1. Electrostatics
Electric Potential
Recall…
Gravitational
Potential
Energy
or
Elastic Potential Energy
Now…
+
+ + +
+ +
+ +
+
+ +
-
Electric
Potential
Energy
(EPE)
EPE is a type of mechanical energy, like…
++
+
Kinetic Energy (KE) = ½ mv2
Rotational Kinetic Energy (KER) = ½ I2
Gravitational Potential Energy (PEgrav) = mgh
Elastic Potential Energy(PEelast) = ½ kx2
= Total Mechanical Energy (E)
is conserved if there are no non-conservative
forces present (ie friction).
+
m
a
q
h
g

b
Gravitational field
F = mg
W = mgh
∆Ug = -W
x
E
Electric field
F = qE
W= - (Fcos)s
W = qEx
∆Ue = -W
Two points are said to differ in electric potential if work is
done to move a charge from one point to another point in
an electric field.
Work = -Δ Electric Potential Energy
Work is required to move two point charge closer together.
F
r
F
This work is converted to potential energy.
This electric potential energy of two
point charges:
Q1Q 2
U e  k
r
[3] An electron starts from rest 32.5 cm from a
fixed point charge with Q = -0.125 μC How fast
will the electron be moving when it is very far
away?
PEinitial  KEfinal
Q1Q 2 mv 2
k

r
2

2kQ1Q 2
 v
mr


2 8.99 x 109 N  m 2 /C2 1.6 x 10 19 C 1.25 x 10  7 C
v
9.1 x 10  31 kg0.325 m 
v  3.49 x 107 m/s

Is the potential energy per
V
unit charge.
Units ~ J/C
W
EPE

or V 
qo
qo

q
a
Ua
Va 
q
b
Ub
Vb 
q
x
E

EPEB EPEA
W
VBA  VB  VA 

  AB
qo
qo
qo
V 
EPE 
qo
WAB

qo
Electrical Potential can also be described by the terms; potential difference,
voltage, potential drop, potential rise, electromotive force, and EMF
Units for Electric Potential
U  U B  U A  qVB  VA 
Difference in
potential energy
Difference in
electric potential
• Electric potential V has units of Joules/Coulomb
which is defined as a Volt:
1 Volt = 1 Joule/Coulomb
• One Joule is the work done in moving one Coulomb
of charge through a potential difference of one Volt.
• Electric field has units of Newtons/Coulomb or
Volts/meter:
1N/C = 1 J/(m C) = 1 V/m
Potential V is the analog of
height/level/altitude/elevation h.
If a charged particle gains kinetic energy in an electric field, it
loses an equivalent amount of potential energy.
Point B is at a lower
electric potential than is
point A.
Points B and C are at
the same electric
potential.
The gain in kinetic energy depends only on the potential
difference and not the path taken.
Example: Kinetic energy of an electron accelerated
through a potential difference of 6,000
Volts (6 kV):
For a conservative force, the kinetic energy gained or lost is
equal to the difference in potential energy:
19
U  qV  (1.60 10 C )( 6,000 V )  9.6 10
 6,000 eV  6 keV
16
1 electron-volt (eV) is the kinetic energy gained by an
elemental charge accelerated by a potential difference of 1
Volt.
1 eV = 1.602 x 10-19 J
J
[1] An electron acquires 7.45 x 1017 J of kinetic energy when
it is accelerated by an electric field from plate A to plate B.
What is the potential difference and which plate is at the
higher potential?
A
B
W  ΔKE
W 7.45 x 10 17 J

V
1.60 x 10 19 C
q
V  466 V
E
Plate B has a positive charge
and is at a higher potential.
q
x
x =1.00 cm
q = 1.60 x 1019 C
E = 2000 N/C m = 9.1 x 1031 kg
m
E
[2] a) Find the speed of the
charge at the lower plate.
ΔKE  W
b) Find the potential through
mv 2
2qEx

qEx
v
which the charge moves.
2
m
W  qΔV
qEx  qΔV
2 1.6 x 10 19 C 2000 N/C0.01 m 
v
ΔV   Ex
9.1 x 10  31 kg
ΔV  2000 N/C 0.01 m 
v  2.65 x 106 m/s
ΔV  20V


q
x
x =1.00 cm
q = 1.60 x 1019 C
E = 2000 N/C m = 9.1 x 1031 kg
m
E
d) Find the acceleration of the charge.
F 3.2 x 10 16 N
a 
m 9.1 x 10  31 kg
c) Find the force on the charge
as it moves.
F  qE
or
F  1.6 x 1019 C2000 N/C 
F  3.2 x 10 16 N
a  3.5 x 1014 m/s 2
v 2  v o2  2ax
2
6
v2
2.65 x 10 m/s
a

2x
20.01 m 

a  3.5 x 1014 m/s 2

Make sure that we understand the difference
between Potential and Electric Potential Energy:
E
V (in Volts) = Potential
a property of a certain position in
an Electric Field with or without
charges placed there
EPE (in Joules) = Electric Potential Energy
a property of charges placed at a certain
position in an external Electric Field
+
-
E
V
V
E
d
V   Ed
E
Let E = 100 N/C
d = 10 cm
V  100 N/C  0.10 m 
10 V
10 cm
7.5 V
5d
V
5 cm
2.5 V
0V
0 cm
V  10 V
Using potentials instead of fields can make solving problems
much easier – potential is a scalar quantity, whereas the
field is a vector.
Electric Potential 17
Electric Field:
Q
E  ke 2 (radially outward)
r

ds
Electric Potential:
V  Vb  Va
1 1
 keQ   
 rb ra 
Convention: V=0 at infinite r
Q
V  ke (electric potential)
r
Electric potential at a
distance r from a
positive charge Q
Electric potential at a
distance r from a
negative charge Q
The electric potential due to a point charge
r
+r
k  Q 
V
r
kQ
V
r
V
V
r
k  Q 
V
r
kQ
V
r
+r
Electric Potential 17
For a system of point charges Qi at distances
ri from a point P:
Q1
r1
Qi
V  ke 
ri
i
P
r2
Q2
r4
Q4
r3
Q3
… an algebraic sum of scalars!
q
r
What is the Potential
at this point?
q
Vk
r
Notes:
1) Include the sign of q in your calculation! (+ or -)
2) Potential Difference can also be calculated:
q
q
k
V = V2 – V1  k
r2
r1
3) The equation can also be used for a charged sphere:
++
+
+
+
++ ++
+
r
q
Vk
r
Total charge
Distance from center
4) Electric Potential is a scalar not a vector
V = V 1 + V2 + V 3 + …
(an algebraic sum, not a vector sum)
qi
V

4o i ri
1
[4]What is the electric potential at a distance of
2.5 x 10-15 m away from a proton?
19 C
1.60
x
10
Q
 8.99 x 109 Nm 2 /C2
Vk
r
2.5 x 10 15 m

+q
d
P

 5.8 x 105 V
[5] Find the potential V at point P due to
the four charges.
d
-q
d
-q
d
+q
Web Link: Complex Electric Field
[6] The +4 μC charge is located
at 4 m on the x-axis and the -6
μC is located at +2 m on the yaxis as shown below.
a) Calculate the magnitude and
determine the direction of the electric
fields at the origin due to the +4 μC
charge and due to the -6 μC charge.
b) Calculate the electric potential at the
origin.
c) Calculate the work done to bring a +4
μC from infinity to the origin.
Ans: 13700 N/C, 100o from +ve x axis; -18,000 V; -0.072 J
Example: Approaching a Charged
Sphere
A proton is fired from far
away at a 1.0 mm diameter glass
sphere that has a charge of
q=+100 nC.
What is the initial speed the
proton must have to just reach
the surface of the glass?
K f  U f  Ki  U i
v
2 Kq p qs
m p rs

0 K
q p qs
rs
 12 m p v 2  0
2 Keqs
 1.86 107 m/s
m p rs
Example: Moving Through a
Potential Difference
A proton with a speed of vi = 2.0x105 m/s enters a region of space where
source charges have created an electric potential.
What is the proton’s speed after it has moved through a potential
difference of V=100 V?
K f  qV f  K i  qVi
K f  Ki  qVi  qV f  Ki  qV
1
2
mv f  mvi  qV
1
2
2
2
2e
v f  vi  V  1.44 105 m/s
m
2
What is vf if the proton is
replaced by an electron?
2e
v f (electron)  vi 
V
me
2
 5.93 106 m/s
Electrostatic Precipitator
The electrostatic
precipitator is highly
effective in
removing tiny
particulates (e.g.
carbon and metals)
from the flue gases
of coal-burning
power plants.
• A strong electric field produces ionization of gases entering
the device.
• Most of the tiny particulates present in the flue gas become
negatively charged, and stick to the walls that are at a
positive electric potential.
http://ap-physics.david-s.org
Cathode-Ray Tube (CRT)
–
+
+
–
The electron beam is scanned
in a raster pattern across the
phosphors on the CRT screen.