Prof. David R. Jackson ECE Dept. Fall 2014 Notes 7 ECE 2317 Applied Electricity and Magnetism Notes prepared by the EM Group University of Houston 1.

Prof. David R. JacksonECE Dept.

Fall 2014

Notes 7

ECE 2317 Applied Electricity and Magnetism

Notes prepared by the EM Group University of Houston

1

Coulomb’s Law

1 22 2

0

120

N

F/m

ˆ4

8.854187818 10

q qF r

r

Experimental law:

c = speed of light = 2.99792458108 [m/s] (defined)

0 0

1c

7

0 H/m4 10 ( ) exact

Charles-Augustin de Coulomb

2

Here is how we can calculate 0 accurately:

(from ECE 3317)

0 20

1

c

x

y

z

r

q1

q2

r

Coulomb’s Law (cont.)

Hence:

But

1 22 2

0

ˆ4

q qF r

r

E1 = E due to q1

2 2 1F q E r

Note: There is no self-force on charge 2 due to its own electric field.

3

r = location of observer

11 2

0

ˆ4

qE r r

r

x

y

z

r

q1

q2

r

r1= (0, 0, 0)

r2= r

Generalization (q1 not at the origin):

Coulomb’s Law (cont.)

r1= (x1, y1, z1)

r2= (x2, y2, z2)

2 1

12 2 2 2

2 1 2 1 2 1

R R r r

x x y y z z

121 2

0

ˆ4

qE r R

R ˆ R

RR

2 1R r r q2 (x2, y2, z2)

x

y

zR R

r2

r1

q1 (x1, y1, z1)

4

2 1 2 1 2 1ˆ ˆ ˆR x x x y y y z z z

Example

q1 = 0.7 [mC] located at (3,5,7) [m]

q2 = 4.9 [C] located at (1,2,1) [m]

E1 (r2) = electric field due to charge q1, evaluated at point r2

22 2 1F q E r

Find: F1, F2

For F2:

F1 = force on charge q1

F2 = force on charge q2

Rq2

q1

5

Example (cont.)

121 2

0

2 1

2 2 2

m

m

ˆ4

ˆ ˆ ˆ1 3 2 5 1 7

ˆ ˆ ˆ2 3 6 [ ]

2 3 6 7 [ ]

2 3 6ˆ ˆ ˆ ˆ7 7 7

qE r R

R

R r r x y z

x y z

R R

RR x y z

R

q1 = 0.7 [mC] located at (3,5,7) [m]

q2 = 4.9 [C] located at (1,2,1) [m]

22 2 1F q E r

6

Rq2

q1

Example (cont.)

3

21 212

4

62 22 2 1 1

V/m

N

0.7 10 2 3 6ˆ ˆ ˆ

7 7 74 8.854 10 7

ˆ ˆ ˆ1.834 10 2 3 6 [ ]

4.9 10

ˆ ˆ ˆ0.08988 2 3 6 [ ]

E r x y z

x y z

F q E r E r

x y z

(Newton’s Law)

2

1

N

N

ˆ ˆ ˆ0.180 0.270 0.539 [ ]

ˆ ˆ ˆ0.180 0.270 0.539 [ ]

F x y z

F x y z

1 2F F

7

General Case: Multiple Charges

q1 : r1 = (x1, y1, z1)

R1 = r - r1

q2 : r2 = (x2, y2, z2)

...

qN : rN = (xN, yN, zN)

R2 = r - r2

...

RN = r - rN

E=E1+E2+…+EN (superposition)

1 21 22 2 2

0 1 0 2 0

ˆ ˆ ˆ...4 4 4

NN

N

qq qE r R R R

R R R

x

y

z

R1

q2

q1 R2 RNR3

q3 qN

r = (x, y, z)

8

Field from Volume Charge

1/22 2 2

ˆ ˆ ˆ

ˆ

R x x x y y y z z z

R R x x y y z z

RR

R

9

r = (x, y, z)

x

y

z

R , ,r x y z

R r r

dV´

, ,v vr x y z ˆ

ˆ

r

r

No

The

The bl

red d

ue dot is as

ot is associ

soc

ate

iated wit

d wit

h

h

te :

Field from Volume Charge (cont.)

20

20

ˆ4

, , ˆ4

v

dQdE R

R

x y z dVR

R

2

0

ˆ4

v

V

rE r R dV

R

10

x

y

zr = (x, y, z)

R

dV´

, ,r x y z

vdQ r dV

R r r R R r r

2

0

ˆ4

s

S

rE r R dS

R

Field from Surface Charge

11

r = (x, y, z)

x

y

z

R

dSdS´

, ,r x y z

sdQ r dS

R r r R R r r

2

0

ˆ4

l

C

rE R dl

R

Field from Line Charge

12

x

y

z r = (x, y, z)

R

dl

, ,r x y z

ldQ r dl

R r r R R r r

Exampleq1 = +20 [nC] located at (1,0,0) [m]

q2 = -20 [nC] located at (0,1,0) [m]

Find E (0,0,1)

R1 = (0,0,1) - (1,0,0)

R2 = (0,0,1) - (0,1,0)

1

1

1

1,0,1

2

1ˆ 1,0,12

R

R

R

2

2

2

0, 1,1

2

1ˆ 0, 1,12

R

R

R

r = (0,0,1)

x

y

z

R1

q2 = -20 [nC]

R2

q1 = +20 [nC]

Solution:

13

1 21 22 2

0 1 0 2

ˆ ˆ4 4

q qE R R

R R

Example (cont.)

1 21 22 2

0 1 0 2

9 9

2 212 12

ˆ ˆ4 4

20 10 1 20 10 11,0,1 0, 1,1

2 24 8.854 10 2 4 8.854 10 2

63.55 1,0,1 0, 1,1

63.55 1, 1,0

q qE R R

R R

V/mˆ ˆ63.55E x y

14

Example

2

ˆ ˆ ˆ

ˆ

ˆ ˆ

R x x x y y y z z z

R z h z

R h z h z h z

R z

Semi-infinite uniform line charge

dl dz

15

2

0

ˆ4

l

C

rE R dl

R

x

y

z

r = (0, 0, h)

R

l = l0 [C/m]

Find E (0, 0, h)

0,0,r z

Example (cont.)

00

20

00

20

0

0

0

ˆ

4

ˆ 1

4

ˆ 1

4

l

l

l

zE dz

h z

zdz

h z

z

h z

0

0

V/mˆ [ ]4

lE zh

Note: The upper limit must be greater than the lower limit to keep dl positive.

16

Example

Find E (0, 0, z)

2

0C

d a d

17

l = l0 [C/m] (uniform)

x

y

z

a

R

r = (0, 0, z)(rectangular coordinates)

, ,0 ( )r a cylindrical

Note: The upper limit must be greater than the lower limit to keep dl positive.

0d so

xa

y

d a d

d

2

0

ˆ4

l

C

rE R dl

R

Example (cont.)

ˆ ˆ ˆ0 cos 0 sin 0

ˆ ˆ ˆcos sin

R x a y a z z

a x y z z

ˆ ˆ ˆcos sinx y Note:

ˆ ˆ ˆR x x x y y y z z z

ˆ ˆR a z z

Hence

y

x

z

a

R

r = (0, 0, z)

18

x

y

Example (cont.)

2 2

2 2

ˆ ˆ

ˆ ˆˆ

R a z z

R a z

a zzR

a z

We can also get these results geometrically, by simply looking at the picture.19

y

x

z

a

R

r = (0, 0, z)

ˆa

ˆz z

2

0

ˆ4

l

C

rE R dl

R

Example (cont.)

20

20

12 22 200 2

2 20

32 20 0 02

ˆ4

ˆ ˆ1

4

ˆ ˆ4

l

C

l

l

rE R dl

R

a zza d

a za z

aa d z z d

a z

20

Reminder: The upper limit must be greater than the lower limit to keep dl positive.

Example (cont.)

0

32 2 2

0

ˆ 2

4

l aE z z

a z

2

0

ˆ 0d

ˆ ˆ ˆcos sinx y

0

32 20 2

V/mˆ [ ]2

l a zE z

a z

2

0

2d

Also,

Hence

21

a

x

y

Example (cont.)

Limiting case: a 0 (while the total charge remains constant)

0 0 03 23

20 0 02

0

20

20

1ˆ ˆ ˆ

2 2 2

2 1ˆ 1 0, 0

4

ˆ [V/m]4

l l l

l

a a az zE z z z

zzz

az z z

z

Qz

z

when when

(point-charge result)

0 2lQ a (total charge on ring)

Note: If we wish Q to remain fixed, than l0 must increase as a gets smaller.

22

where