Electric Potential

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Electric Potential

• Chapter 25

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ELECTRIC POTENTIAL DIFFERENCE
The fundamental definition of the electric
potential V is given in terms of the electric
field
B
?VAB - ?AB E dl
A
?VAB Electric potential difference between the
points A and B VB-VA. This is not
the way we will usually calculate electric
potentials, but we will explore this in a couple
of simple examples to understand it better.
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Constant electric field
E
A
dl

?
L

B
The electric potential difference does not depend
on the integration 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 the angle between E and
dl is constant.
E dl E dl cos(p-?) ? ?VAB -E cos(p-?) ?
dl E L cos ?
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Constant electric field
E
d
A
C


?
q
L

B
This line integral is the same for any path
connecting the same endpoints. 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 ? ?
Notice the electric field points downhill
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Equipotential Surfaces (lines)
A
E
For a constant field E all of the points along
the vertical line A are at the same potential.
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Equipotential Surfaces (lines)
A
E
For a constant field E all of the points along
the vertical line A are at the same potential.
Pf DVbc-?Edl0 because E ? dl. We can say
line A is at potential VA.
b
c
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Equipotential Surfaces (lines)
A
x
E
For a constant field E all of the points along
the vertical line A are at the same potential.
Pf DVbc-?Edl0 because E ? dl. We can say
line A is at potential VA.
The same is true for any vertical line all
points along it are at the same potential. For
example, all points on the dotted line a distance
x from A are at the same potential Vx, where ?VAx
E x
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Equipotential Surfaces (lines)
A
x
E
For a constant field E all of the points along
the vertical line A are at the same potential.
Pf DVbc-?Edl0 because E ? dl. We can say
line A is at potential VA.
The same is true for any vertical line all
points along it are at the same potential. For
example, all points on the dotted line a distance
x from A are at the same potential Vx, where ?VAx
E x
A line (or surface in 3D) of constant potential
is known as an Equipotential
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Equipotential Surfaces

• We can make graphical representations of the
  electric potential in the same way as we have
  created for the electric field

Lines of constant E
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Equipotential Surfaces

• We can make graphical representations of the
  electric potential in the same way as we have
  created for the electric field

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Equipotential Surfaces
It is sometimes useful to draw pictures of
equipotentials rather than electric field lines
Equipotential plots are like contour maps of
hills and valleys. The electric field is the
local slope, and points downhill.
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Equipotential Surfaces
How do the equipotential surfaces look for (a) A
point charge?
E

(b) An electric dipole?
-

Equipotential plots are like contour maps of
hills and valleys. The electric field is the
local slope, and points downhill.
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The Electric Potential
Electric Potential of a Point Charge

• Point Charge q ?

b
What is the electrical potential
difference between two points (a and b) in the
electric field produced by a point charge q?
a
q
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The Electric Potential
Electric Potential of a Point Charge

• Place the point charge q at the origin.
• The electric field points radially outwards.

b
c
Choose a path a-c-b.
DVab DVac DVcb DVab 0 because on this path
a
q
DVbc
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The Electric Potential
Electric Potential of a Point Charge
b
c
From this its 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
a
q
When the source charge is q, and the electric
potential is evaluated at the point r.
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The Electric Potential
Electric Potential of a Point Charge
Never do this derivation again. Instead, know
this simple result by heart
r
This is the most important thing to know about
electric potential the potential of a point
charge q.
Remember this is the electric potential with
respect to infinity we chose V(8) to be zero.
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Potential Due to a Group of Charges

• The second most important thing to know about
  electric potential is how to calculate it given
  more than one charge
• For isolated point charges just add the
  potentials created by each charge (superposition)
• For a continuous distribution of charge

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Potential Produced by aContinuous Distribution
of Charge
In the case of a continuous charge distribution,
divide the distribution up into small pieces and
then sum (integrate) the contribution from each
bit
A
?
r
dVA k dq / r
?
VA ? dVA ? k dq / r
dq
Remember k1/(4??0)
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A charge density per unit length l 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.
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A charge density per unit length l 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
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A charge density per unit length l 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
so
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A charge density per unit length l 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
so
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A charge density per unit length l 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
so
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Example a disk of charge

• Suppose the disk has radius R and a charge per
  unit area s.
• 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.

dq s2pwdw
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Field and Electric Potential
Remember from calculus that integrals are
antiderivatives.
By the fundamental theorem of calculus you can
undo the integral
Given
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Field and Electric Potential
Remember from calculus that integrals are
antiderivatives.
By the fundamental theorem of calculus you can
undo the integral
Given
Very similarly you can get E(r) from derivatives
of V(r).
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Field and Electric Potential
Remember from calculus that integrals are
antiderivatives.
By the fundamental theorem of calculus you can
undo the integral
Given
Very similarly you can get E(r) from derivatives
of V(r).
Choose V(r0)0. Then
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Field and Electric Potential
Remember from calculus that integrals are
antiderivatives.
By the fundamental theorem of calculus you can
undo the integral
Given
Very similarly you can get E(r) from derivatives
of V(r).
Choose V(r0)0. Then
?
? is the gradient operator
The third most important thing to know about
potentials.
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Force and Potential Energy
This is entirely analogous to the relationship
between a conservative force 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.
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Example a disk of charge

• Suppose the disk has radius R and a charge per
  unit area s.
• 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.

dq s2pwdw
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Example a disk of charge
By symmetry one sees that ExEy0 at P. Find Ez
from
This is easier than integrating over
the components of vectors. Here we
integrate over a scalar and then take partial
derivatives.
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Example point charge
Put a point charge q at the origin.
Find V(r) here this is easy
r
q
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Example point charge
Put a point charge q at the origin.
Find V(r) here this is easy
r
q
Then find E(r) from the derivatives
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Example point charge
Put a point charge q at the origin.
Find V(r) here this is easy
r
q
Then find E(r) from the derivatives
Derivative
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Example point charge
Put a point charge q at the origin.
Find V(r) here this is easy
r
q
Then find E(r) from the derivatives
Derivative
So
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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)
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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.
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Energy of a Charge Distribution
CASE I Two Charges
Bringing the second charge requires to perform
work against the field of the first charge.
W Q2 V1 with V1 (1/4??0) (Q1/r)
? W (1/4??0) (Q1 Q2 /r) U
U potential energy of two point charges
U (1/4??0) (Q1 Q2 /r)
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Energy of a Charge Distribution
CASE II Several Charges
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?
Q
Q
Q
Q
a
To answer it is necessary to add up the potential
energy of each pair of charges ? U ? Uij
U12 (1/4??0) (Q1 Q2 /r)
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Energy of a Charge Distribution
fields cancel
A
CASE III Parallel Plate Capacitor
fields add
E
d
fields cancel
Electric Field ? E ? / ?0 Q / ?0 A (?
Q / A)
Potential Difference ? V E d Q d / ?0 A
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Energy of a Charge Distribution
fields cancel
A
CASE III Parallel Plate Capacitor
fields add
E
d
fields cancel
Now, suppose moving an additional very small
positive charge dq 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
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Energy of a Charge Distribution
fields cancel
A
CASE III Parallel Plate Capacitor
fields add
E
d
fields cancel
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
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Energy of a Charge Distribution
fields cancel
A
CASE III Parallel Plate Capacitor
fields add
E
d
fields cancel
The work done in charging the plates ends up as
stored potential energy of the final charge
distribution
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Energy of a Charge Distribution
fields cancel
A
CASE III Parallel Plate Capacitor
fields add
E
d
fields cancel
The energy U is stored in the field, in the
region between the plates.
E Q / (?0 A)
U d Q2 / 2 ?0 A (1/2) ?0 E2 A d
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
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Energy of a Charge Distribution
Electric Energy Density
CASE IV Arbitrary Charge Distribution
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)
The integral covers the entire region in which
the field E exists
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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 ?.