Chapter 19
Electric Potential Energy
and the
Electric Potential
Van de Graaff generator used as a particle accelerator. An electron is
accelerated vertically starting from rest across a +100,000 V potential difference
produced by a Van de Graaff. Find a) the change in electric potential energy of
the electron, b) the kinetic energy gained by the electron (neglecting gravity)
and c) the final speed of the electron. For fun, compare the result in a) with the
change in the gravitational potential energy of the electron assuming the Van de
Graaff is 1 m tall.
a)  EPEB - EPEA = qVB - qVA = q(VB - VA)
= (-1.6 x 10-19)(105 - 0) = -1.6 x 10-14 J
EPE decreases
VB =
+105 V
b) Conservation of energy: KEB + EPEB = KEA + EPEA
KEB = KEA - (EPEB - EPEA) = 0 - (-1.6 x 10-14)
= 1.6 x 10-14 J = 105 eV = 100 KeV
VA = 0 V
electron
q = -1.6 x 10-19 C , me = 9.11 x 10-31 kg
vA = 0 m/s
Van de Graaff example -- continued
c) KEB = 1/2 mevB2
vB = [2KEB/me]1/2 = [2(1.6 x 10-14)/(9.11 x 10-31)]1/2
= 1.87 x 108 m/s
The speed of light is 3.0 x 108 m/s, so this calculation would tell us that the
electron is traveling at ~ 2/3 of the speed of light! To do part c) correctly, we
really should use Einstein’s Theory of Special Relativity (gives 1.64 x 108 m/s)
Fun calculation: compare EPEB - EPEA from a) with GPEB - GPEA for electron.
GPEB - GPEA = meghB - meghA = meg(hB - hA)
= (9.11 x 10-31)(9.8)(1 - 0)
= 8.93 x 10-30 J
From a), EPEB - EPEA = -1.6 x 10-14 J
(1.6 x 10-14)/(8.93 x 10-30) = 1.79 x 1015 , GPE can be neglected for an electron!!
LHC ring in
Geneva area
proton+proton collisions at 7 TeV
CERN Large Hadron Collider
ALICE setup
ITS
TOF
FMD
PMD
HMPID
TRD
TPC
PHOS
MUON SPEC.
19.3 The Electric Potential Difference Created by Point Charges
A positive test charge q0 is being repelled
by another positive point charge q. Since the
force on q0 is not constant between A and B,
integral calculus must be used to find the
work done by the force acting from A to B,
WAB
kqqo kqqo
=
−
rA
rB
The potential difference between A and B is,
− WAB kq kq
VB − VA =
=
−
qo
rAB rAB
If rB goes to infinity, and defining
VB = 0 at infinity, and letting
rA -> r and VA -> V,
Potential of a
point charge
kq
V=
r
19.3 The Electric Potential Difference Created by Point Charges
Example 5 The Potential of a Point Charge
Using a zero reference potential at infinity,
determine the amount by which a point charge
of 4.0x10-8C alters the electric potential at a
spot 1.2m away when the charge is
(a) positive and (b) negative.
19.3 The Electric Potential Difference Created by Point Charges
(a)
kq
=
r
8.99 ×109 N ⋅ m 2 C 2 + 4.0 ×10 −8 C
1.2 m
= +300 V
V=
(
)(
)
A positive point charge raises the potential.
(b)
V = −300 V
A negative point charge lowers the potential.
19.3 The Electric Potential Difference Created by Point Charges
When two or more charges are present, the potential due to all of the charges
is obtained by adding together the individual potentials.
Example 6 The Total Electric Potential
At locations A and B, find the total electric potential.
19.3 The Electric Potential Difference Created by Point Charges
VA
(8.99 ×10
=
VB
9
)(
) (
)(
)
N ⋅ m 2 C 2 + 8.0 ×10 −-98 C
8.99 ×109 N ⋅ m 2 C 2 − 8.0 ×10 −-98 C
+
= +240 V
0.20 m
0.60 m
(8.99 ×10
=
9
)(
) (
)(
)
N ⋅ m 2 C 2 + 8.0 ×10 −-98 C
8.99 ×109 N ⋅ m 2 C 2 − 8.0 ×10-9−8 C
+
=0V
0.40 m
0.40 m
19.3 The Electric Potential Difference Created by Point Charges
Conceptual Example 7 Where is the Potential Zero?
Two point charges are fixed in place. The positive charge is +2q and the
negative charge is –q. On the line that passes through the charges, how
many places are there at which the total potential is zero?
19.4 Equipotential Surfaces and Their Relation to the Electric Field
An equipotential surface is a surface on
which the electric potential is the same everywhere.
kq
V=
r
Equipotential surfaces for
a point charge (concentric
spheres centered on the
charge)
The net electric force does no work on a charge as
it moves on an equipotential surface.
19.4 Equipotential Surfaces and Their Relation to the Electric Field
The electric field created by any charge
or group of charges is everywhere
perpendicular to the associated
equipotential surfaces and points in
the direction of decreasing potential.
In equilibrium, conductors are always equipotential surfaces.
19.4 Equipotential Surfaces and Their Relation to the Electric Field
Lines of force and equipotential surfaces for a charge dipole.
19.4 Equipotential Surfaces and Their Relation to the Electric Field
The electric field can be shown to
be related to the electric potential.
If you move a distance Δs and
the electric potential changes
by an amount ΔV, the component
of the electric field in the
direction of Δs is the negative
of the ratio of ΔV to Δs,
or more compactly,
ΔV
E=−
Δs
ΔV/Δs is called the potential gradient.
19.4 Equipotential Surfaces and Their Relation to the Electric Field
Example 9 The Electric Field and Potential
Are Related
The plates of the capacitor are separated by
a distance of 0.032 m, and the potential difference
between them is VB-VA=-64V. Between the
two equipotential surfaces shown in color, there
is a potential difference of -3.0V. Find the spacing
between the two colored surfaces.
19.4 Equipotential Surfaces and Their Relation to the Electric Field
E=−
Δs = −
ΔV
− 64 V
= +
= 2.0 ×103 V m
Δs 0.032 m
ΔV
− 3.0 V
−3
=−
=
1
.
5
×
10
m
3
E
2.0 ×10 V m
The gradient equation can be simplified for a parallel plate capacitor
with voltage V across it and with plates separated by
distance d as:
E = V/d