W9_parallel_resonance.eps

Week 1: Discrete Charge and the Electrostatic Field 45

r

Etot

E

E

−a

+a

−q

+q

y

x x

θ

θ

Figure 5: Two charges±qon they-axis produce a field that is (still) pretty easy to evaluate at a
points on thex-axis.

(where we are writing down thepositivequadrant 1 values and will handle the signs needed from
the picture). Using these, we can find the components:

Ex = |E~|cos(θ) =

keq

(x^2 +a^2 )

·

x

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

=

keqx

(x^2 +a^2 )^3 /^2

(19)

and

Ey = −|E~|sin(θ) =−

keq

(x^2 +a^2 )

·

a

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

= − keqa

(x^2 +a^2 )^3 /^2

(20)

This is for a single charge (+q). The other charge has components that are the samemagnitude
but itsExobviouslycancelswhile itsEyobviouslyadds. The total field is thus:

E~tot(x,0) =− 2 keqa

(x^2 +a^2 )^3 /^2

yˆ (21)

In terms of theelectric dipole momentfor this arrangement of charges:

~p= 2qayˆ (22)

the field can be expressed as:

E~tot(x,0) =− ke|~p|

(x^2 +a^2 )^3 /^2

yˆ (23)

It is worthwhile to look at the general shape of the dipole field. It is already familiar to any
student who has done the simple experiment of sprinkling iron filings onto a sheet of paper sitting
on a small bar magnet – the resemblance is not a coincidence, as we shall see.

The electric field and electric potential of a dipole will be of great interest to us over the course
the next few weeks. In many cases, the physical dimensions of thedipole (2ain this case) will be
smallcompared tox, the distance of the point of observation to the dipole. In this limit, the field or
potential produced is that of anidealdipole, or apointdipole. We can find the field in the limit that
x≫avery easily by factoring out the larger of the two quantities from the denominator, expressing