Electric Potential
Chapter 25
ELECTRIC POTENTIAL DIFFERENCE
The fundamental definition of the electric potential V is given interms of the electric field:
VAB = - AB E · dl
VAB = Electric potential difference between the points A and B
= VB-VA.
This is not the way we will usually calculate electric potentials, butwe will explore this in a couple of simple examples to understandit better.
A
B
Constant electric field
VAB = - AB E · dl
•
A
B
E
•
E · dl = E dl cos(-) VAB = -E cos(-) dl = E L cos
The electric potential difference does not depend on theintegration path. So pick a simple path.
One possibility is to integrate along the straight line AB.
This is easy in this case because E is constant and theangle between E and dl is constant.
L
dl
•
•
A
B
E
•
C
d
This line integral is the same for any path connecting the sameendpoints. For example, try the two-step path A to C to B.
VAB = VAC + VCB VAC = E d VCB = 0 (E dl)
Thus, VAB = E d but d = L cos
VAB = E L cos
L
VAB = - AB E · dl
Constant electric field
Notice: the electric field points downhill
Equipotential Surfaces (lines)
For a constant field E all of the pointsalong the vertical line A are at the samepotential.
E
A
Equipotential Surfaces (lines)
For a constant field E all of the pointsalong the vertical line A are at the samepotential.
Pf: Vbc=-∫E·dl=0 because E dl.
We can say line A is at potential VA.
E
A
b
c
Equipotential Surfaces (lines)
For a constant field E all of the pointsalong the vertical line A are at the samepotential.
Pf: Vbc=-∫E·dl=0 because E dl.
We can say line A is at potential VA.
E
A
x
The same is true for any vertical line: all points along it are at thesame potential. For example, all points on the dotted line adistance x from A are at the same potential Vx, where VAx = E x
Equipotential Surfaces (lines)
For a constant field E all of the pointsalong the vertical line A are at the samepotential.
Pf: Vbc=-∫E·dl=0 because E dl.
We can say line A is at potential VA.
E
A
x
A line (or surface in 3D) of constant potential is known as an
Equipotential
The same is true for any vertical line: all points along it are at thesame potential. For example, all points on the dotted line adistance x from A are at the same potential Vx, where VAx = E x
We can make graphical representations ofthe electric potential in the same way aswe have created for the electric field:
Equipotential Surfaces
Lines of constant E
We can make graphical representations ofthe electric potential in the same way aswe have created for the electric field:
Equipotential Surfaces
Lines of constant E
Lines of constant V
(perpendicular to E)
Equipotential Surfaces
Equipotential plots are like contour maps of hills and valleys.The electric field is the local slope, and points downhill.
Lines of constant E
Lines of constant V
(perpendicular to E)
It is sometimes useful to draw pictures of equipotentialsrather than electric field lines:
Equipotential Surfaces
How do the equipotential surfaces look for:
(a) A point charge?
(b) An electric dipole?
+
+
-
E
Equipotential plots are like contour maps of hills and valleys.The electric field is the local slope, and points downhill.
Point Charge q
The Electric Potential
b
a
q
What is the electrical potential difference
between two points (a and b) in the electric
field produced by a point charge q?
Electric Potential of a Point Charge
Place the point charge q at the origin.
The electric field points radially outwards.
The Electric Potential
c
b
a
q
Choose a path a-c-b.
Vab = Vac + Vcb
Vab = 0 because on this path
Vbc =
Electric Potential of a Point Charge
The Electric Potential
From this it’s natural to choose
the zero of electric potential
to be when ra
Letting a be the point at infinity, and dropping
the subscript b, we get the electric potential:
When the source charge is q,
and the electric potential is
evaluated at the point r.
c
b
a
q
Electric Potential of a Point Charge
The Electric Potential
This is the most important thing to knowabout electric potential: the potential of apoint charge q.
q
Remember: this is the electric potential withrespect to infinity – we chose V(∞) to be zero.
Electric Potential of a Point Charge
Never do this derivation again. Instead,know this simple result by heart:
r
Potential Due to a Group of Charges
• The second most important thing to know about electricpotential is how to calculate it given more than one charge
• For isolated point charges just add the potentials created byeach charge (superposition)
• For a continuous distribution of charge …
Potential Produced by aContinuous Distribution of Charge
dq
A
dVA = k dq / r
r
VA = dVA = k dq / r
Remember:
k=1/(40)
In the case of a continuous charge distribution, divide thedistribution up into small pieces and then sum (integrate) thecontribution from each bit:
A charge density per unit length stretches a length L.Find the electric potential at a point d from one end.
Example: a line of charge
Break the charge into little bits: say a length dx at position x.
x
L
d
x
r = d+L-x
dq =dx
P
A charge density per unit length stretches a length L.Find the electric potential at a point d from one end.
Example: a line of charge
Break the charge into little bits: say a length dx at position x.
The contribution due to this bit at P is:
x
L
d
x
r = d+L-x
dq =dx
P
A charge density per unit length stretches a length L.Find the electric potential at a point d from one end.
Example: a line of charge
Break the charge into little bits: say a length dx at position x.
The contribution due to this bit at P is:
x
L
d
x
r = d+L-x
dq =dx
P
so
A charge density per unit length stretches a length L.Find the electric potential at a point d from one end.
Example: a line of charge
Break the charge into little bits: say a length dx at position x.
The contribution due to this bit at P is:
x
L
d
x
r = d+L-x
dq =dx
P
so
A charge density per unit length stretches a length L.Find the electric potential at a point d from one end.
Example: a line of charge
Break the charge into little bits: say a length dx at position x.
The contribution due to this bit at P is:
x
L
d
x
r = d+L-x
dq =dx
P
so
Example: a disk of charge
Suppose the disk has radius R and a charge per unit area .
Find the potential at a point P up the z axis (centered on the disk).
Divide the object into small elements of charge and find the
potential dV at P due to each bit. For a disk, a bit (differential
of area) is a small ring of width dw and radius w.
dw
P
r
R
w
z
dq = 2wdw
Field and Electric Potential
Remember from calculus thatintegrals are antiderivatives.
By the fundamental theorem ofcalculus you can “undo” the integral:
Given
Field and Electric Potential
Very similarly you can get E(r) from derivatives of V(r).
Remember from calculus thatintegrals are antiderivatives.
By the fundamental theorem ofcalculus you can “undo” the integral:
Given
Field and Electric Potential
Very similarly you can get E(r) from derivatives of V(r).
Remember from calculus thatintegrals are antiderivatives.
By the fundamental theorem ofcalculus you can “undo” the integral:
Given
Choose V(r0)=0. Then
Field and Electric Potential
Very similarly you can get E(r) from derivatives of V(r).
is the gradient operator
Remember from calculus thatintegrals are antiderivatives.
By the fundamental theorem ofcalculus you can “undo” the integral:
The third most important thing to know about potentials.
Given
Choose V(r0)=0. Then
Force and Potential Energy
This is entirely analogous to the relationship between a conservativeforce and its potential energy.
can be inverted:
In a very similar way the electric potential and field are related by:
can be inverted:
The reason is that V is simply potential energy per unit charge.
Example: a disk of charge
•Suppose the disk has radius R and a charge per unit area .
•Find the potential and electric field at a point up the z axis.
•Divide the object into small elements of charge and find the
•potential dV at P due to each bit. So here let a bit be a small
•ring of charge width dw and radius w.
dw
P
r
R
w
z
dq = 2wdw
This is easier than integrating over the
components of vectors. Here we integrate
over a scalar and then take partial derivatives.
Example: a disk of charge
dw
P
r
R
w
z
By symmetry one sees that Ex=Ey=0 at P.
Find Ez from
Example: point charge
Put a point charge q at the origin.
q
Find V(r): here this is easy:
r
Example: point charge
Put a point charge q at the origin.
q
Find V(r): here this is easy:
r
Then find E(r) from the derivatives:
Example: point charge
Put a point charge q at the origin.
q
Find V(r): here this is easy:
r
Then find E(r) from the derivatives:
Derivative:
Example: point charge
Put a point charge q at the origin.
q
Find V(r): here this is easy:
r
Then find E(r) from the derivatives:
Derivative:
So:
Energy of a Charge Distribution
How much energy ( work) is required to assemble a
charge distribution ?.
CASE I: Two Charges
Bringing the first charge does not require energy ( work)
Energy of a Charge Distribution
How much energy ( work) is required to assemble a
charge distribution ?.
CASE I: Two Charges
Bringing the first charge does not require energy ( work)
Bringing the second charge requires to perform work
against the field of the first charge.
r
Q1
Q2
Energy of a Charge Distribution
CASE I: Two Charges
Bringing the second charge requires to perform work against thefield of the first charge.
r
Q1
Q2
W = Q2 V1 with V1 = (1/40) (Q1/r)
W = (1/40) (Q1 Q2 /r) = U
U = (1/40) (Q1 Q2 /r)
U = potential energy of
two point charges
Energy of a Charge Distribution
CASE II: Several Charges
U12 = (1/40) (Q1 Q2 /r)
U12 = potential energy of
a pair of point charges
a
Q
Q
Q
Q
How much energy is stored in this square charge
distribution?, or …
What is the electrostatic potential energy of the
distribution?, or …
How much work is needed to assemble this
charge distribution?
To answer it is necessary to add up the potential energy of
each pair of charges U = Uij
CASE III: Parallel
Plate Capacitor
Energy of a Charge Distribution
-Q
+Q
fields cancel
fields cancel
fields
add
d
E
A
Electric Field E = / 0 = Q / 0 A ( = Q / A)
Potential Difference V = E d = Q d / 0 A
CASE III: Parallel
Plate Capacitor
Energy of a Charge Distribution
-Q
+Q
fields cancel
fields cancel
fields
add
d
E
A
Now, suppose moving an additional very small positive chargedq from the negative to the positive plate. We need to do work.How much work?
dW = V dq = (q d / 0 A) dq
We can use this expression to calculate the total work needed to
charge the plates to Q, -Q
CASE III: Parallel
Plate Capacitor
Energy of a Charge Distribution
-Q
+Q
fields cancel
fields cancel
fields
add
d
E
A
dW = V dq = (q d / 0 A) dq
The total work needed to charge the plates to Q, -Q, is given by:
W = dW = (q d / 0 A) dq = (d / 0 A) q dq
W = (d / 0 A) [Q2 / 2] = d Q2 / 2 0 A
CASE III: Parallel
Plate Capacitor
Energy of a Charge Distribution
-Q
+Q
fields cancel
fields cancel
fields
add
d
E
A
W = U = d Q2 / 2 0 A
The work done in charging the plates ends up as
stored potential energy of the final charge distribution
Where is the energy stored ?
The energy is stored in the electric field
CASE III: Parallel
Plate Capacitor
Energy of a Charge Distribution
-Q
+Q
fields cancel
fields cancel
fields
add
d
E
A
U = d Q2 / 2 0 A = (1/2) 0 E2 A d
The energy U is stored in the field, in the region between the plates.
E = Q / (0 A)
The volume of this region is Vol = A d,
so we can define the energy density uE as:
uE = U / A d = (1/2) 0 E2
Energy of a Charge Distribution
.
uE = U / A d = (1/2) 0 E2
Electric
Energy
Density
Although we derived this expression for the uniform field of a
parallel plate capacitor, this is a universal expression valid for
any electric field.
When we have an arbitrary charge distribution, we can use uE
to calculate the stored energy U
dU = uE d(Vol) = (1/2) 0 E2 d(Vol) U = (1/2) 0 E2 d(Vol)
CASE IV: Arbitrary
Charge Distribution
[The integral covers the entire region in which the field E exists]
A Shrinking Sphere
A sphere of radius R1 carries a total charge Q distributed evenly
over its surface. How much work does it take to shrink the sphere
to a smaller radius R2 ?.