1. Contact & Non-contact
folk in 1700’s had trouble wrapping brain
around non-contact forces
Non-contact
Michael Faraday (British: 1791-1867)
Field  area around an object that affects
other similar objects
Test charge  charge that’s in an electric
field

2. Satellit
e Proton
Plan Electro
•
et field by
Gravitational • n
Electric field by
Newton Faraday
Satellite & electron experience field forces

3. • Non-contact
– Remember gravity?
2 types of
GM m
F = ra 2
• What is the field?
• a (or g)
GM
a= r2

4. • Non-contact
– Remember gravity?
– What’s the field for a charge?
2 types of
F= ma kQ q
F = rE 2
• What is the field? What is the
•
• a (or g) field?
•E kQ
GM E=
a= r 2
r2

5. Two equations for electric fields
2 types of
F = Eq
F = force on test charge
q = charge of test charge
E = Electric Field (N/C)
kQ Q creates the field
E=
r 2

8. • Drawing Electric field lines
– Rules
Electric Field
• lines tell direction a + test charge will
go
• Out of (away from) positive
Into (toward) negative
• Lines always come out the object
perpendicular to the surface.
• more lines = stronger field.
+ --

9. + +
+ -

11. 1. Excess charges gather only on
the outside of the surface, never
inside.
conductors in
Properties of
electrostatic
2. The electric field is zero
everywhere inside the
conductor. -- --
3. Electric field coming Σ E = 0 N/C --
--
in or out of the
conductor is -- --
perpendicular to
the surface.

12. 4. On an irregularly shaped surface, the charge
will accumulate more where the radius of
curvature is smallest. -- --
This happens because the --
--
electrons will repel --
each other until they ----
reach equilibrium. --
Grounding
 An easy path for e’s to flow to or from
an infinite reservoir of e’s. Usually
from the Earth to the electrical object.

14. • What is the electric field
intensity of a 16 μC
particle 5 cm away from
the charge?
Sample
kQ
E= r2
E= (9x10 )(16x10-6)
9
0.052
E= 5.75x107 N/C

15. • What is the electric field intensity of
a 16 μC particle 5 cm away from the
charge?
– If a 5 μ C test charge is placed at this
spot, how much force will be exerted on
Sample
it? Qq
k
F= r2
F = (9x10 )(16x10-6) (5x10-6)
9
0.052
F= 288 N
or
F = Eq = 5.75 x 107(5x10-6) =

17. • What is the electric field at the
Sample Problem
origin for the three charge
configuration? (electric field is a
vector, don’t forget to find the
angle)
q1 = -1.00 µC r1 = 3.50 m
q2 = +2.00 µC r2 = 5.00 m
q3 = -1.50 µC r3 = 2.00 m
q2 q3 q1
5.00 m 2.00 m 3.5 m

18. q2 q3 q1
5.00 m 2.00 m 3.5 m
kQ (9x10 )(1x10 ) 9 -6
Sample
Eq1 = 2 = (3.5 ) = 734 N/C
r 2
Same
+
Direction
kQ (9x10 )(2x10 ) 9 -6
Eq2 = 2 = (5 ) = 719 N/C
r 2
-
kQ (9x10 )(1.5x10 )
9 -6
Eq3 = 2 = = 3370 N/C
r (2 ) 2
What is the force on a -3 μC test -1920 N/C
charge placed on the origin?
F=Eq=(-1920)(-3x10-6) = 5.76x10-2 N