Transcript Lecture 6 : Potential - University of Central Florida

Electric Potential
Chapter 25
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.
Constant electric field
E
A
VAB = -  AB E · dl
• 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 
Constant electric field
E
d
A
•


L
•
•
VAB = -  AB E · dl
C
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
Thus, VAB = E d
VAC = E d
but d = L cos  
VCB = 0 (E  dl)
VAB = E L cos 
Notice: the electric field points downhill
Equipotential Surfaces (lines)
A
E
For a constant field E all of the points
along the vertical line A are at the same
potential.
Equipotential Surfaces (lines)
A
b
c
E
For a constant field E all of the points
along the vertical line A are at the same
potential.
Pf: Vbc=-∫E·dl=0 because E  dl.
We can say line A is at potential VA.
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: Vbc=-∫E·dl=0 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
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: Vbc=-∫E·dl=0 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
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
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 V
(perpendicular to E)
Lines of constant E
Equipotential Surfaces
It is sometimes useful to draw pictures of equipotentials
rather than electric field lines:
Lines of constant V
(perpendicular to E)
Lines of constant E
Equipotential plots are like contour maps of hills and valleys.
The electric field is the local slope, and points downhill.
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.
ElectricThe
Potential
ElectricofPotential
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?
q
a
ElectricThe
Potential
ElectricofPotential
a Point Charge
b
Place the point charge q at the origin.
The electric field points radially outwards.
c
Choose a path a-c-b.
Vab = Vac + Vcb
Vab = 0 because on this path
rb
Vbc = 
E dr
rb
 E ( r )  d r    E(r)dr
rc

V ab
ra
a
q
rb
 kq
 1
1 
 kq   
rb ra 

ra
dr
r
2

kq
r
rb
ra
ElectricThe
Potential
ElectricofPotential
a Point Charge
V ab
 1
1 
 kq   
rb ra 
b
c
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:
V (r ) 
kq
r
q
When the source charge is q,
and the electric potential is
evaluated at the point r.
a
ElectricThe
Potential
ElectricofPotential
a Point Charge
Never do this derivation again. Instead,
know this simple result by heart:
V (r ) 
kq
r
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(∞) to be zero.
q
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 …
Potential Produced by a
Continuous 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

dq
dVA = k dq / r
VA =  dVA =  k dq / r
Remember:
k=1/(4p0)
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.
P
dq = s2pwdw
1
dV 
4p0 r
V 
V 
dq

s
2 0
dV 

1
4p0
s
w  z
2
R
2 0
 (w
0
( R  z  z)
2
s 2 p wdw
2
2
1
2  2
 z )
z
2
r
wdw
dw
w
R
Field and Electric Potential
Remember from calculus that
integrals are antiderivatives.
Given f ( x ) 

x
x0
g( y ) dy

By the fundamental theorem of
calculus you can “undo” the integral:
f '( x )  g( x )

g( x )  f ' ( x )
Field and Electric Potential
Remember from calculus that
integrals are antiderivatives.
Given f ( x ) 

x
x0
g( y ) dy
By the fundamental theorem of
calculus you can “undo” the integral:
f '( x )  g( x )

g( x )  f ' ( x )
Very similarly you can get E(r) from derivatives of V(r).

Field and Electric Potential
Remember from calculus that
integrals are antiderivatives.
Given f ( x ) 

x
x0
g( y ) dy
By the fundamental theorem of
calculus you can “undo” the integral:
f '( x )  g( x )

g( x )  f ' ( x )
Very similarly you can get E(r) from derivatives of V(r).
Choose V(r0)=0. Then
V (r )  

r
r0
E dl
Field and Electric Potential
Remember from calculus that
integrals are antiderivatives.
Given f ( x ) 

x
x0
g( y ) dy
By the fundamental theorem of
calculus you can “undo” the integral:
f '( x )  g( x )

g( x )  f ' ( x )
Very similarly you can get E(r) from derivatives of V(r).
Choose V(r0)=0. Then
V (r )  

r
r0
E dl

E (r )   V (r )
 V


V

V
ˆi 
ˆj 
  
kˆ 
y
 z 
  x
 is the gradient operator
The third most important
thing to know about potentials.

Force and Potential Energy
This is entirely analogous to the relationship between a conservative
force and its potential energy.
U (r )  

r
r0
F  dl
can be inverted:
F (r )  U (r )
In a very similar way the electric potential and field are related by:
V (r )  

r
r0
E dl

can be inverted:
E (r )  V (r )
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 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
P
ring of charge width dw and radius w.
dq = s2pwdw
1
dV 
4p0 r
V 
V 
dq

s
2 0
dV 

1
s 2 p wdw
4p0
s
w  z
2
R
2 0
 (w
0
z
2
2
1
2  2
 z )
r
wdw
dw
w
R
( R  z  z)
2
2
Example: a disk of charge
V ( z) 
s
2 0
( R  z  z)
2
2
By symmetry one sees that Ex=Ey=0 at P.
Find Ez from
Ez  
V
z
 
s 

z
2  0  R 2  z 2

1

dw
This is easier than integrating over the
components of vectors. Here we integrate
over a scalar and then take partial derivatives.
P
z
r
w
R
Example: point charge
Put a point charge q at the origin.
Find V(r): here this is easy: V ( r )  k
q
r
r
q
Example: point charge
Put a point charge q at the origin.
Find V(r): here this is easy: V ( r )  k
q
r
r
q
Then find E(r) from the derivatives:
E (r )   ( ˆi 
x
 ˆj 
y
 kˆ 
z
)V(x, y, z)
Example: point charge
Put a point charge q at the origin.
Find V(r): here this is easy: V ( r )  k
q
r
r
q
Then find E(r) from the derivatives:
E (r )   ( ˆi   x  ˆj  y  kˆ   z )V(x, y, z)
Derivative:
 1
x r


1
x
x  y  z
2
2
2

1
2x
2 x  y  z
2
2
2

3 2
Example: point charge
Put a point charge q at the origin.
Find V(r): here this is easy: V ( r )  k
q
r
r
q
Then find E(r) from the derivatives:
E (r )   ( ˆi   x  ˆj  y  kˆ   z )V(x, y, z)
Derivative:
So:
 1
x r


1
x
x  y  z
E ( r )  kq
2
2
2

x ˆi  y ˆj  z kˆ
r
3
1
2x
2 x  y  z
 kq
2
2
r
r
3
 kq
2

3 2
ˆr
r
2
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)
Q2
Q1
r
Bringing the second charge requires to perform work
against the field of the first charge.
Energy of a Charge Distribution
CASE I: Two Charges
Q2
Q1
r
Bringing the second charge requires to perform work against the
field of the first charge.
W = Q2 V1 with V1 = (1/4p0) (Q1/r)
 W = (1/4p0) (Q1 Q2 /r) = U
U = (1/4p0) (Q1 Q2 /r)
U = potential energy of
two point charges
Energy of a Charge Distribution
CASE II: Several Charges
Q
Q
Q
Q
a
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
U12 = (1/4p0) (Q1 Q2 /r)
U12 = potential energy of
a pair of point charges
Energy of a Charge Distribution
-Q
CASE III: Parallel
Plate Capacitor
fields cancel
A
fields
add
E
d
+Q
fields cancel
Electric Field  E = s / 0 = Q / 0 A (s = Q / A)
Potential Difference  V = E d = Q d / 0 A
Energy of a Charge Distribution
-Q
CASE III: Parallel
Plate Capacitor
fields cancel
A
E
d
+Q
fields
add
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
Energy of a Charge Distribution
-Q
CASE III: Parallel
Plate Capacitor
fields cancel
A
fields
add
E
d
+Q
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
Energy of a Charge Distribution
-Q
CASE III: Parallel
Plate Capacitor
fields cancel
A
E
d
+Q
fields
add
fields cancel
The work done in charging the plates ends up as
stored potential energy of the final charge distribution
W = U = d Q2 / 2 0 A
Where is the energy stored ?
The energy is stored in the electric field
Energy of a Charge Distribution
-Q
CASE III: Parallel
Plate Capacitor
fields cancel
A
E
d
+Q
fields
add
fields cancel
The energy U is stored in the field, in the region between the plates.
U = d Q2 / 2 0 A = (1/2) 0 E2 A d
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
CASE IV: Arbitrary
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)
[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 ?.