Download Lesson 7 (1) Definition of Electric Potential Consider the electric field

Lesson 7
(1) Definition of Electric Potential
Consider the electric field E created by a charge distribution. A test charge q placed
at any location experiences the force
F = qE
where E is evaluated at that location. When the test charge undergoes a small
displacement dr , the work done by the electric force is
dW = F × dr = qE × dr
E
B
A
dr
If the test charges moves from the point A to B along a path, the work done can be
calculated by dividing the path into small displacements and adding the work done
along the displacements. In the limit when these displacements go to zero, we find
the work done along the path from A to B in the form of a line integral:
B
W = å dW = q ò E × dr
A
It turns out that the electrostatic force is conservative: the work done in going from
one point to the other is independent of the path between the two points. Further,
since the quantity W q is independent of the test charge, being a property of the
electric field alone, we can define a property of the electric field at any point called
the electric potential so that if VA and VB denote the potentials at A and B, the
difference is
VA -VB =
B
ò E × dr
A
We can only define the potential difference between two points. The absolute value
depends on choosing a reference point where we can assign an arbitrary value to V .
In terms of the potentials, we can write
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Work done by electric field = q (VA -VB )
The unit of potential is the Volt (V), which is Joule per Coulomb (J/C).
Customarily, in going from A to B, we define the potential difference as
DV = VB -VA
Therefore
B
DV = - ò E × dr
A
We can imagine applying to a test charge an external force of the same amount as
the electric force but in the opposite direction (push against the electric influence).
Such a force is -qE , and it can be considered to do work against the electric field.
We can write
Work done against electric field = qDV
Thus, in moving a positive charge from a point of low potential to one of high
potential, the external force has to do positive work. However, if the charge is
negative, the external force does negative work.
In mechanics, the potential energy of a particle in a gravitational field increases
when the external force working against gravity does positive work. By analogy, we
define the electric potential energy difference of a test charge as
DU = qDV
With choice of a reference, the potential energy itself is
U = qV
(2) Uniform Electric Field
Consider a uniform electric field in the x-direction:
E = Exiˆ
and two points A and B on the x-y plane with coordinates ( xA , yA ), ( xB , yB )
respectively. To calculate the work done by the electric field on a test charge of 1C
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moving from A to B, divide the path into small displacements dr . The elemental
work is
E × dr = Exiˆ × dr = Ex dx
Note that the change in y-coordinate does not enter. The total work is
B
xB
A
xA
ò E × dr =
òE
x
dx = Ex ( xB - x A )
which demonstrates the path independence of the line integral. Thus,
B
DV = - ò E × dr = -Ex Dx
A
where Dx = xB - xA . The argument is the same in three dimensions. From the above
relation because electric field and potential difference, we can use volt per meter
(V/m) as the unit of electric field instead of N/C.
Locations where the potentials are the same are said to form an equipotential
surface. The equipotential surfaces for a uniform electric field are planes
perpendicular to the field lines. The electric field points from an equipotential
surface of high potential to one of low potential.
B
A
(3) Point Charge
For a point charge q , the electric potential at a point P is expected to depend only on
its distance r from the point charge because of symmetry. The potential at P can be
calculated from a line integral over a straight line starting at infinity and ending on
P. Choose this straight line to be the x-axis with the point charge at the origin.
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x
P
r
q
dr
¥
Over the elemental displacement dr at a distance x from the point charge,
E × dr =
kq ˆ
kq
i × dr = 2 dx
2
x
x
Therefore
P
r
¥
¥
VP -V¥ = - ò E × dr = -kq ò
dx kq
=
x2 r
Choosing as a reference V¥ = 0 , we obtain the formula
V=
kq
r
for the potential of a point charge at a distance r .
We can perform the line integral over an arbitrary path from infinity to P and obtain
the same result as will be demonstrated in the class
The equipotential surfaces are spherical surfaces with the point charge at the
center,, and the field lines are perpendicular to these surfaces.
Thus, a positive charge causes positive potential in its vicinity and a negative charge
causes negative potential.
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For a point charge on the x-axis with coordinate x = a , the potential at a point P on
the x-axis with x-coordinate x is
V=
kq
x-a
(4) Dipole
The potential due to multiple point charge is the sum of the potentials due to each.
For a dipole on the x-axis with charge q at x = -a and -q at x = a , the potential at a
point on the x-axis with coordinate x is therefore
æ 1
1 ö
÷÷
V = kq çç
è x+a x-a ø
A plot of this function is as shown, where the potential is seen to vanish at x=0.
-a
0
-a
In fact, in 3-D, the whole plane at x=0 is an equipotential surface with zero potential,
because every point on this plane is equidistant from the two point charges. A graph
of the equipotentials and field lines of a dipole is as shown.
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In the limit x >> a , using
1
1
1æ aö
=
» ç1- ÷
x + a x (1+ a x ) x è x ø
1
1
1æ aö
=
» ç1+ ÷
x - a x (1- a x ) x è x ø
we find, using the dipole moment p = 2qa
V »-
2kqa
kp
=- 2
2
x
x
The potential at a point off the dipole axis and far away can also be found. Let the
point P be at a distance r from the center O of the dipole and let the angle between
OP and the axis of the dipole be q . Start with the equation
æ1 1 ö
V = kq ç - ÷
è r1 r2 ø
where r1, r2 are the distance of P from the positive and negative charge respectively.
P
r2
r
a
q
o
When r >> a , the approximations
r2 = r + acosq
r1
r1 = r - acosq can be used. Then
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é
ù
ê
ú
1
1
kpcosq
ê
ú»V = kq
æ
ö
æ
ö
r2
ê r 1+ a cosq r 1- a cosq ú
ç
÷
ç
÷
êë è r
ø è r
ø úû
This shows that the potential of a dipole falls off as 1/distance2 in all directions.
(5) Field Lines and Equipotential Surfaces
Field lines and equipotential surfaces are perpendicular to each other. To prove this,
at any point on an equipotential surface, construct a small displacement vector dr
that lies on the surface. The change of potential DV along this displacement is zero.
Therefore
E × dr = -DV = 0
This shows that E and dr are perpendicular to each other. Since dr points in
arbitrary direction on the surface, E is therefore perpendicular to the surface.
E
dr
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