Download Electromagnetic Fields Lecture-5: Static Electric Fields 2

Electromagnetic Fields Lecture-5:
Static Electric Fields 2
Chadi Abou-Rjeily
Department of Electrical and Computer Engineering
Lebanese American University
[email protected]
February 1, 2017
Chadi Abou-Rjeily
Electromagnetic Fields Lecture-5: Static Electric Fields 2
Electric Potential (1)
We have seen in lecture-03 that following from the null
identity ∇ × (∇V ) ≡ 0 for an arbitrary scalar V , curl-free
vectors can be expressed as the gradients of some scalar
quantities.
~ is curl-free (or irrotational):
Since, E
~ =0
∇×E
Then we define a scalar electric potential V such that:
~ = −∇V
E
V is expressed in volts (V).
Chadi Abou-Rjeily
Electromagnetic Fields Lecture-5: Static Electric Fields 2
Electric Potential (2)
The reason for the inclusion of the negative sign is as
follows:
~ is
Since a point charge q placed in an electric field E
~
subject to the force q E, then the work done by the field in
moving the charge from point P1 to P2 is given by:
WE~ = q
Z
P2
~ ~l
E.d
P1
Therefore, in moving the charge form P1 to P2 , work must
be done against the field and is equal to:
W = −WE~ = −q
Chadi Abou-Rjeily
Z
P2
~ ~l
E.d
P1
Electromagnetic Fields Lecture-5: Static Electric Fields 2
Electric Potential (3)
Consequently, the potential difference between two points
P1 and P2 is defined as:
V2 − V1 = −
Z
P2
~ ~l
E.d
(V)
P1
RP
~ ~l does
Note that that we have previously proven that P12 E.d
not depend on the path between P1 and P2 , consequently,
V2 − V1 only depends on the positions of P1 and P2 .
The zero-potential point is often taken at infinity.
Note that, by definition, the direction of ∇V indicates the
direction of the maximum rate of increase of V . Since
~ = −∇V , then the direction of E
~ indicates the direction of
E
~
the maximum rate of decrease of V . Consequently, E
originates from a region of high potential towards a region
of lower potential.
Chadi Abou-Rjeily
Electromagnetic Fields Lecture-5: Static Electric Fields 2
Electric Potential (4)
~
Note that V increases in moving against E.
For example, when a d-c battery is connected between two
~ is directed from positive to
parallel conducting plates, E
negative charges while V increases in the opposite
direction.
Chadi Abou-Rjeily
Electromagnetic Fields Lecture-5: Static Electric Fields 2
Electric Potential (5)
Note that, by definition, the direction of the gradient ∇V is
normal to the surfaces of constant V .
These surfaces of constant potential are called
equipotential surfaces. In two-dimensional situations, these
are referred to as equipotential lines.
Consequently, the electric field lines are everywhere
perpendicular to the equipotential surfaces.
For example, the electric field intensity of a point charge is
radial and the equipotential surfaces are spheres that are
centered at the origin (in the above figure, the equipotential
lines are represented by the dashed circles).
Chadi Abou-Rjeily
Electromagnetic Fields Lecture-5: Static Electric Fields 2
Electric Potential due to discrete charge distribution (1)
The electric potential of a point at a distance R from a point
charge q is:
Z R
~ ~l
V − V∞ = −
E.d
∞
Z R
q
~aR . ~aR dR
⇒ V =−
2
4πǫ0 R
∞
Note that the potential is referred to the potential at infinity
(V∞ = 0).
Note that in the above integral, the components of d~l in the
~aθ and ~aφ directions will cancel out when we perform the
~
dot product with the radial electric field intensity E.
Solving the above integral results in:
q
V =
(V)
4πǫ0 R
Chadi Abou-Rjeily
Electromagnetic Fields Lecture-5: Static Electric Fields 2
Electric Potential due to discrete charge distribution (2)
The potential difference between any two points P1 and P2
at distances R1 and R2 from q is:
q
1
1
V21 = VP2 − VP1 =
−
4πǫ0 R2 R1
Note that this results holds even if P1 and P2 are not on the
same radial line.
Note that concentric circles are equipotential lines.
Since P1 and P3 belong to the same circle, then VP1 = VP3 .
Consequently:
VP2 − VP1 = (VP2 − VP3 ) + (VP3 − VP1 ) = VP2 − VP3
Chadi Abou-Rjeily
Electromagnetic Fields Lecture-5: Static Electric Fields 2
Electric Potential due to discrete charge distribution (3)
In the above figure, no work is done from P1 to P3 .
~ = qE
~ = qER ~aR is radial.
Along this path, the force F
This force is perpendicular to d~l = ~aφ R1 dφ resulting in
~ ~l = 0.
E.d
~ due to a system of n discrete
The electric potential at R
~ ′,...,R
~ ′ is:
charges q1 , . . . , qn located at R
n
1
n
1 X
qk
V =
4πǫ0
~ ~′
k =1 R − Rk Chadi Abou-Rjeily
(V)
Electromagnetic Fields Lecture-5: Static Electric Fields 2
Electric Potential due to discrete charge distribution (4)
Example: Determine the
electric field intensity of an
electric dipole by first
calculating the electric
potential.
The problem will be solved in spherical coordinates. The
electric potential V at point P is given by:


1
1
q 
−

V =
4πǫ0 R
~ − ~d R
~ + ~d 2
2
Chadi Abou-Rjeily
Electromagnetic Fields Lecture-5: Static Electric Fields 2
Electric Potential due to discrete charge distribution (5)
The first term on the right-hand-side can be written as:
−1 "
!
!#−1/2
~
~
~
d
d
d
~
~ −
~ −
R
. R
R − =
2
2
2
−1/2
d2
~ ~
= R 2 − R.
d+
4
"
#−1/2
h
i−1/2
~ ~
R.
d
2
−1
~ d
~
≈ R − R.
≈R
1− 2
R
"
#
~ ~
1 R.
d
d cos θ
−1
−1
≈R
1+
≈R
1+
2 R2
2R
In the same way:
−1
~
d
d cos θ
~
−1
1−
R + ≈ R
2
2R
Chadi Abou-Rjeily
Electromagnetic Fields Lecture-5: Static Electric Fields 2
Electric Potential due to discrete charge distribution (6)
Combining the above equations results in:
qd cos θ
V =
4πǫ0 R 2
In terms of the dipole moment ~p = q ~
d:
~p.~aR
(V)
4πǫ0 R 2
~ can be calculated from:
In spherical coordinates, E
V =
~ = −∇V
E
∂V
∂V
∂V
= − ~aR
+ ~aθ
+ ~aφ
∂R
R∂θ
R sin θ∂φ
p
2 cos θ~aR + sin θ~aθ
=
3
4πǫ0 R
Note that the same result was obtained before by direct
~ . However, determining E
~ from V is
calculations of E
simpler.
Chadi Abou-Rjeily
Electromagnetic Fields Lecture-5: Static Electric Fields 2
Electric Potential due to discrete charge distribution (7)
The next figure shows the electric field lines of a dipole (in
dashed lines) and the equipotential lines (in solid lines):
Chadi Abou-Rjeily
Electromagnetic Fields Lecture-5: Static Electric Fields 2
Electric Potential due to continuous distributions (1)
The electric potential due to a continuous distribution of
charges can be obtained by integration over the charges
region:
For a volume charge distribution:
Z
1
ρ ′
dv
V =
4πǫ0 V ′ R
For a surface charge distribution:
Z
1
ρs ′
V =
ds
4πǫ0 S′ R
For a line charge distribution:
V =
Chadi Abou-Rjeily
1
4πǫ0
Z
L′
ρl ′
dl
R
Electromagnetic Fields Lecture-5: Static Electric Fields 2
Electric Potential due to continuous distributions (2)
Example: Determine the
electric field intensity on the
axis of a circular disk of
radius b that carries a
uniform surface charge
density ρs .
Working with cylindrical coordinates:
Z
1
ρs ′
V =
ds
4πǫ0 S′ R
Z 2π Z b
ρs
1
=
(r ′ dr ′ dφ′ )
1/2
2
′2
4πǫ0 0
0 (z + r )
1/2 b
1/2
ρs
ρs 2
2
′2
2
=
z +r
=
z +b
− |z|
2ǫ0
2ǫ0
0
Chadi Abou-Rjeily
Electromagnetic Fields Lecture-5: Static Electric Fields 2
Electric Potential due to continuous distributions (3)
Note that V is only a function of z. Consequently:
~ = −∇V
E
∂V
= −~az
 ∂zh
i
 ρs 1 − z z 2 + b 2 −1/2 ~az ,
z > 0;
2ǫ0 h
i
=
−1/2
ρ
 − s 1 + z z 2 + b2
~az , z < 0.
2ǫ0
Chadi Abou-Rjeily
Electromagnetic Fields Lecture-5: Static Electric Fields 2
Electric Potential due to continuous distributions (4)
Example: Determine the
electric field intensity along
the axis of a line of length L
that carries a uniform line
charge density ρl .
Consider a point P(0, 0, z) with z > L2 :
Z
1
ρl ′
V =
dl
4πǫ0 L′ R
Z L/2
1
ρl
=
dz ′
4πǫ0 −L/2 z − z ′
ρl
z + (L/2)
L
=
ln
; z>
4πǫ0
z − (L/2)
2
Chadi Abou-Rjeily
Electromagnetic Fields Lecture-5: Static Electric Fields 2
Electric Potential due to continuous distributions (5)
Note that V is only a function of z. Consequently:
~ = −∇V
E
∂V
= −~az
∂z
=
4πǫ0
ρl L
~az
− (L/2)2 ]
[z 2
Note that for very large values of z, z 2 ≫ (L/2)2 resulting
in:
~ = ρl L ~az = Q ~az
E
4πǫ0 z 2
4πǫ0 z 2
where Q = ρl L is the total charge of the line.
Hence, when the point of observation is very far away from
~ approximately follows the inverse
the charged line, E
square law as if the total charge was concentrated at a
point.
Chadi Abou-Rjeily
Electromagnetic Fields Lecture-5: Static Electric Fields 2