Math Review 20 minutes. Lecture 2 Electric Fields Ch. 22 Ed. 7. Cartoon  Analogous to gravitational field Topics Electric field = Force per unit Charge Electric Field Lines Electric field from more than 1 charge Electric Dipoles Motion of point charges in an electric field
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r
q1
q0
+ q0
r
r
q1
q1
If q1 were positive
The Coulomb force is
The force per unit charge is
Then the electric field at r is 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. This is called a field line.
Example of field lines for a point negative charge. Place a unit positive test charge at every point r and draw the direction that it would move
q1
r
The blue lines are the field lines.
The magnitude of the electric field is
The direction of the field is given by the line itself
q1
Important F= Eq0, thenma=q0E,
and then a = q0E/m
F1
E
F2
y
x
y
x
This is called an electric dipole.
The text shows a few examples. Here are the drawing rules.
This is called a uniform electric field.
In order to get a better idea of field lines try this Physlet.
Demo: Show field lines using felt, oil, and 10 KV supply
r
.
 q
E
1 cm away from 1 nC of negative charge
Typical Electric Fields (SI Units)
P
Fair weather atmospheric electricity =
downward 100 km high in the ionosphere
+ + + + + + + + + +
+ + + + + + + + +
E
Earth
+

r
Hydrogen atom
We have at the origin, at x=4 m and y=0.
What is the magnitude and direction of the electric field E at y=3 m and x=0?
Example of finding electric field from two charges lying in a plane
P
Find the x and y components of the electric field for
each charge and add them up.
Find the magnitude of the
field for q1
P
q2 =15 nc
q1=10 nc
Find the magnitude of the field for q2
P
5
φ
φ
Find the x and y components
q2 =15 nc
q1=10 nc
Now add all components
y
φ
q1=10 nc
q2 =15 nc
x
φ1
Magnitude of net electric field is
Direction of the total electric field is
Using unit vector notation we can
also write the electric field vector as:
71.9
y
y
.
.
P
P
3
3
4
4
4
4
x
x
+15 nc
+15 nc
+15 nc
+15 nc
5
ϕ
Example of two identical charges of +15 nC on the x axis. Each is 4 m from the origin. What is the electric field on the y axis at P where y= 3 m? Example using symmetry to simplify.
Find x and y components of the electric field at P due to each charge
and add them up. Find x component first.
By symmetry
y
.
3
4
4
x
+15 nc
+15 nc
y
5
φ
Now find y component
φ
Example of two opposite charges on the x axis. Each is 4 m from the origin. What is the electric field on the y axis at P where y= 3 m? Example using symmetry to simplify.
y
Ex
P
5
3
φ
4
4
x
15 nc
+15 nc
Find the x and y component of the electric field
Ey=0 by symmetry
Electric Dipole
y
E1
θ
E1x
L
L
q
=
=
cos
2
a
a
2
a
a
θ
E2
For large r
This is called an Electric Dipole: A pair of equal and opposite charges q separated by a displacement L. It has an electric dipole moment p=qL.
P
when r is large compared to L
p=qL = the electric dipole moment
r
Note inverse cube law
q
+q
The dipole moment p is defined as a vector directed
from  to +
+

L
x
A uniform external electric field exerts no net force on a dipole, but it does exert torque that tends to rotate the dipole in the direction of the field (align with )
Torque about the com = τ
So,
When the dipole rotates through q, the electric field does work:
And q = 10 e and q = +10e
GIven p = 6.2 x 10  30 C m
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
Work done equals
The minus sign arises because the torque opposes any increase in θ.
Setting the negative of this work equal to the change in the potential energy, we have
Integrating,
Potential Energy
Ans. The energy is minimum when aligns with
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
a = F/m = q E/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 problem constant 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, y =1/2at2 we get y =1/2(qE/m)t2
Example illustrating the motion of a charged particle in an electric field: An electron is projected perpendicularly to a downward uniform electric field of E= 2000 N/C with a horizontal velocity v=106 m/s. How much is the electron deflected after traveling a distance d=1 cm. Find the deflection y.
y
v
electron
d
E
E
y
v
electron
d
E
E
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?
++++++++++++++++++++++++++++++++
Typical field for the electrostatic smoke remover
y
dE
.
.
r
+x
x
 L/2
L/2
dq
Find electric field due to a line of uniform + charge of length L with linear charge density equal to λ at a point on the y axis that bisects the rod.
λ= Q / L
y
++++++++++++++++++++++++++++++++
dE
dE
r
θ
x
+x
 L/2
0
L/2
dq
dE = k dq /r2 Apply Coulombs law to a slug of charge dq
dq = λdx
dEy= k λ dx cos θ/r2
dE = k λdx /r2
dEy= dE cos θ
y
dEy
dE
r =y sec θ r2 =y2sec2θ
r
x= y tanθ dx = y sec2 θ dθ
y
θ
x
+x
x
L/2
0
L/2
dx / r2=dθ / y
dq
dq=λds
s=r θ
ds=r dθ
0 due to symmetric element of charge above and below the x axis
dq=λds
s=r θ
ds=r dθ
0 due to symmetric element of charge above and below the x axis
dEx= k dq cos θ /r2
dq=λds
s=r θ
ds=r dθ
dEx= k λds cos θ /r2
r2 =z2+R2
dq = λds
cos θ =z/r
=0 at z=0
=0 at z=infinity
=max at z=0.7R
A disk of radius 2.7 cm has a surface charge density of 4.0 µC/m2 on its upper face. What is the magnitude of the electric field produced by the disk at a point on its central axis at distance z = 12 cm from the disk?
At some instant the velocity components of an electron moving between two charged parallel plates are vx = 1.7 x 105 m/s and vy = 2.8 x 103 m/s. Suppose that the electric field between the plates is given by E = (120 N/C) j.
(a) In unit vector notation what is the electron's acceleration in the field?
(b) What is the electron's velocity when its x coordinate has changed by 2.7 cm?
An electric dipole consists of charges +2e and 2e separated by 0.74 nm. It is in an electric field of strength 3.4 x106 N/C. Calculate the magnitude of the torque on the dipole when the dipole moment has the following orientations.
(a) parallel to the field
(b) perpendicular to the field
(c) antiparallel to the field