Download The Electric Field

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project

Document related concepts
no text concepts found
Transcript
W1 D3
More on Chapter 14
New Groups are posted on our website.
The Electric Field – W1D3
No QUIZ today. First quiz next Friday
CLICKERS: MONDAY START
Last Time
Coulombs' Law
Vector Notation: <a,b,c>
Superposition
Symmetry
Today – The concept of the Electric Field
Possibly some discussion about the electric dipole
Next Major Topic: Charge and Matter
Note: There is a NEW WebAssign that will appear
sometime today. Topic: E-Field
F
1
q1q2
4 0 r2  r1
r

r
2
 (Unit Vector in the (r2 - r1 )direction)
r
tinu
(r2 - r1 )unit
r2 - r1

r2 - r1
In the figure, what are the horizontal and vertical components of the
resultant electrostatic force on the charge in the lower left corner of the
square if q = 1.5 x 10-7 C and a = 4.0 cm? (Assume the positive
directions are upward and to the right.)
r1  a, 0, 0 
y
2
3
r3
r2
r2  a, a, 0 
r3  0, a, 0 
r1 = r3  a
r2  (a)2  (a)2  (0)2  2a 2  a 2
1
r1
x
Note that r goes from the origin to the point where we are computing
the force. For our example, r=0.
runit
r
=
r
r1  a, 0, 0 
r2  a, a, 0 
r3  0, a, 0 
r1 = r3  a
r2  (a)2  (a)2  (0)2  2a 2  a 2
q  2 x 2qr1unit 2qr2unit 2qr3unit
F



2
2
4 0 
r1
r2
r32
q2

4 0
 4r1unit 2r2unit
r3unit

2 2

2
2
r2
r3
 r1
q2

4 0
 4r1 2r2 2r3 
q2
 3  3  3  `
r2
r3  4 0
 r1






 4r1
2r3 
2r2
 3 
 3  3
2a 2 a 
 a
r1  a, 0, 0 
 2qr1unit qr2unit qr3unit



2
2
2
r
r
r

3
2
1
2q 2  2r1unit r2unit r3unit 
 2 



2
2
4 0  r1
r3 
r2
2q
F
4 0
2q 2

4 0



r2  a, a, 0 
r3  0, a, 0 
 2r1 r2 r3  2q 2  2r1
r3 
r2





 3
 3
3 
3 
3
3

4
a
a
r
r
r
2
a
2
 1

0 
3 
2

2q 2

4 0 a 3


a

0
,
a

0,



0
,

,
a




0
0,
,
a


2



2
2


2q 2

4 0 a 3


a

0
,
a

0,



0
,
,
a



0
0,
,
a
2



2
2


2q 2

4 0 a 2


1

0
1,

0,



0
,
1,



0
0,
2,

  0.6, 0.16, 0  NEWTONS

2 2


A proton is placed a distance d above a table and is then released.
Describe its motion. The room is in a vacuum.


“wt_?”
d
d
Observed
F  qE
F
E
q
E(r)
1
1
Related documents