Download CH 16 – Electric Potential

Electric Potential
This chapter covers electric potential energy, electric potential, and electrical capacitance.
Work done by an electric field
A charge in an electrical field experiences a force given by F = qE. If the charge moves,
then the field does work if the field has a component parallel to the displacement. If E is
uniform and has a component in the direction of the displacement, then the work done is
W  Fx x  qE x x
If E makes an angle  with respect to the displacement, then this can be written as
W  Fx cos   qEx cos 
The electrical force is conservative, so we can associate a potential energy with this force.
If the work done by E is positive, then the potential energy decreases; if it is negative,
then the potential energy increases. Specifically,
PE  W  qE x x
According to the work-energy theorem, W = KE. This means that the total energy
(kinetic plus potential) is constant.
KE  PE  0
Or,
KE  PE  cons tant
Example:
A proton is released from rest between oppositely charged parallel plates where the
electric field strength is E = 500 N/C. What is the speed of the proton after it has moved
1 cm?
W  qEx  ( 1.6 x10 19 C )( 500 N / C )( 0.01m )  8 x10 19 J
KE  1 mv 2  W
2
v
2W
2( 8 x10 19 J )

 3.1x10 4 m / s
 27
m
( 1.67 x10
kg )
1
Electric Potential
The electric potential difference V between two points A and B is defined as the electric
potential energy difference of a charge q between these two points divided by the charge.
V  VB  V A 
PE
q
(unit = J/C =volt = V)
Since PE  qE x x (uniform E), then
V   E x x
Electric potential is a way of characterizing the space around a charge distribution.
Knowing the potential, then we can determine the potential energy of any charge that is
placed in that space. This is similar to the concept of electric field. The electric field is
another way of characterizing the space around a charge distribution. If we know the
field, then we can determine the force on any charge placed in that field. Electric
potential is a scalar quantity (magnitude and sign (+ or -), while electric field is a vector
(magnitude and direction). Electric potential, just like potential energy, is always defined
relative to a reference point (zero potential). The potential difference between two points,
V, is independent of the reference point.
Example:
A uniform field of magnitude E = 200 N/C is
created by two oppositely charged parallel plates,
as shown. What is the potential difference V = VC
– VA between points C and A? The distance from
A to B is 0.3 m and the distance from B to C is 4 m.
C
A
B
Solution:
VCA  VC  V A  VCB  VBA
Since E is perpendicular to the displacement from B to C, then B and C are at the same
potential. So,
VCA  VBA   E x ( x B  x A )  ( 200 N / C )( 0.3m )  60V
The negative sign means that point C is a lower potential than point A.
2
Example:
If the potential difference between the positive and negative plates were 1000 V and the
separation of the plates were 10 cm, what would be the magnitude of the electric field
between the plates?
Solution:
Since V   E x x , then
Ex  
V
( 1000V )

 10,000V / m (  10,000 N / C )
x
0.1m
The positive sign means that Ex points to the right, in the direction of decreasing potential.
Potential Energy of Two Point Charges
qq
The force between two point charges is F  k e 1 2 . This force can do work by
r2
pushing the charges apart (if q1 and q2 have the same sign) or pulling them together (if q1
and q2 have opposite sign). There is a potential energy associated with these two charges.
The change in potential energy is the negative of the work done during the displacement.
Since the force is not constant, then we must calculate this work from the area under the
force versus displacement curve, or by using integral calculus. Potential energy always
depends on the choice of where the potential energy is assumed to be zero. For point
charges, the convention is to assume that PE = 0 when r = . The result is
qq
PE  k e 1 2 .
r
Note that this equation is similar to the force formula except that PE varies inversely with
r instead of r2. Also, PE is a scalar (it can be + or -), whereas force is a vector
(magnitude and direction).
Example:
Two protons are released from rest when they are 1 cm apart. What is their speed when
they have moved very far apart?
3
Solution:
e2
At r = , PE = 0, so PE  PE final  PEinitial  k e
.
r
KE  PE
2
1 mv 2  1 mv 2  k e ( both protons are moving )
e
2
2
r
v
kee 2
( 9 x10 9 )( 1.6 x10 19 )2

 3.7 m / s
mr
( 1.67 x10  27 )( 0.01 )
Electric Potential of a Point Charge
PE
, then the potential of a single point charge is
q
q
V  ke ,
r
Since V 
where it is assumed that V = 0 when r = .
Example:
What is the electric potential 50 cm from a point charge q = 1 x 10-6 C?
Solution:
V  ke
q
1x10 6
 ( 9 x109 )
 1.8x10 4 V  18 kV
r
0.5
Example:
P
What is the electric potential at point P in the diagram
to the right?
5m
3m
4m
Solution:
q1 = 1.5 µC
q
q
V  V1  V2  k e 1  k e 2
r1
r2
( 1.5 x10  6 )
( 1x10  6 )
 ( 9 x10 9 )
(5)
(3)
 2700V  3000V  300 V
 ( 9 x10 9 )
4
q2 = -1 µC
Equipotential Surfaces
If a charge is moved perpendicular to an electric field, then no work is done and there is
no change in potential energy and no change in electric potential. Thus, any surface over
which the electric field is perpendicular is at constant potential and is called an
“equipotential” surface. For a point charge, equipotential surfaces would be spheres with
the charge at the center.
electric field
lines
equipotential
surfaces
Equipotential surfaces for a positive/negative charge pair:
The surface of a charged conductor in equilibrium is an equipotential surface since the
electric field is everywhere perpendicular to the surface. Also, the volume of a conductor
is at constant potential. This is true since V = -Exx and since E = 0 everywhere inside
a conductor.
Capacitance
A capacitor consists of two metal electrodes which can be given equal and opposite
charges. If the electrodes have charges Q and –Q, then there is an electric field between
them which originates on Q and terminates on –Q. There is also a potential difference
between the electrodes which is proportional to Q. The capacitance of the configuration
is defined as
C
Q
V
(unit = C/V = farad = F)
5
The capacitance is a measure of the capacity of the electrodes to hold charge for a given
potential difference.
A geometrical simple capacitor would consist of two parallel metal plates. If the
separation of the plates is small compared with the plate dimensions, then the electric
field between the plates is nearly uniform.
At the end of chapter 15 we saw that the electric field between two oppositely charged
plates is given by E = /0, where  is the charge per unit area ( = Q/A) on the plates.
Also, the potential difference between the plates is V = Ed, where d is the separation of
the plates. Thus, the capacitance is
C
Q
A  0 EA
,


V Ed
Ed
Or
 A
C 0
d
(parallel plate capacitor)
Example:
A capacitor consists of two plates 10 cm x 10 cm with a separation of 1 mm. What is the
capacitance? If the capacitor is charged to a potential difference of 100 V, how much
charge is on the capacitor?
Solution:
 A ( 8.85 x10
C 0 
d
12
C 2 / N  m 2 )( 0.1m ) 2
 8.85 x10 11 F (  88.5 pF )
( 0.001m )
Q  CV  ( 8.85 x10 11 F )( 100V )  8.85 x10 9 C (  8.85 nC )
6
Combinations of Capacitors
Circuits can have multiple capacitors. In the simplest configurations, the capacitors
would be either in parallel, in series, or in a combination of series and parallel.
Capacitors in parallel
In the parallel circuit, the electrical potential across the capacitors is the same and is the
same as that of the potential source (battery or power supply). This is because the
capacitors and potential source are all connected by conducting wires which are assumed
to have no electrical resistance (thus no potential drop along the wires). The two
capacitors in parallel can be replaced with a single equivalent capacitor. The charge on
the equivalent capacitor is the sum of the charges on C1 and C2. (This is the meaning of
equivalent.) Thus,
Q  Q1  Q2
C||V  C1V1  C2 V2
Since V = V1 = V2, then
C||  C1  C2
V
C1
Q1
C2
Q2
V
Q
C||
Capacitors in series
In the series circuit, the potential of the source is the sum of the potentials across C1 and
C2. This is a consequence of conservation of energy. (A test charge moved across the
source must have the same change in potential energy as a test charge moved across C1
and then across C2.) The charges on each capacitor are the same (since each charge
comes from the potential source). Also, the charge on each capacitor is the same as the
charge on the single equivalent capacitor, Cseries. Thus,
V  V1  V2
Q Q1 Q2


C s C1 C 2
Or, since Q = Q1 = Q2,
7
1
1
1


C s C1 C 2
Another way of writing this equation is
Cs 
C1C 2
C1  C 2
Q1
C1
V
C2
Q2
V1
V
Cs
Q
V2
For more than two capacitors in parallel or in series, the addition equations are
C||  C1  C2  C3  ...
1
1
1
1



 ...
C s C1 C 2 C3
Energy Stored in a Capacitor
The energy stored in a charged capacitor is given by
U  1 QV ,
2
where Q is the charge on the capacitor and ∆V is the voltage (potential) across the
capacitor. If we moved a small charge q through a potential difference ∆V, the change in
potential energy would be U = q∆V. The reason for the factor of ½ in the above equation
is because the potential varies from 0 to ∆V as the capacitor is charged, so we must take
the average to find the total work done in charging it.
Using Q = C∆V, we can alternatively write the energy stored as
2
2 1Q
1
U  C( V ) 
2
2 C
8
Example:
A 3-µF and a 6-µF capacitor are connected in parallel to a 12-V battery. Find the charge
on each capacitor, the equivalent parallel capacitance, and the total energy stored.
Solution:
Q1  C1V  ( 3F )( 12V )  36 C  36 x10 6 C
Q2  C 2 V  ( 6F )( 12V )  72 C  72 x10  6 C
C||  C1  C 2  3F  6F  9 F
U  1 C||( V )2  1 ( 9 x10  6 F )( 12V )2  6.48 x10  4 J  648 J
2
2
Example:
The capacitors in the above example are now connected in series to the 12-V battery.
Find the charge on each capacitor, the voltage across each capacitor, the equivalent
parallel capacitance, and the total energy stored.
Solution:
The equivalent capacitance and the charge on the equivalent capacitance are
Cs 
C1C 2
( 3 )( 6 )

 2 F
C1  C 2
36
Qs  C s V  ( 2F )( 12V )  24 C
We note that for series capacitors, Q1 = Q2 = Qs. So, the voltages across the capacitors
are
Q
24C
V1  1 
 8V
C1
3F
V2 
Q2
C2

24C
 4V
6 F
Note that the smaller voltage is across the larger capacitor and vice-versa. Also, note that
V1  V2  12V , as is required.
The energy stored is
U  1 C s ( V )2  1 ( 2F )( 12V )2  144 J
2
2
9
Example:
In the circuit below, find the total equivalent capacitance, the charge on each capacitor,
and the voltage across each capacitor.
C1 =
3 µF
V =
24 V
C2 =
2µF
C3 =
4 µF
Solution:
We will not give the complete solution here, but you should learn how to work it.
Suggestion: Find the parallel equivalent of C2 and C3. This reduces the circuit to two
capacitors in series. Then solve this using the previous example.
10