ELECTRIC POTENTIAL ENERGY

1. You are already familiar with the concept of gravitational potential energy. When you
lift a weight, you have to do work against the downwards pull of the Earth. That work
is stored as potential energy. Suppose a force acts on something and displaces it by .
Then the work done is . The work done in going from point to point (call it )
is then got by adding tog
ab
F ds
F⋅ds a b W
􀁇 􀁇
􀁇 􀁇
ether the little bits of work, = . The change in
potential energy is defined as . Remember always that we can only
define the potential at a point if the force is co
b
ab a
b a ab
W Fds
U U U W
U
⋅
Δ = − = −
∫
􀁇 􀁇
nservative.
2. The electrostatic force is conservative and can be represented by a potential. Let us see
how to calculate the potential. So consider two charges separated by a distance as below.
Let us take the point very far from the fixed charge , and the unit charge at the point
to be at a distance from . Then the work you did in bringing the unit charge from infinity
t
a q b
R q
2
0 0 0
o is, ( ) 1 1 1 1 1 . Since the charges
4 4 4
repel each other, it is clear that you had to do work in pushing the two charges closer
together. So where
R R
ab
R W qE dr drq q q
πε r πε R πε R ∞ ∞
= − = − = − ⎛⎜⎝ − ∞⎞⎟⎠= − ∫ ∫
did the negative sign come from? Answer: the force you exert on the
unit charge is directed the charge , i.e. is in the negative direction. This is why
( ) . Now ( ) - ( )
towards q
F⋅ds = −qE dr ΔU =U R U ∞
􀁇 􀁇
0
0
1 . If we take the potential at to be
4
zero, then the electric potential due to a charge at the point is ( ) 1.
4
Remember that we know how to calculate the force given the poten
q
R
q r U r q
r
πε
πε
= ∞
=
2
0
1 2
tial: . Apply
this here and you see that 1 , (which also has the correct repulsive sign).
4
3. From the above, it is quite obvious that the potential energy of two charges , is,
F dU
dr
F q
r
q q
U
πε
= −
=
( ) 1 2 ( ) 1 2
0
1 . Compare this with the formula for gravitational energy, .
4
What is the difference? From here, you can see that the gravitational force is always
negative (which means attrac
r q q Ur G mm
πεr r
= =−
tive), whereas the electrostatic force can be both attractive
or repulsive because we have both + and - charges in nature.
3. The electric potential (or simply potential) is the energy of a unit charge in an electric field.
So, in our MKS units, the unit of potential is 1 Joule 1 Volt. Another useful unit is
Coulomb
"electron volt" or eV. The definition is:
One electron - volt = energy gained by moving one electron charge through one Volt
=
( 19 ) 19
3
6
1.6 10 1 1.6 10
It is useful to note that 1 Kev = 10 (kilo-electron-volt)
1 Mev = 10 (million-electron
C V J
eV
eV
= × − × = × −
9
12
-volt)
1 Gev = 10 (giga-electron-volt)
1 Tev = 10 (tera-electron-volt)
4. Every system seeks to minimize its
eV
eV
potential energy (that is why a stone falls down!).
So, positive charges accelerate toward regions of lower potential, but negative charges
accelerate toward regions of higher potential. Note that only the potential difference
matters - even if a charge is placed in a region where there is a high potential, it will not
want to move unless there is some other place where the potential is higher/lower.
5. Given a system of charges, we can always compute the force - and hence the potential -
that arises from them. Here are some important general statements:
a)Potentials are more positive in regions which have more positive charge.
b)The electric potential is a scalar quantity (a scalar field, actually).
c)The electric potential determines the force through F = − , and hence the electric
field because .
d)The electric potential exists only because the electrostatic force is conservative.
dU
dr
F=qE
1 2
1 0 1
6. To compute the potential at a point, the potentials
arising from charges 1, 2, N must be added up:
1 . Here is
4
the distance of the i'th charge fr
N N
i
N i i
i i i
V V V V V q r
= πε = r
⋅ ⋅ ⋅
= + + ⋅ ⋅ ⋅ =Σ = Σ
( ) 1 2 1 3 2 3
0 12 0 13 0 23
om the point where
the potential is being calculated or measured. As an
example, the potential from the three charges is:
V 1 1 1 .
4 4 4
r q q q q q q
πε r πε r πε r
= + +
1 q
2 q 12 r
13 r
23 r
3 q
R
x
y
z
P
λds
2 1
1 2
0 1 2 0 12
7. Let us apply these concepts to the dipole system considered earlier. With two charges,
1 . We are particularly interested in the situation
4 4
where
P
V V V q q q r r
r r rr
r d
πε πε
⎛ − ⎞ −
= + = ⎜ + ⎟=
⎝ ⎠
>> 2
2 1 12
2 2
0 0
. from the diagram you can see that cos and that . Hence,
cos 1 cos . So we have calculated the potential at any with such
4 4
little difficulty. Note that 0 at
r r d rr r
V q d p
r r
V
θ
θ θ
θ
πε πε
− ≈ ≈
≈ =
= .
2
π
θ =
P
θ
1 r
2 r
d r
+q
−q
2 1 r −r =dcosθ
8. Now let us calculate the potential which comes
from charges that are uniformly spread over a
ring. This is the same problem as in the previous
lecture, but simpler. Give the small amount
2
0
2 2 2
0 0
of
potential coming from the small amount of charge
some name, 1 . Then obviously
4
1 1 .
4 4
dq ds dV dq
r
V dV dq q
r R z
λ
πε
πε πε
= =
= = =
+ ∫ ∫
+ + + + + + + + + + + + + + +
d Gaussian surface