Post on 20-Jan-2018
description
Prof. David R. JacksonECE Dept.
Spring 2016
Notes 14
ECE 3318 Applied Electricity and Magnetism
1
Potential Calculation from Field
2
In this set of notes we explore calculating the potential function (x,y,z) (assuming that we already know the electric field).
Note: The electric field is found first, either from Coulomb’s law (superposition) or from Gauss’s law.
Choose:
Potential Formula
B
ABA
V E dr
A B
R
r
r R E dr
r
R
r R E dr 3
or
A r B R
r
R
Then we have:
Potential Formula (cont.)
r
R
r R E dr
Cr
R
4
1) Pick a reference point (if it is not already given).2) Integrate in a coordinate system of your choice (remember that any path can be chosen in statics).
Recipe:
Path Independence of Voltage Drop
20
0
0
ˆ ˆˆ ˆ sin
4
4
1 14
B
A
B
A
B
A
B
AB rA
r
rr
r
r
r
r
A B
V r E r dr r d r d
E dr
q drr
qr
qr r
Consider first a point charge:
By superposition, the potential from any set of charges is path independent in statics.
The integral is path independent (the answer only depends on the endpoints).
5
y
z
q
Point charge
x
A
CB
Example
Given: R = , (R) = 0
20
ˆ4
qE rr
Choose a radial path for convenience.
Find the potential function
r
R
r R E dr 0r
R
E dr
y
z
q
r
RC
Point charge
x
6
Example (cont.)
20
0
ˆ ˆˆ ˆ sin
4
4
r
rR
r
rR
r
r
r E r dr r d r d
E dr
q drr
qr
0
V4
qrr
7
Hence, we have
Summary for a Point Charge
0
V , 04
qrr
8
20
V/mˆ4
qE r rr
y
z
q
r
x
rThese results for a point charge form the building blocks for all
other charge distributions (using superposition).
This is done in Notes 15.
ExampleGiven: R is at = b, (R) = 0
0
0
ˆ2
E
Find the potential function
Choose a radial path
0
ˆˆ ˆ ˆ
r
R
r
R
b
E dr
E d z dz d
E d
Infinite line charge
9
x
y
z
l0 [C/m]
r
R ( = b)
C
b
r
R
r R E dr
Example (cont.)
0
0
0
0
0
0
2
ln2
ln2
b
b
d
b
Note: b cannot be chosen as (there is an infinite voltage drop between = and = ).
0
0ln V
2b
10
In 2D problems (infinite in z) the reference point cannot be put at infinity.
Example
x
y
z
a
l0 [C/m]
r (0,0,z)
Given: () = 0
Note: We only know E on the z axis (from a previous example), so we MUST choose a path on the z axis.
Find the potential function on the z axis
Circular ring of line charge
11
r
R
r R E dr
Example (cont.)
() = 0
C
x
y
z
r
R ()
Choice of path:
12
r
R
r R E dr
Example (cont.)
From Coulomb’s law, we know that on the z axis the field is
ˆ ˆˆ ˆr
zR
z
z
r z E xdx y dy z dz
E dz
03/22 2
0
0, 0,2z
a zE zz a
13
0r
R
r E dr
r
R
r R E dr
so
Example (cont.)Hence,
03/22 2
0
03/22 2
0
1/22 20
0
0, 0,2
2
2
z
z
z
a zz dzz a
a z dzz a
a z a
02 2
0
10, 0, V2
azz a
The result is:
14
Adding a Constant to
Proof:
2
22 2
r
R
r R E dr
1
11 1
r
R
r R E dr
1
2
1
2
1 21 2 1 2
1 21 2
Rr
R r
R
R
r r R R E dr E dr
R R E dr
constant
C1
R1
r
C2
R2
15
1 1 01 2 2 02,R R
Assume two solutions:
Two different solutions for the potential function can only differ by a constant.
Adding a Constant to (cont.)
2 1, , , ,x y z x y z C
Conclusion:
Valid potential functions can only differ by the addition of a constant.
Adding a constant to a valid potential function gives a another valid potential function (this does not change the electric field).
16
ExampleGiven: (z = 1 [m]) = 10 [V]
Find the potential function on the z axis
02 2
0
10, 0,2
az Cz a
Start with our previous solution, which has zero volts at infinity, and add a constant to it:
Circular ring of line charge
x
y
z
al0 [C/m]
r (0, 0, z)
R (0, 0, 1)10 [V]
17
Example (cont.) 0
2 20
10, 0,2
az Cz a
0, 0,1 10 [V]
02 2
0
1 102 1
a Ca
Set:
x
y
z
a
l0 [C/m]
r (0, 0, z)
R (0,0,1)10 [V]
18
02 2
0
110 [V]2 1
aCa
Hence we have:
02 2 2 2
0
1 10, 0, 10 [V]2 1
azz a a
The solution is thus: