When an object is
dropped on earth,
the work done by the
gravitational force is
equal to the change in
gravitational potential
energy:
W = mghinitial - mghfinal
Similarly, when a charge
is moved in an electric
field, the work done on
the charge by the field is
equal to the change in
electrical potential energy:
W = EPEinitial - EPEfinal.
The electric force on the
charge in the field is F = q0E.
It is useful to express this
work per-unit-charge, by
dividing both sides of the
previous equation by q0.
W/q0 = EPEinitial/q0 EPEfinal/q0
The value EPE/q0
is called electric
potential or
the potential, or
potential difference.
The electric potential V at a
given point is the electric
potential energy EPE of a small
test charge q0 situated at that
point divided by the charge
itself:
V = EPE/ q0
The unit is the joule/coulomb = volt (V).
The electric potential
difference between two
points is:
Vfinal - Vinitial =
EPEfinal/q0 - EPEinitial/q0 = -W/q0
or,
∆V = ∆(EPE)/q0 = -W/q0
We only measure the
differences of electrical
potential V and
electrical potential
energy EPE, not their
absolute amount.
Ex. 1 - The work done by the
electrical force on a test charge
(q0 = +2.0 x 10-6 C) as it moves
from A to B is WAB = +5.0 x 10-5 J.
(a) Find the difference ∆(EPE)
between these points.
(b) Determine the potential
difference, ∆V, between the points.
A positive charge accelerates
from a region of higher
electric potential energy
(or higher potential) toward
a region of lower electric
potential energy (or lower
potential).
A negative charge
accelerates from a
region of lower
potential toward a
region of higher
potential.
Ex. 2 - Three points, A, B, C,
are located on a horizontal line.
A positive test charge is released
from rest at point A and accelerates
toward point B. Upon reaching B, the
test charge continues to accelerate
toward C. Assuming that only motion
along the line is possible, what will a
negative test charge do when it is
released from rest at B?
An automobile battery has a
potential that is 12 V higher at
the positive terminal than at the
negative terminal, V+ - V- =
12 V. The battery takes the
positive charge that reaches the
negative terminal and moves it
to the positive terminal through
the battery.
This requires work to
be done to the charge
by the battery; this
energy comes from the
battery’s reserve of
chemical energy.
Ex. 3 - Determine the number
of particles, each carrying a
-19
charge of 1.60 x 10 C, that
pass between the terminals
of a 12-V car battery when
a 60.0-W headlight burns for
one hour.
The word “volt” is also used in
the context of an energy unit
called the electron volt. One
electron volt is the change in
potential energy of an electron
(q0 = 1.60 x 10-19 C) when the
electron moves through a
potential difference of one volt.
This change in potential
energy is q0•∆V =
(1.60 x 10-19 C)x(1.00 V) =
-19
1.60 x 10
J.
-19
So, 1 eV = 1.60 x 10
J.
One MeV = 10+6 electron volts.
One GeV = 10+9 electron volts.
The total energy of an
object includes all
forms of kinetic and
potential energy.
2
Etotal = 1/2•mv +
2
1/2•Iw + mgh +
2
1/2•kx + EPE
Ex. 4 - (a) A particle has a mass of
m = 1.8 x 10-5 kg and a positive
charge of q0 = +3.0 x 10-5 C. It is released
from rest at point A and accelerates
horizontally until it reaches point B.
Angular speed w is zero. The only force
acting on the particle is an electric force,
and the electric potential at A is 25 volts
greater than that at B. (VA - VB = 25 V)
What is the speed vB of the particle when
it reaches point B?
(b) If the same
particle had a
negative charge and
were released from
rest at B, what would
be its speed vA at A?
When a positive test charge
+q0 is moved away from a
positive point charge +q,
work is done by the force
between the charges. The
magnitude of this force is
given by Coulomb’s law:
F = kq0q/r2.
Since the distance from the
charge +q varies, so does the
force F. Therefore, the work is
not simply F x d.
Integral calculus can be used to
find the work WAB when the
particle +q0 is moved from point
A to point B.
WAB = kqq0/rA - kqq0/rB
Therefore:
VB - VA = kq/rB - kq/rA
At an infinite distance rB,
VB and kq/rB become
zero. The equation
becomes:
V = kq/r.
This V = kq/r is the
potential of a point charge
compared to the potential at
infinity. This V has no value in
the absolute sense, it refers
to the difference from the
potential an infinite distance
away.
When q is positive, V is also
positive, indicating that a
positive charge raises the
potential around it above
zero. Conversely, a negative
q decreases the potential
below the zero reference
value.
Ex. 5 - Using a zero
reference potential at infinity,
determine the amount by
which a point charge of
4.0 x 10-8 C alters the electric
potential at a spot 1.2 m
away when the charge is
(a) positive and (b) negative.
When two or more
charges are present,
the potential due to
all the charges is
obtained by adding
together the individual
potentials.
Ex. 6 - Charges of +8.0 x 10-9 C
and -8.0 x 10-9 C are separated by
a distance of 0.80 m. What
is the total electric potential at
(a) a point exactly halfway between
the two charges (0.40 m from each)
and (b) between the two charges,
0.20 m from the positive charge
and 0.60 m from the negative.