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ELECTRIC POTENTIAL ENERGY
AND ELECTRIC POTENTIAL
• POTENTIAL ENERGY
• ELECTRIC POTENTIAL
• WORK-ENERGY THEOREM
• CAPACITANCE
• COMBINATIONS OF CAPACITORS
• STORED ENERGY
Written by Dr. John K. Dayton
POTENTIAL ENERGY IN A UNIFORM ELECTRIC FIELD:
WAB  F cos   x
F  qE
WAB  qE x
U AB  WAB
U AB   qE x
In this example a charged particle is moved
from point A to point B in a uniform field by
the electrostatic force. The work done by the
force and the change in potential energy of
the particle can be calculated in the usual
way. Remember, energy is a scalar quantity.
ELECTRIC POTENTIAL:
Electric potential is defined as the electric potential energy per unit
charge.
V
U
q
V 
U
q
In a uniform electric field, where U = -qE x, the change in electric
potential will be V = -Ex.
The SI unit for electric potential is the volt, V. V=J/C.
To find a particle’s change in potential energy, use:
 U  q V
Electric potential is a scalar quantity.
EXAMPLE: What is the change in a particle’s potential energy if it
moves from a position of 100 volts to a position of 150 volts? Let the
particle have an electric charge of -3mC.
Click For Answer
U  qV
U   3mC   150V  100V 
U  150m J
THE ELECTRIC POTENTIAL OF A POINT CHARGE:
q
V k
r
VA B  VB  VA
VA B
q
q
k k
rB
rA
VA B
1 1
 kq   
 rB rA 
If q is a point charge (positive or
negative), then the electric potential a
distance r from q is given by the top
equation. Note, V will be negative if q is
negative.
The difference between the potentials at
two points is also shown.
r
V
V vs r for a
positive charge
V vs r for a
negative charge
r
V
THE SUPERPOSITION OF ELECTRIC POTENTIALS OF
SEVERAL POINT CHARGES:
Let P be a point in space near several point charges such that it is a
distance r1 from q1, r2 from q2, and r3 from q3. Then the net electric
potential at P is:
VP  VP ,1  VP ,2  VP ,3
q1
q2
q3
VP  k  k  k
r1
r2
r3
EXAMPLE: Calculate the electric potential on the x axis at 15.0 cm
produced by two point charges; q1 = -4.0 mC on the origin and q2 = +3.0
mC at 10.0cm on the y axis.
Click For Answer
q2 = +3mC
+
r2 = .1803m
10cm
-
r1 = .15m
15cm
r2 
.1m   .15m 
2
2
 0.1803m
VP  VP ,1  VP ,2
P
This diagram is the
first step in a well
planned solution.
q1 = -4mC
Solution continues on next slide.
continued from previous slide…
kq1 kq2
VP 

r1
r2
VP 
8.99  10
Superposition of point charge potentials
9 Nm 2
C2
   4  10 C 
.15m
VP  9.03  104V
6

8.99  10
9 Nm 2
C2
6

3

10
C

.1803m
Electric potential is a scalar so is much easier to work with
than the electric field. No directions are involved. Don’t
forget to use the sign of the charge when calculating an electric
potential.
THE WORK-ENERGY THEOREM FOR ELECTROSTATICS:
A particle of charge q and mass m
moves under the influence of an
electric field from point A to
point B.
K  U  0

1
2
In General:
mv B2  12 mv A2    q  V A B   0
 V A B  V B  V A
For a single point charge:  V A B
For a uniform field E:  V A B
For an infinite plane of charge, E 

2òo
:
For a charged, infinite conducting sheet,

E :
òo
 1
1
 kqo   
 rB rA 
  E  x B  bA 
 V A B


 xB  x A 
2 òo
 V A B

   xB  x A 
òo
EXAMPLE: A proton is released from rest 5.0 cm from the surface of a
charged sphere of radius 10.0 cm and charge Q = 4.0 mC. What will the
proton’s speed be when it has moved 1 meter?
The proton will move between ri = .15m and rf = 1.15m.
Work-Energy Theorem for
0
 K f  Ki   q V f  Vi  Click
For Answer
Electrostatics
 kQ kQ 
Working Equation with
2
1

0

2 mv f  0  e 
Potentials of a spherical charge
 r
ri 
 f


2ekQ  1 1
vf 
 
m  ri rf
vf 
2 1.6  10
v f  7.21  106 ms
19



Solved for final velocity
C  8.99  10
9 Nm 2
C2
1.67  1027 kg
Final Answer
  4  10 C  
6
1
1 



 .15m 1.15m 
CAPACITANCE AND THE CAPACITOR:
A capacitor is comprised of two charged, conducting bodies maintained
at a potential difference. The charge in the capacitor is actually a charge
separation.
Capacitance is defined as the ratio charge to potential difference within a
capacitor.
Q
C
V
C = capacitance in SI units of coulombs/ volts. This combination is
called the farad, F.
Q = charge on capacitor. One conductor has +Q while the other has -Q.
V = voltage difference between conductors.
A capacitor’s capacitance depends on its size and shape, not on the
charge separation and not on the voltage difference.
THE PARALLEL PLATE CAPACITOR:
The parallel plate capacitor is comprised of
two metal plates that face each other with
each plate connected to a battery terminal.
The inside area of each plate is A and they are
separated by a distance d.
d
A
E
+Q
V
-Q
The plate connected to the positive battery
terminal will have a charge of +Q on its inside
surface. The plate connected to the negative
terminal will have -Q on its inside surface.
The voltage difference between the plates will
be the battery voltage, V. The electric field
between the plate will be uniform given by
E   / òo pointing from the positive plate to the
negative plate.
Q  A  E òo A
V  Ed
Q E òo A
C 
V
Ed
òo A
C
d
EXAMPLE: Calculate the capacitance of a parallel plate capacitor
made of two circular disks of radius 6.0 cm and separated by 0.5mm.
How large a radius should they have if the capacitor is to be 1.0 F?
(a)
òo A òo r 2
C

d
d
8.85  10

C
Click For Answer
(b)
  .06m 
12 C 2
Nm 2
3
2
.5  10 m
C  2.00  1010 F  200 pF
òo A òo r 2
C

d
d
r
dC
 r
òo
3
.5

10
m  1F 

8.85  10
12 C 2
Nm 2
r  4.24  103 m

CAPACITORS CONNECTED IN SERIES:
C1
C2
+
V
Ceq
+
-
V
-
Beginning with a group of capacitors
connected in series, find the single,
equivalent capacitor.
Q
Q
V1 
V2 
C1
C2
In series each capacitor has the same
charge on it: Q1 = Q2 = Qeq = Q.
Q
Q Q


Ceq C1 C2
In series the voltages across each
capacitor add to the battery voltage:
V1 + V2 = Veq = V.
1
1
1


Ceq C1 C2
V  V1  V2
Q
V
Ceq
EXAMPLE: C1 = 4.0 mF and C2 = 6.0 mF are connected in series to a
24V battery. What is the stored charge in C1?
C1
C2
+
V
Ceq
Click For Answer
-
1
1
1
1
1




Ceq C1 C2 4 m F 6m F
+
V
Use series equation
Ceq  2.4 m F
Charge stored in
Qeq  CeqV   2.4 m F  24V   57.6mC C .
eq
Capacitors in series each have the
Q1  Q2  57.6mC
same charge.
CAPACITORS CONNECTED IN PARALLEL:
C1
+
+
-
Ceq
V
C2
Beginning with a group of capacitors
connected in parallel, find the single,
equivalent capacitor.
In parallel each capacitor has the same
voltage across it:
V1 = V2 = Veq = V.
The individual charges add to the charge
on the equivalent capacitor:
Q1 + Q2 = Qeq.
-
V
Q1  VC1 Q2  VC2
Qeq  VCeq
Qeq  Q1  Q2
VCeq  VC1  VC2
Ceq  C1  C2
EXAMPLE: C1 = 4.0 mF and C2 = 6.0 mF are connected in parallel
to a 24.0 V battery. What is the stored charge in C1?
C1
Ceq
+
+
-
V
C2
V
Click For Answer
-
Ceq  C1  C2  4 m F  6m F  10m F
Equation for parallel
Qeq  CeqV  10 m F  24V   240mC
Charge in Equivalent
Capacitor
V1  V2  Veq  24V
Capacitors in parallel
have the same voltage
Q1  C1V1   4 m F  24V   96mC
CAPACITORS CONNECTED IN GENERAL:
In Parallel:
In Series:
Ceq   Ci
1
1

Ceq
Ci
C3
In the diagram C1 and C2 are
not in series; they cannot be
directly combined. C2 and C3
are in parallel and can be
combined. Once C2,3 is known
it can be combined with C1
in series.
C1
C2
+
V
-
ENERGY STORED IN A CAPACITOR:
Capacitors store energy within their electric fields. Assume the average
voltage during the charging process is one-half the final voltage, V/2.
Thus the charge that separates crosses this voltage and the net change in
potential energy is QV/2. This is the energy stored in the capacitor:
U  12 QV  12 CV 2 
Calculate the energy density within a
parallel plate capacitor:
òo A
C
d
Vol  Ad
V  Ed
The final expression is good for any
capacitor.
1
2
Q2
C
2
U
CV
u
 12
Vol
Vol
2
 òo A 

  Ed 
d 
u  12 
Ad
u  12 òo E 2
EXAMPLE: Calculate the equivalent capacitance of the circuit and the
energy stored in each of the original capacitors.
C1
C2,3
C3 = 2mF
C1 = 2mF
+
+
Click For Answer
C2 = 3mF
-
Ceq
V
-
V
+
V
A series of reduced circuits. C2
and C3 can be combined first
because they are in parallel.
Solution Continues on Next Slide
continued from previous slide…
C2,3  C2  C3  3m F  2 m F  5m F
1
1
1
1
1
 


Ceq C1 C2.3 2 m F 5m F
Compute equivalent
capacitor
Ceq  1.4286m F
Qeq  CeqVBat  1.4286m F  24V   34.2857 mC
Q1  Q2,3  34.2857 mC
(Because they are in series)
Q1 34.2857 mC
V1 

 17.1429V
C1
2m F
Q2.3 34.2857 mC
V2,3 

 6.8571V
C2.3
5m F
Distribute charge and
voltage up one circuit
level
continued from previous slide…
V2  V3  V2.3  6.8571V
(Because they are in parallel)
Q2  C2V2   3m F  6.8571V   20.5713mC
Q3  C3V3   2 m F  6.8571V   13.7142 mC
 2 m F 17.1429V   293.879 m J
2
2
1
1
U 2  2 C2V2  2  3m F  6.8571V   70.5297 m J
2
2
1
1
U 3  2 C3V3  2  2 m F  6.8571V   47.0198m J
U1  C V 
2
1 1
1
2
2
1
2
total  411.429 m J
U eq  C V
1
2
2
eq Bat

1
2
1.4286m F  24V 
2
A
N
S
W
E
R
S
 411.429 m J
The total of the individual stored energies must equal the energy
stored in the equivalent capacitor.
End Of Presentation