W3007 – Electricity and Magnetism
1
The electric field of a dipole
We know that the dipole contribution to the electrostatic potential is
Vdip (r) =
1 r̂ · p
1 r · p
=
,
2
4π0 r
4π0 r 3
(1)
where p is the charge distribution’s dipole moment, which is an intrinsic (vector) property of the
source and does not depend on r. What is the corresponding electric field? It is
dip .
dip = −∇V
E
(2)
To take the gradient, it is convenient to use the “cartesian tensor” notation, with Einstein’s
convention that whenever an index appears twice in a product, summation over all possible values
that index can take is understood. So, for instance, the potential is
Vdip =
where ri pi ≡
3
i=1 ri
1 ri pi
,
4π0 r 3
(3)
pi = r · p . With this notation, the j-th component of the electric field is
Ejdip = −
∂ ri ∂
1
Vdip = −
pi
.
∂rj
4π0 ∂rj r 3
(4)
To compute the partial derivative, we first decompose it into
As to the first term, we have
∂ 1
∂ ri ∂ri 1
+
r
.
=
i
∂rj r 3
∂rj r 3
∂rj r 3
(5)
∂ri
= δij ,
∂rj
(6)
where δij is the Kronecker-delta: it is one for i = j, and zero otherwise. Eq. (6) is just a fancy
way of writing
∂x
∂x
=1,
=0,
etc.
(7)
∂x
∂y
in a compact notation. As to the second term in eq. (5), we have
1
r̂ r ∂ 1
rj
=
∇
=
−3
=
−3
= −3 5 .
3
3
4
5
∂rj r
r j
r j
r j
r
Eq. (5) thus reduces to
ri rj ∂ ri 1
= 3 δij − 3 2 .
∂rj r 3
r
r
(8)
(9)
2
The electric field of a dipole
As a final step, to compute the electric field according to eq. (4), we have to multiply eq. (9) by
pi and, following the summation convention, sum over i = 1, 2, 3. We have
Ejdip = −
1 1
(pi ri )rj p
.
δ
−
3
i
ij
4π0 r 3
r2
(10)
Now, pi δij = pj , because the Kronecker-delta combined with the summation over all possible i’s
selects the term with i = j. Also pi ri = p · r. We finally get
Ejdip =
p · r)rj
1 1 (
3
−
p
j ,
4π0 r 3
r2
(11)
or, going back to vector notation:
p · r)r
1 1 (
3
−
p
4π0 r 3
r2
1 1
=
3(
p
·
r̂)r̂
−
p
4π0 r 3
dip =
E
(12)
(13)