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:

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 = 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 to be zero when the dipole is at right angles to the field (= 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 = = C/m2

Now instead of q we talk about (e.g.), dq = ds

The charge of one small part of a linear distribution

Field from Distribution

dE = (1/(40)) (ds/r2) Note:

We need to integrate with respect to ds for the entire distribution

Rings

For a uniform charged ring:

E = qz / (40(z2+R2)3/2)

for large distances this reduces to the point charge expression

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

E =/20

Infinite sheet

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