UCSD Physics 2B
Summer Session
Unit 2 Lecture Notes
Chapter 25
Section 1
Electric Potential
Energy, Work & Fields
Finding the force on one charge due to one, or even several other charges was
fairly straightforward. Vector sums were involved for multiple sources but,
even then, the problem was at least digestible.
Once we began considering continuous charge source distributions, we found it
much easier to define
E=
The Electric Field
F
q
and then use this to find the force on any particular charge once that field had
been calculated:
F = qE
Force on a charge
In this way we didn’t have to drag around all the source charges every time we
did when using Coulomb’s force equation.
Now recall the analogy made between the electric & gravitational fields. We
similarly defined a gravitational field g by calculating the force on an arbitrary
mass and then dividing by that very mass:
F = mg ⇒ g =
F
m
2
Notice that g has the units of acceleration (which happen to be 9.8 m/s at the
surface of the Earth. In a similar way, the Electric Field E represents the force
on a unit charge which, when divided by its mass, is also acceleration.
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Section 2
Electric Potential
Recall that in Mechanics we defined mechanical work as force times distance,
and generalized this, for non-constant force over a path, as the line integral
W = ∫ F • dL
definition of work
where F is the force and dL is differential length increment along the path.
We used this to compute Gravitational Potential Energy of a mass using the
work we do to get it there. Near the earth’s surface where the weight mg is
nearly constant and dL is straight up, we got the formula PE = mgh .
However: This is the negative of the work that the field itself does, since g ( r )
points down while we’re moving the mass up – we work against the field to
increase PE. So let’s use the field vector g and define this potential energy as
(
PE = m − ∫ g ( r ) • dL
)
Note that the mass here is constant and was moved outside the integral. Since
all the effort goes into computing the integral, we can define Gravitational
Potential (rather than the potential energy) by simply dividing out the mass
UG =
PE
= − ∫ g ( r ) • dL
m
where g ( r ) is the field at the point r
Notice two important things:
1. The Potential, like potential energy, is a scalar – not a vector and hence
much easier to work with.
2. The Electric Potential, like its gravitational counterpart, should be pathindependent since both electrostatic and gravitational fields are
conservative (hint: the force equations both have the same mathematical
form.) Hence we can always choose the easiest path available, say, one
where E is always parallel along dL
Since this looks like a vast improvement, let’s define
ΔVab = − ∫ E ( r ) • dL
b
a
Electric Potential Difference
Now the work needed to move a charge from one point to the other is just the
charge times the potential difference:
Wab = qΔVab
Work to move a charge from a to b
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Units of Electric Potential
Force
Newtons
=
the Electric Potential
Charge Coulomb
Newton ⋅ meter
Joules
=
.
has the units of E • L or
Coulomb
Coulomb
Since the Electric Field E has units
1 Volt = 1
Joule
Coulomb
Units of Electric Potential
Note that we can now go back and describe the electric field in alternative units
E=
Newtons Volts
=
Coulomb meter
Remember that the Volt is a unit of Potential, not Energy. Volts are not Joules.
Another useful unit of Energy is the electron volt, the energy gained by an
electron which has moved through a potential difference of one Volt.
Section 3
Calculating Potential Differences
Recall that when calculating the gravitational potential energy of a particular
mass at some height h, we had to set an arbitrary altitude as the zero-point. For
example, this might have been the first floor of a building when calculating the
PE of an elevator on a high floor, but the building itself could have been up a
mountain. In other words, the zero-point of potential is arbitrary, so it’s
reasonable to refer to the Potential Difference between two points in space.
KEY POINT HERE
( know how to do this !!! )
Now we can always calculate the Force F on a charge, electric field E or the
potential V from any of the others i.e., F ⇔ E ⇔ V through the equations
FORCE
F = qE
FIELD
R
E = −∇V
(
POTENTIAL
R
)
V = − ∫ E • dL
∂
∂
∂
Where the gradient operator ∇ = ∂x , ∂y , ∂z or “vector derivative”
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Section 4
Calculating the Potential Difference from the Field
Example:
Point Charge or Sphere
Recall the Electric Field from a spherically symmetric source:
E (r ) =
kq
rˆ
r2
where r̂ points radially away from the center.
Hence, dL becomes dr and the potential difference between a and b is
Vab = − ∫ E ( r ) • dr = − ∫
b
b
a
a
b dr
kq
⎛1 1⎞
ˆ
•
d
=
−
kq
=
kq
r
r
⎜ − ⎟
∫a r 2
r2
⎝b a⎠
Here, we integrate along a path radially outward from charge starting at the
distance r = a and ending at r = b . Hence the field and the differential line
element are always co-linear ( cos θ = 1 ) so that the dot product collapses and
the constants k and q move outside the integral.
Example:
Line Charge or Cylinder
Recall the Electric Field from a cylindrically symmetric source:
E (r ) =
2k λ
rˆ
r
where r̂ is perpendicular to the axis.
The potential difference between a and b is
Vab = − ∫ E ( r ) • dr = − ∫
b
b
a
a
b dr
2k λ
⎛b⎞
= −2k λ ln ⎜ ⎟
rˆ • dr = −2k λ ∫
a r
r
⎝a⎠
Here, we’ve integrated along a path radially away from the axis of the line
charge or cylinder and so, again, the dot product collapses and the constants k
and λ move outside the integral.
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Example:
Infinite Sheet of Charge
Recall the Electric Field from an infinite sheet of charge:
E=
σ
rˆ
2ε 0
where r̂ points perpendicularly away from the sheet.
The potential difference between distances from the sheet a and b is
Vab = − ∫ E ( r ) • dr = − ∫
b
b
a
a
σ
σ
rˆ • dr = −
2ε 0
2ε 0
∫
b
a
dr = −
σ
(b − a )
2ε 0
This time we’ve integrated along a path perpendicular to the plane of the sheet.
Again, the dot product collapsed and the constants moved outside the integral.
Example:
Two Parallel Oppositely Charged Sheets
The Electric Field between two infinite sheets ( E = 0 outside) of separation d is
E=
σ
rˆ
ε0
so between them we simply have Vab =
σd
ε0
Calculating the Field from the Potential Difference
Suppose we are given the electrostatic potential which is defined everywhere by
the expression
V ( x, y, z ) = ( 5V ) − ( 7V ) x + (13V ) y 2 − ( 4V ) z 3
Where the coordinates x, y, and z are dimensionless (units reduced out). Let’s
apply the formula for converting this to an electric vector field.
E ( x, y, z ) = −∇V ( x, y, z )
(
=− ∂
(
∂x
,∂
∂y
,∂
⎡ 50V ) − ( 7V ) x + (13V ) y − ( 4V ) z ⎤⎦
∂z ) ⎣(
2
= − − ( 7V ) , (13V )( 2 y ) , ( −4V ) 3 z 2
(
= +7V, −26Vy, +12Vz 2
3
)
)
5
(
)
This represents the three parts of the electric field vector Ex , E y , Ez .
Note that the first term in the potential function is constant which yields a
electric field. In other words, if the function were just V ( x, y, z ) = 5V , then the
electric field would have to be zero, so that no work is done moving a charge.
For another example, consider the following potential function
V ( x, y, z ) = 6V ( xy 2 z 3 )
Again we apply the gradient operator and obtain
E ( x, y, z ) = −∇V ( x, y, z )
(
=− ∂
∂x
,∂
∂y
(
,∂
∂z
) (6Vxy z )
2 3
= − 6Vy 2 z 3 ,12Vxyz 3 ,18 xy 2 z 2
)
We take the x-derivative of the entire function and stick it in the x-position of
the ordered triple Ex , E y , Ez and follow by putting the y- and z-derivatives in
(
)
their proper places as well. This gives us a vector function which is really three
functions in one – one each for the x- y- and z-directions.
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Section 5
Charged Conductors & Equipotentials
Consider the following sketch of various conductors, joined together with wires
which are also conductors:
Since everything above is a conductor, we can consider the entire assembly as
one conductor, because they are all connected together. Now recall that
1. The electrostatic field inside a perfect (ideal) conductor is zero
2. Any excess charge must reside somewhere on the surface
3. The electric field at the surface is perpendicular
4. A charge can travel anywhere on the conductor force-free
5. In other words, since WORK = − ∫ qE • dL = 0
the potential difference between any two points ΔV = 0 .
DEFINITION:
Anyplace or anything which has the same potential V everywhere we call an
EQUIPOTENTIAL
In our gravitational analogy, any locally flat surface is an equipotential. Any
spherical surface around an ideal planet or star is also one. Hence any spherical
surface around a point charge is an equipotential.
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Section 6
Electric Field vs. Potential
Consider a uniform electric field E = E ( x ) ˆi in a particular region of space.
Suppose we want to move a charge q from x = a to x = b as shown below.
E(x)
0
q
q
a
b
x
Since we have only one dimension, V = − ∫ E ( x ) • dx = − ∫ E ( x ) dx
Now suppose we have V(x), but not E(x). We can we compute V from E by
taking the derivative of both sides of the equation:
d
d
V ( x ) = − ∫ E ( x ) dx
dx
dx
dV ( x )
⇒ E ( x) = −
dx
Example:
V ( x ) = ( 50 + 15 x ) Volts
E ( x) = −
d
Volts
( 50 + 15 x ) = −15
dx
meter
In more than one dimension, we need only recall that vectors can be
decomposed into their components, î , ĵ and k̂ or, equivalently in triplet
notation ( x, y, z ) . Hence, taking each component in turn, we can say
⎛ ∂V ˆ ∂V ˆ ∂V ˆ ⎞
⎛ ∂V ∂V ∂V ⎞
,
,
E ( x, y , z ) = − ⎜
i+
j+
k ⎟ = −⎜
⎟ = −∇V
∂
x
∂
y
∂
z
∂
x
∂
y
∂
z
⎝
⎠
⎝
⎠
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Addendum: Absolute Potential
In most cases we can define an “absolute” potential rather than a potential
difference by choosing a suitable reference point, say at distance infinity.
q
r
from ∞
Q
Let’s define the absolute potential as zero at infinity. Then the work required
to bring in and calculate the work the electric field does against us as we bring
the test charge Q in from infinity to a distance r from the charge q:
r
kqQ
kqQ
⎡1⎤
W = ∫ F • dr = ∫ 2 dr = −kqQ ⎢ ⎥ = −
∞
∞ r
r
⎣ r ⎦∞
r
r
Since energy is conserved, the work must have been stored as potential energy.
Hence, after dividing by the value of the charge we must have
V =−
kq
r
“Absolute” Potential of a Point Charge
Note that we cannot do the same thing with an infinite line charge or infinite
sheet, since these configurations are not spatially bound and integrating the
work done by their fields does not produce finite answers. Try this yourself!
Why bother with this extra piece of business?
If we already happen to know the absolute potential at two different points, say
point a and point b, then the potential difference is the difference of the absolute
potentials:
ΔVab = V ( b ) − V ( a )
Trivial? Not really, since, as we saw above the work needed to move a charge
from one point to the other is just the charge times the potential difference:
Wab = qΔVab
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Chapter 26
Electrostatic Energy & Capacitors
SKIM Sections 1 – 3 for the moment
Section 4 Capacitors
Let’s say you want to store water in a tank, but you must pump it in through a
pipe at the bottom. The height of the water in the tank is proportional to the
pressure generated by the pump, for the column height determines the water
pressure at the bottom. How much water can the tank hold? It depends on the
cross-sectional area of the tank, since the volume equals the area times the
height. For a given tank, then, the fluid capacity is proportional to the height
which is proportional to the pump pressure. Put another way, the capacity for a
given pump pressure is proportional to the tank area.
Now consider two oppositely charged plates (finite sheets) separated by a
distance d and each of area A. If the plate area A is large compared with the
separation d, then the configuration can be treated as a section of infinite sheets
and the field is uniform and unidirectional almost everywhere.
Recall that the electric field between the plates is E =
σ
and the surface
ε0
Q
(charge divided by area) so that the electric field with
A
Q
.
respect to the magnitude of the charge on each plate must be E =
ε0 A
Qd
or, put
Now the potential difference between the plates is just V = Ed =
ε0 A
another way, the amount of charge stored per volt is
charge density is σ =
Q
A
= ε0
V
d
We’ll see below that the two other symmetric geometries we’ve investigated
yield a similar result, namely, that the capacity of charge for a given potential is
fixed once the physical parameters have been fixed. Hence, we can define the
capacitance of a particular configuration of conductors as
C=
Q
V
Capacitance in Coulombs per Volt = FARAD
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Example: Concentric Spheres
The potential difference between two concentric spheres of inner radius r = a
and outer radius r = b and carrying equal and opposite charges ±Q is
Vab = kQ (1 a − 1 b )
(where we’ve taken the positive sense)
Here again, the capacitance is simply
C=
Q
Q
1
=
=
V kQ (1 a − 1 b ) k (1 a − 1 b )
Example: Parallel Plates
The potential between two parallel plates of finite area A and separation
distance d carrying equal but opposite total charge ±Q
V=
Qd
ε0 A
The capacitance of this configuration is simply
C=
Q
Q
=
V Qd
ε0 A
= ε0
A
d
Example: Coaxial Cylinders
The potential difference between two coaxial cylinders of inner radius r = a
and outer radius r = b is Vab = 2k λ ln ( b a ) where λ is the linear charge
density. For a configuration of finite length L, the total charge Q = λ L , so that
C=
Q
L
λL
=
=
V 2k λ ln ( b a ) 2k ln ( b a )
Note: In the last two cases we have ignored the finite-length “end effects” and
treated the electric field and the potential as if the configurations were infinite.
This works only when the separation distance is much smaller than the length.
Note also that Capacitance is always taken in the positive sense.
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