The electric field of a dipole - Columbia...
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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
4πε0
r̂ · �pr2
=1
4πε0
�r · �pr3
, (1)
where �p is the charge distribution’s dipole moment, which is an intrinsic (vector) property of thesource and does not depend on �r. What is the corresponding electric field? It is
�Edip = −�∇Vdip . (2)
To take the gradient, it is convenient to use the “cartesian tensor” notation, with Einstein’sconvention that whenever an index appears twice in a product, summation over all possible valuesthat index can take is understood. So, for instance, the potential is
Vdip =1
4πε0
ri pi
r3, (3)
where ri pi ≡∑3
i=1 ri pi = �r · �p . With this notation, the j-th component of the electric field is
Edipj = − ∂
∂rjVdip = − 1
4πε0pi
∂
∂rj
( ri
r3
). (4)
To compute the partial derivative, we first decompose it into
∂
∂rj
( ri
r3
)=
∂ri
∂rj
1
r3+ ri
∂
∂rj
1
r3. (5)
As to the first term, we have∂ri
∂rj= δij , (6)
where δij is the Kronecker-delta: it is one for i = j, and zero otherwise. Eq. (6) is just a fancyway of writing
∂x
∂x= 1 ,
∂x
∂y= 0 , etc. (7)
in a compact notation. As to the second term in eq. (5), we have
∂
∂rj
1
r3=
(�∇ 1
r3
)j= −3
( r̂
r4
)j= −3
( �r
r5
)j= −3
rj
r5. (8)
Eq. (5) thus reduces to∂
∂rj
( ri
r3
)=
1
r3
(δij − 3
rirj
r2
). (9)
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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) bypi and, following the summation convention, sum over i = 1, 2, 3. We have
Edipj = − 1
4πε0
1
r3
(pi δij − 3
(pi ri)rj
r2
). (10)
Now, piδij = pj, because the Kronecker-delta combined with the summation over all possible i’sselects the term with i = j. Also pi ri = �p · �r. We finally get
Edipj =
1
4πε0
1
r3
(3(�p · �r)rj
r2− pj
), (11)
or, going back to vector notation:
�Edip =1
4πε0
1
r3
[3(�p · �r)�r
r2− �p
](12)
=1
4πε0
1
r3
[3(�p · r̂)r̂ − �p
](13)