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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
Example 3: Electric field at the
center of a charged ring
Think first!
(before you start doing calculations)

E 0
Example 4: Find the electric field at the center of
a semi-circle of radius R, if a charge Q is uniformly
spread over the semi-circle.

E
O
Q
2  0 R
2
2

ix
Have a great day!
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