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Transcript
10/27/2014
Electric Field
• Force per Unit test
charge: A Ratio
Electric Field & Potential
Physics 115
Eyres
E=
Fonq0byQ
• From
Coulomb’s
Law
q0
kQq0
2
E= r
q0
E=
Comparing Fields
• Electric Fields
F=
kqby qon
2
r
 kqby 
F =  2 qon
 r 
F = Eqon
kQ
r2
Electric field due to a single point-like
charged object
• Gravitational Fields
F=
Gmby mon
r2
 Gmby 
F =  2 mon
 r 
F = gmon
© 2014 Pearson Education, Inc.
Find Fe
Using the superposition principle
• Find the Electric Force
on q1 caused by the q2
and q3.
• q1= q2 = 4 µC
8m
• q3 = q4= -3 µC
q1
q2
q4
• R12=5 m
• R13=8m
q3
6m
© 2014 Pearson Education, Inc.
1
10/27/2014
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
F=
8.99 x109
Zero
Nm 2
(4 x10 −6 C )(3 x10 −6 C )
C2
82 m 2
Resultant
F on 1 by 2 and 3 =
4.52x10-3N @ 40º above –x axis
-3.46x10-3
Positive
5.75x10-3 cos 37º=
+4.6x10-3
Negative
-1.69x10-3
Find Ee
• Find the Electric Field at
location p caused by the
q2 and q3.
• q2 = 4 µC
8m
• q3 = q4= -3 µC
p
q2
q4
2.91x10-3
• R12=5 m
• R13=8m
q3
6m
Finish the problem
E at 1 by 2 @37 degrees W of N
E=
8.99 x109
Nm 2
( 4 x10 −6 C )
C2
2
2
5 m
E at 1 by 3 @ -90 degrees
E=
8.99 x109
Nm 2
(3 x10 −6 C )
C2
82 m 2
Resultant
E at 1 by 2 and 3 =
E field lines
Field Lines
Electric Field
© 2014 Pearson Education, Inc.
Gravitational Field
Fig. 15.13, p.399
2
10/27/2014
E field lines
Simulations
• http://www.vias.org/si
mulations/simusoft_efi
eld.html
• http://phet.colorado.ed
u/sims/charges-andfields/charges-andfields_en.html
© 2014 Pearson Education, Inc.
Electric potential due to a single
charged object
© 2014 Pearson Education, Inc.
Quantitative Exercise 15.6
• Suppose that the heart's dipole charges −Q and +Q
are separated by distance d. Write an expression for
the V field due to both charges at point A, a distance
d to the right of the +Q charge.
1. Simplify and diagram.
2. Represent mathematically.
−
, =
2
+
, =
1
Add:
=
© 2014 Pearson Education, Inc.
+
Potential difference
• The value of the electric potential depends on
the choice of zero level, so we often use the
difference in electric potential between two
points.
=
© 2014 Pearson Education, Inc.
3
10/27/2014
Checking Understanding
Change in Electric Potential
=
=
!
−
=
−
=
−
=
"
!
"
=
Rank in order, from largest to smallest, the electric
potentials at the numbered points.
Δ
#$
%&
'(
!
"
-
−
#$
%&
')
!
"
−
=Δ
=
HINT:
Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley.
Example
A proton has a speed of 3.5 x 105 m/s at a point where the
electrical potential is 600 V. It moves through a point where the
electric potential is 1000 V. What is its speed at this second point?
Its about Energy! How do
you know?
Δ
=Δ
Equipotential surfaces: Representing
the V field
• The lines represent
surfaces of constant
electric potential V,
called equipotential
surfaces.
• The surfaces are
spheres (they look
like circles on a twodimensional page).
© 2014 Pearson Education, Inc.
Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley.
Contour maps: An analogy for
equipotential surfaces
A Topographic Map
© 2014 Pearson Education, Inc.
Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley.
4
10/27/2014
Graphical Representations of Electric Potential
Δ
Connecting Potential and Field
=Δ
Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley.
Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley.
Potential and Field for Three Important Cases
The Potential Inside a Parallel-Plate Capacitor
Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley.
+,-./ =
Δ
Δ0
Slope = E
Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley.
Example
Example
A proton is released from rest at point a. It then travels past point
b. What is its speed at point b?
A proton has a speed of 3.5 x 105 m/s at a point where the
electrical potential is 600 V. It moves through a point where the
electric potential is 1000 V. What is its speed at this second point?
(How is this similar to the previous problem?)
Its about Energy! How do
you know?
Δ
Its about Energy! How do
you know?
=Δ
Δ
=Δ
Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley.
Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley.
5
10/27/2014
Example: (How is this similar to the previous problems?)
Example
For the situation shown in the figure, find
A parallel-plate capacitor is held at a potential difference of 250 V.
A proton is fired toward a small hole in the negative plate with a
speed of 3.0 x 105 m/s. What is its speed when it emerges
through the hole in the positive plate? (Hint: The electric potential
outside of a parallel-plate capacitor is zero).
A. The potential at points a and b.The potential difference
between a and b.
B. The potential energy of a proton at a and b.
C. The speed at point b of a proton that was moving to the right at
point a with a speed of 4.0 x 105 m/s.
D. The speed at point a of a proton that was moving to the left at
point b with a speed of 4.0 x 105 m/s.
Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley.
Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley.
Example
Source charges create the
electric potential shown.
A. What is the potential at point
A? At which point, A, B, or C,
does the electric field have its
largest magnitude?
B. Is the magnitude of the
electric field at A greater than,
equal to, or less than at point
D?
C. What is the approximate magnitude of the electric field
at point C?
D. What is the approximate direction of the electric field at
point C?
Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley.
6