W9_parallel_resonance.eps

44 Week 1: Discrete Charge and the Electrostatic Field

Example 1.4.1: Field of Two Point Charges

x

+q

+a

−a

−q

y

E

E

−

+

y

y−a

y+a

Figure 4: Two charges±qon they-axis produce a field that is easy to evaluate at points on thex
andy-axis (and not terribly difficult toapproximatelyevaluate at all points in space that are “far”
from the origin relative toa). This arrangement of charges is called anelectric dipoleand is a very
important concept that we will work with extensively below.

Suppose two point charges of magnitude−qand +qare located on they-axis aty=−aand
y= +a, respectively. Find the electric field at an arbitrary point on thexandyaxis.

They-axis is quite simple. The field due to the positive charge points directlyaway from it,
hence in the positiveydirection at a pointy > aand is equal to:

E~+(0, y) = keq

|y−a|^2

yˆ (13)

The field of the negative charge points towards it and is equal to:

E~−(0, y) =− keq

|y+a|^2

yˆ (14)

Hence the total field on theyaxis is just:

E~tot(0, y) =keq

(

1

|y−a|^2

−

1

|y+a|^2

)

yˆ (15)

The field on thex-axis is atinybit more difficult. Here the field produced byeachcharge has
bothcomponents. To find the vector field, we must first find the magnitude of the field, then use
thegeometry of the pictureto find itsxandycomponents.

Note that the distance from the charge to the point of observation drawn above isr= (x^2 +a^2 )^1 /^2.
Then the magnitude of the electric field vector of either charge is just:

|E~(x,0)|=

keq

r^2

=

keq

(x^2 +a^2 )

(16)

Look at the right triangle formed byx,aandr. By definition:

cos(θ) =

x

r

=

x

(x^2 +a^2 )^1 /^2

(17)

sin(θ) = a

r

= a

(x^2 +a^2 )^1 /^2

(18)