Download Electric Potential Difference

p. 22
Electric Potential Difference
Remark. The term difference is used to mean the difference between a final value and an initial value:
difference = final value – initial value
Differences are often denoted with the Greek letter . Thus V  V final  Vinitial . The word change means the
same thing as difference. Thus V  change in V .
Example
A charge of +2.82 C sits in a uniform electric field of 12.0 N/C directed at an angle of 60 above the +x axis.
The charge moves from the origin (point A) to the point (1.40 m, 0) (point B) on the x-axis.
a.
b.
c.
d.
a.
Find the force exerted on the charge by the electric field.
Find the work done on the charge by the electric field as the charge moves from A to B.
Find the change in the charge’s electric potential energy as it moves from A and B.
Find the electric potential difference between points A and B.
y
E
F  qE
  2.82 C 12.0 N/C 
F  33.8 N at 60 above the +x axis.
A
60
x
B
b.
WAB  Fs cos 
  33.8 N 1.40 m  cos60
WAB  23.7 J
c.
WAB  EPE A  EPE B
so
EPE A  EPE B  23.7 J
Is this the answer? Why not?
EPE  EPEB  EPEA (Final value - Initial value) . Thus EPE = -WAB and EPE = -23.7 J.
d.
V 
EPE
q0
23.7 J
2.82 C
V  8.40 volts

We see that the electric potential at A is higher than the electric potential at B.
p. 23
Remark. When subject solely to electric forces, the acceleration of a positive charge will always be directed
from a higher electric potential to a lower electric potential.
In general, V 
 WAB
.
q0
Recall that the existence of an electric force requires the existence of a test charge. However, it is assumed the
electric field has an existence independent of the test charge, even though a test charge was (temporarily)
introduced to define it.
*Similarly, the existence of electric potential energy requires the existence of a test charge. However, it is
assumed that the electric potential has an existence independent of the test charge, even though a test charge
was (temporarily) introduced to define it.
Electric force and electric potential energy require the presence of a test charge. The electric field and
the electric potential can exist in space without a test charge.
The Electron Volt
Definition One electron volt is the change in electric potential energy of an electron when the electron moves
through a potential difference of one volt.
Note that the electron volt (eV) is a unit of energy.
1eV = 1.60 x 10-19 J.
Conservation of Energy
Changes in electric potential energy (if any) must be taken into account when solving problems using
conservation of energy.
Example
Two rectangular copper plates are oriented horizontally with one directly above the other. They are separated by
a distance of 25 mm. The plates are connected to the terminals a 5.0 volt flashlight battery. The positive plate
(the one at the higher electric potential) is at the bottom; the negative plate (the one at the lower electric
potential) is at the top.
5.0 volts
+
If an electron is released from rest from the upper plate, with what speed will it strike the lower plate?
Use conservation of energy.
p. 24
2
Ebottom  12 mvbottom
 EPE bottom
2
Etop  12 mvtop
 EPE top
Ebottom  Etop , vtop  0
1
2
2
mvbottom
 EPE bottom  EPE top
1
2
2
mvbottom
 (EPE bottom  EPE top )
1
2
2
mvbottom
 qV
(Note that EPE = qV)
V = Vbottom – Vtop = +5.0 volts and q = -e so
1
2
2
mvbottom
   e  5.0 V 
1
2
2
mvbottom
 5.0 eV
 J 
5.0 1.60 1019 C  
C

m
2
19
8.0 10 kg  ms2

 8.78  1011 m 2 / s 2
31
9.1110 kg

2
bottom
v
2
vbottom

vbottom  9.37 105 m/s
The Electric Potential Difference Due to a Point Charge
Let q be a source charge and q0 be a small positive test charge. As the test
charge moves from point A to point B the electric field of the source charge
does work on the test charge. Let rA be the distance from the test charge to the
source charge when the test charge is at point A; let rB be the distance from the
test charge to the source charge when the test charge is at point B. Using
calculus it can be shown that
WAB 
kqq0 kqq0

rA
rB
Note that if q is positive, WAB is positive if rA < rB, as in the figure.
Using the result V  VB  VA  
WAB
we get
q0
VB  VA 
This is Equation 19.5 in the text.
kq kq

rB rA
B
A
q0
q
p. 25
Since only potential differences are important we may set the electric potential equal to zero at any convenient
location. When dealing with point charges it is convenient to set the electric potential equal to zero at infinity. If
we let rB   and VB  0 in Equation 19.5 we get
V 
kq
r
where we have dropped the subscript “A” on V and r. This is Equation 19.6 in the text.
One important aspect of the electric potential is that it is a scalar, not a vector. Like the electric field, the
electric potential obeys the Principle of Superposition. However, the Principle of Superposition is much easier
to apply to the electric potential: since they are scalars, electric potentials add arithmetically. Electric fields, on
the other hand, add vectorially.
Example
Two positive charges sit in an (x, y)-coordinate system. The first one
has charge q1 = 0.40 C and sits at (- 0.30 m, 0). The second one has
charge q2 = 0.30 C and sits at (0, +0.30 m). Find the electric
potential at the origin.
V1 
V1 
kq1
,
r1
V2 
8.99  109
V  V1  V2 ;
q2
kq2
r2
N  m2
C2
 0.40 C
0.30 m
V1  1.2  10 volts
4
y
x
q1
V2 
8.99  109
N  m2
C2
 0.30 C
0.30 m
V2  0.90  104 volts
V  2.1  104 volts
Example
a. Find the electric potential energy of the system of two charges shown in the figure above.
b. Find the electric potential energy of the system if a third charge q3 = -0.10 C is placed at the origin.
a. First, remove all charges from the system (take them all “to infinity”). The electric potential of the system is
then zero. Move charge q1 to the point (-0.30 m, 0) on the x-axis. Since the electric potential of the system
was zero, this first charge has zero electric potential energy (EPE1 = 0). Next move charge q2 to the point (0,
+0.30 m) on the y-axis. Now the electric potential of the system is not zero since charge q1 is already
present. First calculate the potential due to q1 at the position of q2:
kq1
2
2
; r   0.30 m    0.30 m   0.42 m
r
2
8.99  109 NCm2  0.40 C
V
 8.6  103 volts
0.42 m
V
EPE 2  q2V

  0.30 C  8.6  103 volts
EPE 2  2.6 mJ

p. 26
b. Now we bring in the third charge q3 “from infinity”. We calculate the electric potential energy of this third
charge at the origin. We do this as before, using EPE3 = q3V, where V is the electric potential at the origin.
But this value was calculated in the previous example! Hence
EPE3  q3V

 (0.10 C) 2.1  10 volts
EPE3  2.1 mJ
4

Total EPE  EPE1  EPE 2  EPE 3
 0  2.6 mJ  2.1 mJ
EPE = 0.50 mJ
The total EPE is just the work required to assemble the system of charges.