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)

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)