Transcript Would move right and feel twice the force as an electron at B

Electric Dipole
PH 203
Professor Lee Carkner
Lecture 3
EB
PAL # 2
Electric Field
q
EA q
+2q
 Distance to point P is 5 cm (hypotenuse of a 3-4-5 right
triangle)
 Top angle of triangle, sin q = 3/5, q = 37 deg.
 EA = kq/r2 = (8.99X109)(5X10-6) / (0.05)2 = 1.8X107 N/C
 qA = 270 - q = 233 deg
 EB = (8.99X109)(2)(5X10-6) / (0.05)2 = 3.6X107 N/C
 qB = q + 90 = 127 deg
E
PAL # 2
Electric Field
+2q
 EAX = EA cos qA = (1.8X107)(cos 233) = -1.1X107 N/C
 EAY = EA sin qA = (1.8X107)(sin 233) = -1.4X107 N/C
 EBX =EB cos qB = (3.6X107)(cos 127) = -2.2X107 N/C
 EBY = EB sin qB = (3.6X107)(sin 127) = 2.9X107 N/C
 EX = EAX + EBX = -3.3X107 N/C
 EY = EAY + EBY = 1.5X107 N/C
 E2 = EX2 + EY2
 E = 3.6X107 N/C
 q = arctan (EY/EX) = 24 deg above negative x-axis
 q = 180-22 =156 deg from positive x-axis
The Dipole

 Dipole moment = p = qd

 z is the distance from the
center of the dipole to
some point on the dipole
axis
Dipole Field
At a distance z that is large compared to d, the
electric field reduces to:
E = (1/(2pe0)) (p/z3)
Note that:

Since the two opposite and equal charges of the
dipole cancel each other out

Doubling charge is the same as doubling distance
between charges
Dipole in an External Field
Assumptions:

The dipole’s structure is rigid and unchangeable
Since the two charges are equal and opposite,
the two forces are equal and opposite

However, since the charges are not in the same
place, there is a net torque on the dipole
A dipole in an external field will have no
translation motion, but will have rotational
motion
Dipole Torque

 Torque is then Fd sin q =
Eqd sin q
 Remember that p is
direction from negative
charge to positive charge
How will the dipole spin?

The dipole wants to
align the dipole
moment with E
Dipole has the most
torque when
perpendicular to the
field (q = 90)
Dipole Energy
We set the potential energy to be zero when the
dipole is at right angles to the field (q= 90)

Dipole perpendicular to field: PE = 0

Can write potential energy (U) as:
The work (done by an external force) to turn a
dipole in a field is thus:
W = Uf - Ui
Charge Distribution

For a distribution of charge we need to know
how the charge is distributed over space
The charge density:

Surface = s = C/m2

Now instead of q we talk about (e.g.), dq = l ds
The charge of one small part of a linear distribution
Field from Distribution

dE = (1/(4pe0)) (lds/r2)
Note:

We need to integrate with respect to ds for
the entire distribution
Rings
For a uniform charged ring:
E = qz / (4pe0(z2+R2)3/2)

for large distances this
reduces to the point charge
expression
Disks
For a uniform charged disk:
E = (s/2e0)(1 –[z/(z2+R2)½]

The field depends not on the
total charge but the charge
density

E =s/2e0
Infinite sheet
Next Time
Read 23.1-23.4
Problems: Ch 22, P: 42, 52, 53, Ch 23, P:
4, 5
What is true about the magnitude and direction of
the fields from charges A and B at point P?
A)
They are equal in
magnitude and point in the
same direction
B) They are equal in
magnitude and point
towards charges A and B
C) They are unequal in
magnitude and point away
from charges A and B
D) They are unequal in
magnitude and 180 apart in
direction
E) The net field at P is zero
A
B
What is true about the x and y components of the
fields from charges A and B at point P?
A) They both add
B) They both cancel
C) The x components add and
the y components cancel
D) The x components cancel
and the y components
add
E) We can’t tell with out
knowing the magnitude of q
A
B
The above electric field,
A)
B)
C)
D)
E)
increases to the right
increases to the left
increases up
increases down
is uniform
An electron placed at A,
A)
Would move left and feel twice the force as an
electron at B
B) Would move right and feel twice the force as an
electron at B
C) Would move left and feel half the force as an electron
at B
D) Would move right and feel half the force as an
electron at B
E) Would move right and feel the same force as an
electron at B