Download Physics 208

Coulomb’s Law

1 q1q2
FE 
40 x 2
1
40
 9 109 Newton  meter 2 / coulomb 2  9 109 N  m 2 / C 2
 0 is the permittivity of free space
Charge
Charge
q1
q2
Conservation of electric charge
Charge is conserved: in any isolated
system, the total charge cannot
change.
If it does change, then the system is
not isolated: charge either went
somewhere or came in from
somewhere
Principle of Superposition
The presence of other charges does not change the force exerted by
point charges. One can obtain the total force by adding or
superimposing the forces exerted by each particle separately.
Suppose we have a number N of charges scattered in some region.
We want to calculate the force that all of these charges exert on some
test charge q0 .
q3
q2
q1
q7
q8
q5
q0
q4
q6


1  q0 q1rˆ1 q0 q2 rˆ2
1
 2  2  ... 
Fq0 
40  r1
r2
 40
q0 qi rˆi
.

2
ri
i 1
N
Problem 2 page 9
Two equal, positive charges of magnitude q are
positioned along the x-axis as shown. What would
be the force on a positive charge, q0 , placed on
the y-axis a distance H from the x-axis?
y
q0
H
q
q
a
a
x
Principle of Superposition
(revisited)
The presence of other charges does not change the force exerted by
point charges. One can obtain the total force by adding or
superimposing the forces exerted by each particle separately.
Suppose we have a number N of charges scattered in some region.
We want to calculate the force that all of these charges exert on some
test charge q0 .
q3
q2
q1
q7
q8
q5
q0
q4
q6


1  q0 q1rˆ1 q0 q2 rˆ2
1
 2  2  ... 
Fq0 
40  r1
r2
 40
q0 qi rˆi
.

2
ri
i 1
N


1  q0 q1rˆ1 q0 q2 rˆ2
1
 2  2  ... 
Fq0 
40  r1
r2
 40
q0 qi rˆi
.

2
ri
i 1
N
N 
We introduce the charge density or charge per unit volume

q

How do we calculate the total force acting on the test charge
q?
We chop the blob up into little chunks of volume V ; each
chunk contains charge q  V . Suppose there are N
chunks, and we label each of them with some index .
i
q
r̂i
Let r̂i be the unit vector pointing from i th chunk to the
test charge; let ri be the distance between chunk and
test charge. The total force acting on the test charge is
 N 1 q( Vi )rˆi
F 
2
4

r
i 1
0
i
This is approximation!
The approximation becomes exact if we let the number of chunks go to
infinity and the volume of each chunk go to zero – the sum then becomes
an integral:

1 qdV rˆ
F 
2
40 r
V
If the charge is smeared over a surface, then we integrate a surface
charge density
over the area of the surface A:


1 qdA rˆ
F 
2
4

r
0
A
If the charge is smeared over a line, then we integrate a line charge
density
over the area of the length:


1 Qdl rˆ
F 
2
4

r
0
L
Another example on force due to a
uniform line charge
A rod of length L has a total charge Q smeared
uniformly over it. A test charge q is a distance a
away from the rod’s midpoint. What is the force
that the rod exerts on the test charge?

dx
( x  c)
2
3
2
x

c( x  c)
2
1
2

xdx
( x 2  c)
3
2

1
( x 2  c)
1
2
The electric field
y
q3
q0
q4
x
q1
q5


F
E  lim
q0  0 q
0


E has the same direction as F
q2

 Newtons N
E 

Coulomb C
Michael Faraday
1791-1867
“The best experimentalist in the history of science”
Electric field lines
These are fictitious lines we sketch which point in
the direction of the electric field.

1) The direction of E at any point is tangent to the
line of force at that point.
2) The density of lines of force in any region is
proportional to the magnitude of E in that region
Lines never cross.
How to calculate

E
?
1) Put a “test charge” q0 at some point and do not
allow it to move any other charges
2) Calculate
the electric force on

obtain E
q0 and divide by q0
The force that N charges exert on a test charge

1
Fq0 
40
q0:
q0 qi rˆi
.

2
ri
i 1
N
to
We also calculated the force that a blob of charged material
with charge density
exerts on a test charge:


1 q0 dV rˆ
F 
2
40
r
V
We wrote the similar formulas if the charge is smeared out over a
surface with surface density
, or over a line with line density

In all of these cases, the force ends up proportional to the test charge
We might factor it out. This is the electric field!

 F
E
q0
.
q0 .

 F
E
q0
qi rˆi


2
40 i 1 ri
1

V
N
1
dV rˆ
40
r2
(N point charges)
(Charge continuum)
Given an electric field, we can calculate force exerted
on some point charge q :


F  qE
Example 1: Electric field of a point charge is
directly radially away from or toward the charge.

E
1
q
ˆ
r
2
40 r
Example 2: Electric field of a dipole
Have a great day!
Hw: All Chapter 1 problems
and exercises
Reading: Chapter 1, 2