Transcript PowerPoint Presentation - Lecture 1 Electric Charge

Lecture 2 Electric Fields Chp. 23
•
•
•
•
•
•
Cartoon - Analogous to gravitational field
Opening Demo - Bending of water stream with charged rod
Warm-up problem
Physlet
Topics
– Electric field = Force per unit Charge
– Electric Field Lines and Electric Flux
– Electric field from more than 1 charge
– Electric Dipoles
– Motion of point charges in an electric field
– Examples of finding electric fields from continuous charges
List of Demos
– Van de Graaff Generator, workings,lightning rod, electroscope, electric wind
– Smoke remover or electrostatic precipitator
– Kelvin water drop generator
– Transparent CRT with visible electron gun
– Field lines using felt,oil, and 10 KV supply.
Summer July 2004
1
The Electric Field
•
Definition of the electric field. Whenever charges are present and if I
bring up another charge, it will feel a net Coulomb force from all the
others. It is convenient to say that there is field there equal to the force
per unit positive charge. E=F/q0. The direction of the electric field is
along r and points in the direction a positive test charge would move.
This idea was proposed by Michael Faraday in the 1830’s. The idea of
the field replaces the charges as defining the situation. Consider two
point charges:
r
q1
Summer July 2004
q0
2
r
+ q0
q1
The Coulomb force is F= kq1q0/r2
The force per unit charge is E = F/q0
and then the electric field at r is E = kq1/r2 due to the point charge q1 .
The units are Newton/Coulomb. The electric field has direction and is a vector.
How do we find the direction.? The direction is the direction a unit positive test
charge would move.
r
E
If q1 were positive
q1
Summer July 2004
3
Point negative charge
q1
r
E= kq1/r2
Summer July 2004
q1
4
Electric Field Lines
Like charges (++)
Opposite charges (+ -)
This is called an electric dipole.
Summer July 2004
5
Electric Field Lines: a graphic concept used
to draw pictures as an aid to develop
intuition about its behavior.
The text shows a few examples. Here are the drawing rules.
• E-field lines begin on + charges and end on - charges. (or infinity).
• They enter or leave charge symmetrically.
• The number of lines entering or leaving a charge is proportional to
the charge
• The density of lines indicates the strength of E at that point.
• At large distances from a system of charges, the lines become
isotropic and radial as from a single point charge equal to the net
charge of the system.
• No two field lines can cross.
Show a physlet 9.1.4, 9.1.7
Show field lines using felt,oil, and 10 KV supply
Summer July 2004
6
Typical Electric Fields (SI Units)
• 1 cm away from 1 nC of negative charge
– E = kq /r2 = 1010 *10-9/ 10-4 =105 N /C
–
N.m2/C2 C / m2 = N/C
r
q
.E
• Fair weather atmospheric electricity = 100 N/C downward
100 km high in the ionosphere
+++++ •E
--------Earth
• Field due to a proton at the location of the electron in the H
atom. The radius of the electron orbit is 0.5*10-10 m.
– E = kq /r2 = 1010 *1.6*10-19/ (0.5 *10-10 )2 = 4*1011 N /C
Hydrogen atom
Summer July 2004
r
-
1N / C = Volt/meter
+
7
Example of field lines for a
uniform distribution of positive
charge on one side of a very large
nonconducting sheet.
This is called a uniform electric field.
How would the electric field change
if both sides were charged?
How would things change if the sheet
were conducting?
Summer July 2004
8
Methods of evaluating electric fields
• Direct evaluation from Coulombs Law or brute force method
If we know where the charges are, we can find E from E = ∑ kqi/ri2.
This is a vector equation and can be complex and messy to
evaluate and we may have to resort to a computer. The principle
of superposition guarantees the result.
• Instead of summing the charge we can imagine a continuous
distribution and integrate it. This distribution may be over a volume,
a surface or just a line.
– E = ∫dE = ∫ kdqi/r2 r where r is a unit vector directed from charge
dq to
the field point.
– dq = dV , or dq = dA, or dq =  dl
Summer July 2004
9
Example of finding electric field from two charges
We have q =+10 nC at the origin, q = +15 nC at x=4 m.
1
2
What is E at y=3 m and x=0? point P
y
P
3
q1=10 nc
4
x
q2 =15 nc
Use principle of superposition
Find x and y components of electric field due to both charges
and add them up
Summer July 2004
10
Example continued
Recall E =kq/r2
and k=8.99 x 109 N.m2/C2
Field due to q1
E = 1010 N.m2/C2 10 X10-9 C/(3m)2 = 11 N/C
in the y direction.
Ey= 11 N/C
y
E
f
3
q1=10 nc
5
f
4
x
q2 =15 nc
Ex= 0
Field due to q2
E = 1010 N.m2/C2 15 X10-9 C/(5m)2 = 6 N/C
at some angle f
Resolve into x and y components
Ey=E sin f = 6 * 3/5 =18/5 = 3.6 N/C
Ex=E cos f = 6 * (-4)/5 =-24/5 = -4.8 N/C
Summer July 2004
Now add all components
Ey= 11 + 3.6 = 14.6 N/C
Ex= -4.8 N/C
Magnitude
E = E x2  E y2
11
E
Example continued
Ey= 11 + 3.6 = 14.6 N/C
Ex= -4.8 N/C
f1
3
4
q1=10 nc
x
q2 =15 nc
Using unit vector notation we can
also write the electric field vector as:
Magnitude of electric field
E = E x2  E y2
E=

(14.6)  (-4.8)
2
2
= 15.4N /C

E = -4.8i14.6 j
f1 = atan Ey/Ex= atan (14.6/-4.8)= 72.8 deg

Summer July 2004
12
Example of two identical
charges on the x axis. What
is the filed on the y axis?
Example of two opposite
charges on the x axis. What
is the filed on the y axis?
y
y Ey
Ex
3q
4
q2 =15 nc
3q
5
4
f
x
4
q2 = -15 nc
q2 =15 nc
5
4
f
x
q2 =15 nc
E = 1010 N.m2/C2 15 X10-9 C/(5m)2 = 6 N/C
Ey=2*E sin f = 2*6 * 3/5 = 7.2 N/C
Ex=0
Summer July 2004
Ey=0
Ex=2*E cos f = 2*6 * 4/5 = - 9.6 N/C
13
4 equal charges symmetrically spaced along a line. What is the field at
point P? (y and x = 0)
y
P
q4
r1
q1
q1
q
r2 q2 3 r3
q2
4
E y = k qi cosqi /ri2
i=1
Summer July 2004
r4
f
x
q3
q4

E y = kqi cosqi /r
2
i
i=1
14
Find electric field due to a line of uniform + charge of length L
with linear charge density equal to 
y
dE = k dq /r2
dE
dEy
dEx
-x
0
dEy= dE cos q
+x
x
dq
dEy= k  dx cos q /r2
E y = k 
E y = k / y
L /2
cos q dx /r 2
-L / 2
-q cosq dq
Summer July 2004
q0
0
 dE
x
=0
-L / 2
r
y q
-L/2
Ex =
L /2
L/2
dq = dx

Ey= k  q cos q /r2 for a point charge
dx/r2 = dq/y
2k
Ey =
sin q 0
y
x= y tanq
r =y sec q
sin q 0 =
dx = y sec2 q dq
r2 =y2sec2 q
L /2
y 2  L2 /415
What is the electric field from an infinitely long wire
with linear charge density of +100 nC/m at a point 10
away from it. What do the lines of flux look like?
+++++++++++++++++++++++++++++++++++++++++
2k
Ey =
sin q 0
y
.
y =10 cm
Ey
2 *1010 Nm 2 *100 *10-9 C /m
Ey =
sin90 = 2 *10 4 N /C
0.1m
Typical field for the electrostatic smoke cleaner
Summer July 2004
16
Electric field gradient
• When a dipole is an electric field that varies with position, then
the magnitude of the electric force will be different for the two
charges. The dipole can be permanent like NaCl or water or
induced as seen in the hanging pith ball. Induced dipoles are
always attracted to the region of higher field. Explains why wood
is attracted to the teflon rod and how a smoke remover or
microwave oven works.
• Show smoke remover demo.
Summer July 2004
17
Electrostatic smoke precipitator model
• Negatively charged central wire has electric field that varies as
1/r (strong electric field gradient). Field induces a dipole moment
on the smoke particles. The positive end gets attracted more to
the wire
• In the meantime a corona discharge is created. This just means
that induced dipole moments in the air molecules cause them to
be attracted towards the wire where they receive an electron
and get repelled producing a cloud of ions around the wire.
• When the smoke particle hits the wire it receives an electron
and then is repelled to the side of the can where it sticks.
However, it only has to enter the cloud of ions before it is
repelled.
• It would also work if the polarity of the wire is reversed
Summer July 2004
18
Electric Dipoles: A pair of equal and opposite charges q
separated by a displacement d is called an electric dipole. It
has an electric dipole moment p=qd.
IEI ~ 2kqd/r3 when r is large compared to d
p=qd = the electric dipole moment
•P
+ +q
d
p
- -q
Summer July 2004
r
IEI ~2kp/r3 Note inverse cube law
IEI = kq/r2 Note inverse square law
for a single charge.
19
Water (H2O) is a molecule that has a permanent dipole moment.
GIven p = 6.2 x 10 - 30 C m
And q = -10 e and q = +10e
What is d?
d = p / 10e = 6.2 x 10 -30 C m / 10*1.6 x 10 -19 C = 3.9 x 10 -12 m
Very small distance but still is responsible
for the conductivity of water.
When a dipole is an electric field, the dipole
moment wants to rotate to line up with
the electric field. It experiences a torque.
Leads to how microwave ovens heat up food
Summer July 2004
20
H2O in a Uniform Electric Field
There exist a torque on the water molecule
To rotate it so that p lines up with E.
x
Torque about the com = t
F x sin q  F(d-x)sin q = Fdsin q =
qEdsin q = pEsin q = p x E
t=pxE
Potential Energy = U = -W = -pEcosq = - p. E
Is a minimum when p aligns with E
Summer July 2004
21
Motion of point charges in electric fields
• When a point charge such as an electron is placed in an electric
field E, it is accelerated according to Newton’s Law:
a = F/m = qE/m for uniform electric fields
a = F/m = mg/m = g for uniform gravitational fields
If the field is uniform, we now have a projectile motion problemconstant acceleration in one direction. So we have parabolic
motion just as in hitting a baseball, etc except the magnitudes
of velocities and accelerations are different.
Replace g by qE/m in all equations;
For example, In y =1/2at2 we get y =1/2(qE/m)t2
Summer July 2004
22
Example: An electron is projected perpendicularly to a
downward electric field of E= 2000 N/C with a horizontal
velocity v=106 m/s. How much is the electron deflected
after traveling 1 cm.
V •e
d
E
E
Since velocity in x direction does not change, t=d/v =10-2/106 = 10-6
sec, so the distance the electron falls upward is
y =1/2at2 = 0.5*eE/m*t2 = 0.5*1.6*10-19*2*103/10 - 30*(10-8)2 = 0.016m
•Demo Transparent CRT with electron gun
Summer July 2004
23
Back to computing Electric Fields
•
Electric field due to a line of uniform charge
• Electric field due to an arc of a circle of uniform charge.
• Electric field due to a ring of uniform charge
• Electric field of a uniform charged disk
• Next we will go on to another simpler method to calculate
electric fields that works for highly symmetric situations using
Gauss’s Law.
Summer July 2004
24
Field due to arc of charge
L /2
Ey =
 dE
y
=0
-L / 2
dEx= k dq cos q /r2
dEx= k  ds cos q /r2
E x = k
Ex =
L /2
q0
-L / 2
-q 0
2
rd
q
cos
q
/r
= k /r  dq cosq

s=r q
ds=r dq
2k
sinq0
r
What is the field at the center
of a circle of charge?
Ans. 0
Summer July 2004
25
Find the electric field on the axis of a uniformly charged ring with
linear charge density  = Q/2pR.
Ez =
 dEcosq
k cosq
Ez =
r2

dq
ds
dE = k 2 = k 2
r
r
2p
2p
0
0
 ds  ds =  Rdq = R  dq = 2pR

k cosq
Ez =
2pR
2
r
s = Rq


dq = ds
kQz
Ez = 2
(z  R 2 ) 3 / 2

Summer July 2004

r2 =z2+R2

cos q =z/r
=0 at z=0
=0 at z=infinity
=max at z=0.7R26
Warm-up set 2
1. [153709] Can there be an electric field at a point where there is no charge?
Can a charge experience a force due to its own field? Please write a one or two sentence answer
for each question.
2. [153707] An insulator is a material that...
three are correct
is not penetrated by electric fields
none of these
cannot carry an electric charge
cannot feel an electrical force
3. [153708] Which of the following is true of a perfect conductor ?
There can be no electric charge on the surface.
There cannot be an electric field inside.
There cannot be any excess electric charge inside.
There cannot be any electric charges inside.
Two of the choices are correct
Summer July 2004
27
Kelvin Water Drop Generator
Am. J. Phys. 68,1084(2000)
Summer July 2004
28