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Electric Potential
Disclaimer: These lecture notes are not meant to replace the course textbook. The content may be incomplete. Some topics may be unclear. These notes are only meant to be a study aid and a supplement to your own notes. Please report any inaccuracies to the professor.
Work and Potential Energy
Applying a force over a distance requires work:
12
if and are constant
otherwise to move the object from initial point i to final point f
f i
W
W d
F d F d
F s
The work done by a force on an object to move it from point i to point f is opposite to the change in the potential energy:
W = โฮ U = โ ( U f โ Ui )
In other words, if the work expended by the force is positive, the potential energy of the object is lowered. For example, if an apple is dropped from the branch of a tree, the force of gravity does work to move (accelerate actually) the apple from the branch to the ground. The apple now has less gravitational potential energy.
These concepts are independent of the type of force. So the same principal also applies to the electric field acting on an electric charge.
We define the electric potential as the potential energy of a positive test charge divided by the charge q 0 of the test charge.
0
U
V
q
It is by definition a scalar quantity, not a vector like the electric field.
The SI unit of electric potential is the Volt (V) which is 1 Joule/Coulomb. The units of the electric field, which are N/C , can also be written as V/m (discussed later).
Changes in the electric potential similarly relate to changes in the potential energy:
0
U
V
q
So we can compute the change in potential energy of an object with charge q crossing an electric potential difference:
ฮ U = q ฮ V
This motivates another unit for potential energy, since often we are interested in the potential energy of a particle like the electron crossing an electric potential difference. Consider an electron crossing a potential difference of 1 volt:
( )(^ )
ฮ U = q ฮ V = e V ฮ = 1.6 ร 10 โ^19 C 1 V = 1.6 ร 10 โ^19 J = 1 eV
This is a tiny number, which we can define as one electron-volt (abbreviated โeVโ). It is a basic unit used to measure the tiny energies of subatomic particles like the electron. You can easily convert back to the SI unit Joules by just multiplying by the charge of the electron, e.
A common convention is to set the electric potential at infinity (i.e. infinitely far away from any electric charges) to be zero. Then the electric potential at some point r just refers to the change in electric potential in moving the charge from infinity to point r.
ฮ V = V r โ V (^) โโ Vr
The work done by the electric field in moving an electric charge from infinity to point r is given by:
W = โฮ U = โ ฮ q V = โ q V ( r โ V โ)= โ qVr
where the last step is done by our convention. But keep in mind that it is only the differences in electric potential that have any meaning. A constant offset in electric potential or potential energy does not affect anything.
Electric Potential from Electric Field
Consider the work done by the electric field in moving a charge q 0 a distance ds:
dW = F โ
d s = q 0 E โ
d s
The total work done by the field in moving the charge a macroscopic distance from initial point i to final point f is given by a line integral along the path:
0
f
W = q โซ i E โ
d s
This work is related to the negative change in potential energy or electric potential:
Let us choose a radial path. Then E โ
d s = โ E ds since the field points in the opposite
direction of the path. However, if we choose integrating variable d r , then ds = โ dr since r points radially outward like the field. We thus have:
2 2
r r
r r r
V E ds Edr
qdr dr q K Kq Kq K r r r r
โ โ
โ โ (^) โ
Since the electric potential is chosen (and shown here) to be zero at infinity, we can just write for the electric potential a distance r away from a point charge q :
q V r K r
It looks similar to the expression for the magnitude of the electric field, except that it falls off as 1/ r rather than 1/ r^2.
We also could integrated in the opposite sense:
V V V r (^) r d
โ
ฮ = โโ = โ โซ E โ
s
Then E โ
d s = E dr
2 2
r (^) r r
r r (^) r
r
V V d Edr
qdr dr q K Kq Kq K r r r r q V K r
โ โ
โ โ โ
E s
Potential of Many Point Charges
By the superposition principal, the electric potential arising from many point charges is just:
i i i
q V K r
where qi is the charge of the i th^ charge, and ri is the distance from the charge to some
point P where we wish to know the total electric potential. The advantage of this calculation is that you only have to linearly add the electric potential arising from each point charge, rather than adding each vector component separately as in the case of the electric field.
Electric Dipole
Letโs see how to calculate the electric potential at point P due to an electric dipole.
By the superposition principle, the total potential is:
q q V V V K r r
r r V Kq r r
โ +
where r (^) + is the distance from the positive charge to point P, and r (^) - the distance from the negative charge.
Now for large distances, r d โ r โ โ r + โ d cos ฮธ, where d is the separation of the
electric dipole.
2 2
d cos p cos V Kq K r r
where r = r + (^) โ r โand p โก qd is the electric dipole moment.
Potential of Continuous Charge Distributions
Potential between 2 Parallel Plates
Letโs calculate the electric potential difference between 2 large parallel conducting plates separated by a distance d , with the upper plate (denoted โ+โ) at higher electric potential than the lower.
From what we learned by Gaussโs Law and conductors, we know that the electric field
arising from a conductor with a charge density ฯ is 0
E =. It is thus a constant between
the two plates in this example. The electric potential difference is given by a line integral:
E ds
y
d ฮธ rโ x โ q
x
y
z
V E x V E y V E z โ = โ โ โ = โ โ โ = โ โ So the electric field is related to the negative rate of change of the electric potential.
This is a specific manifestation of a more general relation that a force is related to the rate of change of the corresponding potential energy:
F = โโ U ( in one dimension:
dU F dx
For the case of the electric field, F = q E and U = qV , so
q E = โ โ q V โ E = โโ V
Conductors and Electric Potential
Recall that the valence electrons in a conductor are free to move, but that in electrostatic equilibrium they have no net velocity. Another consequence of this is that:
ฮ V = 0 across a conductor
If not, electrons would move from higher to lower potential, and thus not be in static equilibrium. This implies that the surface of the conductor, no matter what shape, is also an equipotential surface. We learned already that the electric field is perpendicular to the surface of a conductor (otherwise charges would accelerate along the surface), and equipotential lines are always perpendicular to the electric field lines.
Example:
Letโs consider as an example 2 conducting spheres connected by a thin conducting wire. One sphere has a smaller radius ( r 1 ) than the other ( r 2 ); and the charges on the two spheres are q 1 and q 2 respectively.
By the above argument, all surfaces are at the same electric potential. Letโs raise the entire system to potential V with respect to a point infinitely far away. The two spheres must have the same potential, so by equating the potential energy of each charged sphere (which is the same as that of a point charge at the center of the sphere) we get:
1 2 1 2 1 1 2 2
q q K K r r q r q r
Equipotential line
Surface of conductor
r 2 q 2
r (^1) q 1
