Download ppt ElecForce

5/13/2009
Vector Review
Add vectors by components
Electric Forces
Physics 122
Eyres
F1 = 35N @ 57 degrees E of N
F2 = 72N F2
72N
@ ‐116 degrees
Resultant
F 1 + F 2 =
35 cos(33)=
+29.35
35 sin(33)=
+19.06
72 cos(64)=
‐31.56
72 sin(64)=
‐64.71
‐2.21
‐45.65
45.7N @ _87 deg below –x axis_
Charge
• Properties
– Quantized: multiple of a fundamental unit of charge, e
• The SI unit of charge is the Coulomb (C)
• e = 1.6 x 10‐19 C
–
–
–
–
Exceptions: see fundamental particle chart
Positive: proton charge +e Negative: electron charge ‐e
Repel (Like) Attract (Unlike) • Neutral objects: charge separation
Charging an Object
• Contact
– A charged object (rod) is placed in contact with a neutral object (sphere)
– Some electrons on the Some electrons on the
rod move to the sphere
– When the rod is removed, the sphere is left with a charge
– The object being charged with same type charge as rod
1
5/13/2009
Charging an Object
• A negatively charged rubber rod is brought near an uncharged sphere
• The charges in the sphere are redistributed
• A grounded conducting wire is connected to the sphere
• The wire to ground is removed, the sphere is left with excess positive charge
• The positive charge evenly distributes
Electric Force
• Coulomb’s Law
– q is charge
– r is distance between centers
– k is a constant
F = ke
Electric Force
• Superposition of Forces
– Remember how to add vectors
– Use tabular form
q1 q2
r2
Find Fe
• Find the Electric Force on q1 caused by the q2 and q3.
• q1= q2 = 4 μC
8 m
• q3 = q4= ‐3 μC
q1
q2
q4
• R12=5 m
• R13=8m
q3
6 m
2
5/13/2009
Finish the problem
F on 1 by 2 @37 degrees W of N
F=
8.99 x109
Negative
5.75x10‐3 sin 37º=
Nm 2
−6
−6
(
4
x
10
C
)(
4
x
10
C
)
‐3.46x10-3
C2
52 m 2
F on 1 by 3 @ ‐90 degrees
Zero
Nm 2
N
8.99 x109 2 (4 x10 −6 C )(3 x10 −6 C )
C
F=
82 m 2
‐3.46x10-3
Resultant
F on 1 by 2 and 3 =
-3
4.52x10 N @ 40º above –x axis
Positive
5.75x10‐3 cos 37º=
+4.6x10-3
Negative
1.69x10
69x10‐3
‐1
2.91x10-3
Electric Field
• Force per Unit test charge: A Ratio
E=
Fonq0byQ
• From Coulomb’s Law
q0
kQq0
2
E= r
q0
E=
Comparing Fields
• Electric Fields
F=
kqbyy qon
2
r
⎛ kqby ⎞
F = ⎜⎜ 2 ⎟⎟qon
⎝ r ⎠
F = Eqon
• Gravitational Fields
F=
kQ
r2
Field Lines
Electric Field
Gravitational Field
Gmbyy mon
r2
⎛ Gmby ⎞
F = ⎜⎜ 2 ⎟⎟mon
⎝ r ⎠
F = gmon
Fig. 15.13, p.399
3
5/13/2009
Find Ee
Finish the problem
E at 1 by 2 @37 degrees W of N
• Find the Electric Field at location p caused by the q2 and q3.
• q3 = q4= ‐3 μC
p
E=
q2
8 m
q4
Nm 2
(4 x10 −6 C )
C2
52 m 2
E at 1 by 3 @ ‐90 degrees
E=
• R12=5 m
• R13=8m
8.99 x10 9
8.99 x109
Nm 2
N
(3 x10 −6 C )
C2
82 m 2
Resultant
E at 1 by 2 and 3 =
q3
6 m
Simulations
• http://www.vias.org/si
mulations/simusoft_efi
eld.html
• http://phet.colorado.ed
u/sims/charges‐and‐
fields/charges‐and‐
fields_en.html
4