Electric Dipole PH 203 Professor Lee Carkner Lecture 3
Feb 04, 2016
Electric Dipole
PH 203
Professor Lee Carkner
Lecture 3
PAL # 2 Electric Field
Distance to point P is 5 cm (hypotenuse of a 3-4-5 right triangle)
Top angle of triangle, sin = 3/5, = 37 deg. EA = kq/r2 = (8.99X109)(5X10-6) / (0.05)2 = 1.8X107 N/C
A = 270 - = 233 deg
EB = (8.99X109)(2)(5X10-6) / (0.05)2 = 3.6X107 N/C
B = + 90 = 127 deg
+2q
EA
EB
PAL # 2 Electric Field
EAX = EA cos A = (1.8X107)(cos 233) = -1.1X107 N/C EAY = EA sin A = (1.8X107)(sin 233) = -1.4X107 N/C EBX =EB cos B = (3.6X107)(cos 127) = -2.2X107 N/C EBY = EB sin B = (3.6X107)(sin 127) = 2.9X107 N/C EX = EAX + EBX = -3.3X107 N/C EY = EAY + EBY = 1.5X107 N/C E2 = EX
2 + EY2
E = 3.6X107 N/C = arctan (EY/EX) = 24 deg above negative x-axis
= 180-22 =156 deg from positive x-axis
+2q
E
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/(20)) (p/z3) Note that: E falls off very rapidly with z
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 = Eqd sin
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 ( = 90)
Dipole Energy
We set the potential energy (U) to be zero when the dipole is at right angles to the field (= 90) Dipole perpendicular to field: U = 0
Can write potential energy as:
The work (done by an external force) to turn a dipole in a field is thus:
W = Uf - Ui
Charge Distribution
The charge density: Linear = = C/m Surface = Volume = =C/m3
For example:dE = (1/(40)) (ds/r2)
Rings
For a uniform charged ring:
E = qz / (40(z2+R2)3/2)
Disks
For a uniform charged disk:
E = (/20)(1 –[z/(z2+R2)½]
The field depends not on the total charge but the charge density
Next Time
Read 23.1-23.4 Problems: Ch 22, P: 42, 52, 53, Ch 23, P:
4, 5
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 magnitude and direction of the fields from charges A and B at point P?
A) They both addB) They both cancelC) The x components add and
the y components cancelD) The x components cancel
and the y components add
E) We can’t tell with out knowing the magnitude of q
A B
What is true about the x and y components of the fields from charges A and B at point P?
The above electric field,
A) increases to the right
B) increases to the left
C) increases up
D) increases down
E) 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