prof. jeffery t. williams dept. of electrical & computer engineering university of houston fall...

Prof. Jeffery T. WilliamsDept. of Electrical & Computer Engineering

University of Houston

Fall 2004

Coulomb’s Law

ECE 2317: ECE 2317: Applied Electricity and MagnetismApplied Electricity and Magnetism

Charles Coulomb

Coulomb’s Force Law

• proportional to the product of the charges,

• inversely proportional to the square of the distance between them,

• directed along the line joining them, and

• repulsive for like charges and attractive for unlike charges.

Charles A. Coulomb, a French military engineer, measured the forces between electric charges, reporting his results in 1785. He found that the electric force between two charged particles was

y

z

x

1q

2q

r

21F

r̂

Coulomb’s LawCoulomb’s Law

1 22 2

0

120

ˆ N4

8.854187818 10 F/m

q q

r

F r

Experimental law:

Note: c = speed of light = 2.99792458108 [m/s] (exactly)

0 0

1c

7

0 4 10 H/m (exactly)

x

y

z

q1

q2

r̂

r

Electric FieldElectric Field

Hence:

But

1 22 2

0

ˆ4

q q

rF r

12

0

ˆ4

q

r1E r

2 2 1 2qF E r

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

x

y

z

q1

q2

r̂

r

( : electric field due to q1 )1E

Generalization (q1 not at the origin):

Generalizing Coulomb’s LawGeneralizing Coulomb’s Law

r1= (x1, y1, z1)

q2 (x2,y2,z2)

x

y

z R̂q1 (x1,y1,z1)

r2= (x2, y2, z2)

2 1

12 2 2 2

2 1 2 1 2 1

R

x x y y z z

R r r

12

0

ˆ4

q

R1E R

R

RR

2 1 R r r

1r2r

R

ExampleExample

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

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

q1

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

2q2 1F E

Find: F1, F2

For F2:

F1 = force on charge q1

F2 = force on charge q2

R

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

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

Rq2

q1

11 2

0

2 1

2 2 2

4

ˆ ˆ ˆ1 3 2 5 1 7

ˆ ˆ ˆ2 3 6

2 3 6 7 [ ]

2 3 6ˆ ˆ ˆ

7 7 7

q

R

R m

E R

R r r x y z

R x y z

R

RR x y z

R

2 2q 1F E

Example (cont.)Example (cont.)

3

1 212

41

62 2 1 1

2

2

1

ˆ ˆ ˆ0.180 0.

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 [ ]

270 0.539 [ ]

ˆ ˆ ˆ0.180 0.270

q

N

N

E x y z

E x y z

F

F x y z

F

F x

x

E

y

E

y z

0.539 [ ]Nz(Newton’s Law)

x

y

z

General Case: Multiple ChargesGeneral Case: Multiple Chargesq1 : r1 = (x1,y1,z1)

q2

q1

q3 qN

q2 : r2 = (x2,y2,z2)

...

qN : rN = (xN,yN,zN)

(superposition)

1 21 22 2 2

0 1 0 2 0

...4 4 4

NN

N

qq q

R R R E R R R

r

1 1 R r r

2 2 R r r

N N R r r1 2 N E E E E

1R

2R3R

NR

x

y

z

Field from Volume ChargeField from Volume Charge

r= (x,y,z)

V (r´) = V (x´,y´,z´)

1/ 22 2 2

ˆ ˆ ˆ' ' '

' ' '

x x y y z z

R x x y y z z

R r r x y z

R

RR

R

' ', ', 'x y zrR

x

y

z

Field from Volume Charge (cont.)Field from Volume Charge (cont.)

r = (x,y,z)

R

dQ = V (r´) dV´

dV´

20

20

4

', ', ' '

4V

dQd

R

x y z dV

R

E R

R

2

0

''

4V

V

r dVR

r

E R

' ', ', 'x y zr

r = (x,y,z)

dQ = S (r´) dS´

2

0

''

4S

S

dSR

r

E R

Field from Surface ChargeField from Surface Charge

x

y

z

R

dSdS´

' ', ', 'x y zr

dQ = l (r´) dl´

2

0

''

4l

C

rdl

RE R

Field from Line ChargeField from Line Charge

x

y

z r = (x,y,z)

R

dl´

' ', ', 'x y zr

ExampleExample

r = (0,0,1)

x

y

z

R1

q2 = -20 [nC]

R2

q1 = +20 [nC]

q1 = +20 [nC] located at (1,0,0)

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

Find E (0,0,1)

Solution:

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

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

1

1

1

1,0,1

2

11,0,1

2

R

R

R

2

2

2

0, 1,1

2

10, 1,1

2

R

R

R


1 21 22 2

0 1 0 2

9 9

2 212 12

4 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 q

R R

E R R

ˆ ˆ63.55 V/m E x y

x

y

z

ExampleExample

r = (0,0,h)

R

l = lo [C/m]

r´= (0, 0, z´ )

20

2

''

4

ˆ '

' ' '

ˆ

l

C

rdl

R

h z

R h z h z h z

E R

R r r

z

R z

Find E(0,0,h)

h>0

semi-infinite uniform line charge


0

20

0

20

0

0

ˆ'

4 '

ˆ 1'

4 '

ˆ 1

4 '

lo

lo

lo

dzh z

dzh z

h z

zE

z

z

0

ˆ [V/m]4

lo

h

E z

ExampleExample

x

a

´

d´

dl´= a d´

l = lo [C/m] (uniform)

Find E(0,0,z)

y

x

y

z

a

R

r = (0,0,z)

r´= (a, ´, 0 )

d´ > 0 so 2

0

' 'C

d a d

2

0

''

4l

C

dR

r

E R


ˆ ˆ ˆ0 cos ' 0 sin ' 0

ˆ ˆ ˆcos ' sin '

a a z

a z

R x y z

x y z

ˆ ˆ ˆ' cos ' sin ' ρ x y

x

´

ˆ 'ρ

Note:

R r r

y

x

y

z

a

R

r = (0,0,z)

r´= (a, ´, 0 )


2 2

2 2

ˆ ˆ'

ˆ ˆ'

a z

R a z

a z

a z

R ρ z

ρ zR

R

a

z

x


20

2

12 22 200 2

2 2

32 20 0 02

''

4

ˆ ˆ1 ''

4

ˆ ˆ' ' '4

l

C

lo

lo

rdl

R

a za d

a za z

aa d z d

a z

E R

ρ z

ρ z


3

2 2 20

ˆ 2

4

lo az

a z

E z

2

0

ˆ ' ' 0d ρ

ˆ ˆ ˆ' cos ' sin ' ρ x y

3

2 20 2

ˆ [V/m]2

lo a z

a z

E z

2

0

' 2d

ˆρ


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

30

0

20

20

ˆ2

2 1ˆ 1 when 0, when 0

4

ˆ4

lo

l

a z

z

az z

z

Q

z

E z

z

z

(point-charge result)

prof. jeffery t. williams dept. of electrical & computer engineering university of houston fall...

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