) What is the potential difference between the two levels?
b) Which level has a higher potential?
q
V k
r
e
V1  k
r1
r1
r2
19
1.6 10
V1  (9 10 )
 27.2V
9
.0529 10
19
1
.
6

10
V2  (9 109 )
 6.79V
9
.212 10
+e
9
r1 is at a higher potential.
potential diff  V  27.2  6.79  20.4V
Example 2
What is the electric potential at the center of the square?
45º
45º
r
r
r
2k
V V 
qq2
kr rr
 0.10 
 6 6

10

5

2
r

9 10  10  
9.01
 
10 
V V 9 910

071
.
.
071


m
r  .071
65 J
10
3410
V V 16..27
C
2
r
r
V  k
i
2
qi
r
Vtotal  1.27 106  (1.27 106 )  6.34 105  6.34 105
J
Vtotal  1.27 10
C
6
Example 3: A proton is moved from the negative plate to the positive plate of a
parallel-plate arrangement. The plates are 1.5cm apart, and the electric field is
uniform with a magnitude of 1500N/C.
a) How much work would be required to move a proton from the negative to the
positive plate?
b) What is the potential difference between the plates?
c) If the proton is released from rest at the positive plate, what speed will it have
just before it hits the negative plate?
W  U E  qEd
N
W  (1.6 10 C )(1500 )(.015m)
C
W  3.6 10 18 J
19
Example 3: A proton is moved from the negative plate to the positive plate of a
parallel-plate arrangement. The plates are 1.5cm apart, and the electric field is
uniform with a magnitude of 1500N/C.
b) What is the potential difference between the plates?
V  Ed
N
V  (1500 )(.015m)
C
J
V  22.5
C
Example 3: A proton is moved from the negative plate to the positive plate of a
parallel-plate arrangement. The plates are 1.5cm apart, and the electric field is
uniform with a magnitude of 1500N/C.
c) If the proton is released from rest at the positive plate, what speed will it have
just before it hits the negative plate?
Use conservation of energy
E0  W  E f
UE  K
1 2
qV  mv
2
2qV
v
m
2(1.6 10 19 )( 22.5)
v
1.67 10  27
m
v  6.57 10 4
s
Example 4:
Compute the energy necessary to bring together the charges in the
configuration shown below:
Calculate the electric potential
energy between each pair of
charges and add them together.
qq12q1qq332
UU13
2312kk
rr
66 6
666


(
4

10
)(

4

10
(
4

10
)(
4

10
))
99
9
UU13
(910
10 ))
2312(9
.2
UU13
72JJ
231200..72
U total  .72 J  (.72 J )  (.72 J )
U total  .72 J
Capacitors
Two parallel conducting plates
used to store charge, which in turn
stores energy
Charge flows through
the circuit until the
capacitor is charged at
which point no more
charge flows.
Capacitance
The amount of charge stored per
volt of potential difference.
Q
C
V
C = Capacitance (C/V = Farad (F)).
Q = The stored charge (C).
V = The potential difference
(voltage) between the plates (V).
Capacitance based on the physical
characteristics of the capacitor
C
0 A
d
Є0 = Permittivity of free space = 8.85 x 10-12
A = Surface area of one plate.
d = distance between plates.
Example 1: Find the Charge stored in the
capacitor.
Q
C
V
Q  CV
12V
20μF
C
Q  (20 10
)(12V )
V
Q  2.4 10  4 C
6
Capacitors in Series
1
1
1
1



 ........
Cs C1 C2 C3
1
1

Cs
i Ci
Example 2: Find the total charge stored in
each the capacitor.
FirstNow
calculate
thetotal
totalcharge
capacitance.
find the
5μF
12V
10μF
30μF
1Q
1
C 
CsV i Ci
Q 1CV 1 1
1
  6 
Q C(s 3.051010F )(30
12V )
5
C

3
.
0

Q  s3.6 10 FC
In a series circuit the charge stored on each capacitor is
the same and the voltage is split.
Capacitors in Parallel
C p  C1  C2  C3  ......
C p   Ci
i
Example 3: Find the total charge stored in
all the capacitors.
Now find
calculate
the
First
the total
total charge.
capacitance.
12V
5μF
10μF
30μF
Q
CC
p 
V
Q  CV
C
i
i
C p  5  10  30
Q  (45 10 6 F )(12V )
C p  454F
Q  5.4 10 C
In a parallel circuit the charge stored on each capacitor
can be different and the voltage must be the same.
Example 4: Find the total capacitance of all
the capacitors.
First find the total capacitance
of the parallel capacitors
12μF
C p  8  4  12F
12V
8μF
4μF
Now, combine this value in
series with 12μF capacitor.
1
Ctotal
1
1


12 12
Ctotal  6 F
Energy Stored in a Capacitor
2
1
Q
1
2
U C  QV 
 CV
2
2C 2
Units: Joules (J)