Download Lecture 5 : Potential

Electric Potential
Chapter 25
Electric Potential Energy
Electric Potential
Equipotential Surfaces
Electric Potential: the Bottom Line
The electric potential V(r) is an easy way to calculate the
electric field – easier than directly using Coulomb’s law.
For a point charge at
the origin:
V (r ) 
1 q
4 0 r
For a collection of charges qi at
1
qi
V
(r
)


positions ri the electric potential
4 0 i r  ri
at r is the scalar sum:

Calculate V and then find the electric field by taking the gradient:
V

V ˆ V
ˆ
E(r )  V (r )   i 
j
x
y

kˆ 
z 
Example: two point charges

1  q
V (x, y) 

2
2
40  x  y


q

2
x  a  y 2 
r

q
V
q  1
2x
E x (x, y)  

 2

3
2
2 2
x
40 
x

y



 


2
x

a


  12
3 
2
2 2 
x  a  y 
 


a
q
CONSERVATIVE FORCES
A conservative force “gives back” work that has been done against it
When the total work done by a force on an object moving around
a closed loop is zero, then the force is conservative
° F·dr = 0
 F is conservative
The circle on the integral sign indicates that the integral is taken over a closed path
The work done by a conservative force, in moving and object
between two points A and B, is independent of the path taken
is a function of A and B only – it is
NOT a function of the path selected.
We can define a potential energy difference as UAB=-W.
W=AB F·dr
POTENTIAL ENERGY
The change UAB in potential energy due to the electric force is thus
UAB = -q AB E·dr
UAB = UB – UA = potential energy difference between A and B
Potential energy is defined at each point in space, but it is only the
difference in potential energy that matters.
Potential energy is measured with respect to a reference point
(usually infinity). So we let A be the reference point (i.e, define UA
to be zero), and use the above integral as the definition of U at
point B.
POTENTIAL ENERGY IN A CONSTANT FIELD E
E
L
A
• dl
•B
The potential energy difference between A and B equals the
work necessary to move a charge +q from A to B
UAB = UB – UA = -  q E·dl
But E = constant, and E.dl = -E dl, so:
UAB = -  q E · dl =  q E dl = q E  dl = q E L
UAB = q E L
ELECTRIC POTENTIAL DIFFERENCE
The potential energy U depends on the charge being moved.
To remove this dependence, we introduce the concept of the
electric potential V. This is defined in terms of the difference V:
VAB = UAB / q = - AB E · dl
Electrical Potential = Potential Energy per Unit Charge
= line integral of -E·dl
VAB = Electric potential difference between the points A and B.
Units are Volts (1V = 1 J/C), and so the electric potential is often
called the voltage.
A positive charge is pushed from regions of high potential to
regions of low potential.
ELECTRIC POTENTIAL IN A CONSTANT FIELD E
E
L
A
• dL
•B
VAB = UAB / q
The electrical potential difference between A and B equals
the work per unit charge necessary to move a charge +q
from A to B
VAB = VB – VA = -  E·dl
But E = constant, and E.dl = -1 E dl, so:
VAB = -  E·dl =  E dl = E  dl = E L
VAB = E L
UAB = q E L
Cases in which the electric field E is not aligned with dl
E
A
VAB = -  AB E · dl
•

•
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 convenient in this case because the field E is
constant, and the angle  between E and dl is constant.
B
E . dl = E dl cos   VAB = - E cos   dl = - E L cos 
A
Cases in which the electric field E is not aligned with dl
E
d

A
•

L
•
•
VAB = -  AB E · dl
C
B
Another possibility is to choose a path that goes from A to C, and
then from C to B
VAB = VAC + VCB
Thus, VAB = E d
VAC = E d
VCB = 0 (E  dL)
but d = L cos  = - L cos 
VAB = - E L cos 
Equipotential Surfaces (lines)
A
E
B
For a constant field E
VAB = E L
x
L
E
Similarly, at a distance x from plate A
VAx = E x
All the points along the dashed line
at x, are at the same potential.
The dashed line is an
equipotential line
L
Equipotential Surfaces (lines)
x
E
It takes no work to move a charge
at right angles to an electric field
E  dl   E•dl = 0  V = 0
L
If a surface(line) is perpendicular to
the electric field, all the points in
the surface (line) are at the same
potential. Such surface (line) is called
EQUIPOTENTIAL
EQUIPOTENTIAL  ELECTRIC FIELD
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
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 plots are like contour maps of hills and valleys.
A positive charge would be pushed from hills to valleys.
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.
Equipotential
plots
are like
contourfrom
mapshills
of hills
and valleys.
A positive charge
would
be pushed
to valleys.
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 =
F dr
rb
a
q
rb
dr
 q tE(r ) dr   q t E(r)dr  kq tq  r 2
rc
ra
ra
ElectricThe
Potential
ElectricofPotential
a Point Charge
b
Place the point charge q at the origin.
The electric field points radially outwards.
c
First find the work done by q’s field when qt
is moved from a to b on the path a-c-b.
W = W(a to c) + W(c to b)
W(a to c) = 0 because on this path F dr
rb
rb
F=qtE
a
q
rb
dr
W(c to b) =  q tE (r ) dr   q t E(r)dr  kq tq  2
r
rc
ra
ra
 1 1 
hence W  kq t q  
ra rb 
ElectricThe
Potential
ElectricofPotential
a Point Charge
b
 1 1 
U(rb )  U(ra )  W  kq tq   
rb ra 
And since
VAB = UAB / qt
VAB = k q [ 1/rb – 1/ra ]
c
F=qtE
q
a
ElectricThe
Potential
ElectricofPotential
a Point Charge
b
VAB = k q [ 1/rb – 1/ra ]
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=kq/r
F=qtE
q
When the source charge is q,
and the electric potential is
evaluated at the point r.
Remember: this is the electric potential with respect to infinity
a
Potential Due to a Group of Charges
• 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 distribution of charge we first
divide the distribution up into small pieces, and then we
sum the contribution, to the electric potential, from each
piece:

dqi
Potential Produced by a
Continuous Distribution of Charge
In the case of a continuous distribution of charge we first
divide the distribution up into small pieces, and then we
sum the contribution, to the electric potential, from each
piece:
In the limit of very small pieces, the sum is an integral
A
r
dV = k dq / r
A

dq
Remember:
VA =  dVA =  k dq / r
vol
vol
k=1/(4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 = s2wdw
1 dq
1 s 2wdw
dV 

r
z
4 r 4 w  z
2
0
2
0
s
V   dV 
(w  z ) wdw

2
R
2
0
V
0
s
( R  z  z)
2
2
0
2
2
 12
dw
w
R