How to use electric fields?
Clicker Question
- field due to a point charge:
E
Q = "source charge"
E-field at point P
distance = r
Compare the electric force on ball 1 from ball 2 and the rod.
A) The rod exerts a larger magnitude force.
B) The ball 2 exerts a larger magnitude force.
C) The rod and ball 2 exert the same magnitude force.
v kQ
E = 2 rˆ
r
Note that sometimes we
exchange one constant for
another. See why later...
k=
1
4πε 0
1
2
Electric Field Strength:
First attempt to visualize Electric fields...
Inside a Current Carrying Wire
|E| ~ 10-2 N/C
Near the Earth's Surface
|E| ~ 102- 104 N/C
Electric Breakdown in Air
|E| ~ 106 N/C
Inside an Atom
|E| ~ 1011 N/C
3
4
- field from several discrete point charges
E
y^
Electric Field from a Dipole
+Q
Find the E
- field at this point.
d
+Q
x^
d
Dipole is defined as equal and opposite separated charges.
5
+Q
x
6
1
Clicker Question
y
+Q
Clicker Question
If we move this point P very far away
(x very large), we expect the E-field
strength to approach...
d
+Q
x^
d
+Q
x
Ball 2 has charge Q. Ball 2 is torn
apart into a rod of 10 smaller balls
each with charge Q/10.
A) |E| = kQ/x2
B) |E| = 3kQ/x2
C) |E| = kQ/x
D) |E| = 0
E) |E| = kQ
Compare the electric force on ball 1 from ball 2 and the rod.
A) The rod exerts a larger magnitude force.
B) The ball 2 exerts a larger magnitude force.
C) The rod and ball 2 exert the same magnitude force.
7
Clicker Question
+Q
What is the E-field at point P?
- Q
s
8
Clicker Question
+2Q
A) |E| = 2kQ / s2
What is the E-field at point P?
- Q
s
A) |E| = 2kQ / s2
B) |E| = sqrt(2) kQ / s2
s
s
P
B) |E| = sqrt(2) kQ / s2
s
C) |E| = kQ / ( sqrt(2) s)2
s
P
D) zero
- Q
s
+Q
C) |E| = kQ / ( sqrt(2) s)2
D) zero
- Q
E) none of the above
s
+Q
E) none of the above
9
In the previous question, what is the accelerator of an
electron if placed at point P?
10
Electric Field from a Continuous Distribution of Charge
v
v
Enet = ∑ Ei
We need some numbers...
Q = 5.0 pC (picoCoulombs) = 5.0 x 10-12 C
s = 1.0 mm
= 1.0 x 10-3 m
k = 9 x 109 Nm2/C2
me = 9.1 x 10-31 kg
i
dq
dE
|E| = 2kQ / s2 = 9 x 104 N/C
|F| = q x |E| = e x 9 x 104 N/C = 1.4 x 10-14 N
|a| = |F| / m = 1.6 x 1016 m/s2 !!!
11
v
v
 kdq 
Enet = ∫ dE = ∫  2 rˆ
 r 
12
2
Clicker Question
Example: Semi- Infinite Line of Charge
dx
x=0
charge = dq=λdx
++++++++++++++++++++++++++++++
x^
x=d
What is the Electric Field at x=0?
Work out on the board.
13
Clicker Question
14
MasteringPhysics comment on Significant Figures
In 1897, the Indiana House Bill #246 was passed legislating
the value of π should equal exactly 3.
If we knew the diameter of a particular circle was 6.0 meters
(and only to two significant figures), would the Indiana Bill
#246 make a difference in our calculation of the circle's
circumference?
A) Yes, it does matter.
B) No, it does not matter.
C) There is no way to determine the answer.
15
16

 kdQ 
dE z =  2  cos θ = 

 r 

Ring of charge
By symmetry, all -E field
contributions in ^x and y^ cancel.
^
Only non
- zero E
- field is along z.
(

h

2 
2
2
R 2 + h 2  R + h
kdQ
)




 khdQ 

dE z = 
 R 2 + h2 3/ 2 


(
v
dE z =| E | cos θ
)
Now integrate the contributions around the ring.
17
18
3
 khdQ  
kh
=
E z = ∫ dE z = ∫ 
 R 2 + h2 3/ 2   R 2 + h2

 
(

khQ
Ez = 
 R2 + h2

(
)
3/ 2
)
(

 dQ
3/ 2 ∫

Clicker Question
)




Electric field a distance h above a ring.
19
20
How about the E
- field from an infinite plane of
charge?
Before our ring was one dimensional.
Charge density λ = [C/m]
Now a good calculation trick is to integrate a whole set
of rings out to infinity.
Now consider ring with infinitesimal width
dr. Thus the inner radius is r and the
outer radius is r + dr.
Area = (2πr) dr
If the charge = dq, then η = dq / (2πrdr)
Charge density η = [C/m2]
21
Clicker Question
Each ring creates an E
- field.
A charge Q is spread uniformly on a rectangle of sides a and b.
The original rectangle (#1) is then broken into smaller
rectangles (#2 and #3). What is correct about the surface
charge densities (η1,η2,η3)?
b
#1
a
b #2
1/3a
22
#3
2/3a
A) η1 = η2 = η3
B) η2 = 1/3η1, η3=2/3η1
C) η2 = 3η1, η3=3/2η1
D) none of the above
23
 kh(dq)   kh(2πrdrη ) 

=
dE z = 
 r 2 + h2 3/ 2   r 2 + h2 3/ 2 

 

(
)
(
)
Then we need to integrate over all rings
from r = 0 to r = infinity to make our infinite
plane.
24
4
∞
 kh(2πrdrη ) 
rdr
 = 2πkhη
E z = ∫ dE z = ∫ 
∫
2
2
 r 2 + h2 3/ 2 
0
0 r +h

∞
(
)
(
)
3/ 2
Ez =
η
2ε 0
Ez
z
You should know how to do such integrals. Try u=r2+h2.
 −1 
E z = 2πkhη 

2
2
 r +h 
infinity
= 2πkη =
0
The electric field strength is independent of the distance from
the infinite plane of charge!
No real plane is infinite, but this calculation is quite good
when z << extent of the plane.
η
2ε 0
25
26
Two Finite Charged Planes are used to accelerate
charged particles.
J.J. Thompson - 1897
27
5