W9_parallel_resonance.eps

Week 3: Potential Energy and Potential 107

(so thatU 0 is zero, if you prefer). We remain free to choose a different zero, however, in any problem
where doing so is computationally convenient.

Using the relations above, it is easy to show that the potential energy of two point charges is:

U=

kq 1 q 2

|~x 1 −~x 2 |

(140)

which again looks very much like that for gravity as might be expected.

One important advantage of working with the potential energy is that it is ascalar. To find the
total potential energy of a collection of charges, we justadd it up pairwise:

Utot=^1

2

∑

i 6 =j

kqiqj

|~xi−~xj|

(141)

Note that in this sum the 1→2 interaction is countedtwice, once asq 1 q 2 and once asq 2 q 1. We
only wish to count it once, so we divide the result by 1/2. Another wayto deal with this issue is to
order the sum so that we simply never do a pair twice:

Utot=

∑

i<j

kqiqj

|~xi−~xj|

(142)

This stands for “sum over allqjand allqisuch thati < j” which excludes all the self-energyi=j
terms. Good thing, too, since they are all infinite!

3.2: Potential

The good thing about potential energy is that it is a scalar and easierto evaluate than thevector
force or field. However, it isn’t terribly easy! It is still a two-body interaction term and requires
us to do a nasty double sum (that becomes an even nastier double integral) when we have a large
collection of charges.

A couple of weeks ago we introduced the idea of thefieldto eliminate two body computations
for electric force and to give us the comfort of an apparent action-at-a-distancecauseof the electric
force. Let us do exactly the same thing here. We will define the electrostaticpotentialto be a scalar
field of “potential energy per unit charge” that is thecauseof a charged particle placed in it having
a potential energy.

The formal definition of the potential is that it is the potential energy of a small test chargeq 0
interacting with all the other charges that create the potential, per unit test charge, in the limit
that this small test charge vanishes:

V(~x) = limq

0 →^0

U(~x)

q 0 (143)

Note that this strange-seeming condition ensures that the test charge itself doesn’t perturb the
charge distribution that produces the potential.

The SI units for potential are:

1 Volt =

1 Joule

1 Coulomb

(144)

If we apply this rule compute the potential at~xproduced by a point chargeqat the origin of
coordinates, we get:

V(~x) = limq

0 →^0

1

q 0

kqq 0

|~x− 0 |

=

kq

r

(145)