The mutual energy of a group of n stationary point charges
We shall denote by 𝜓𝑖 (𝐫) the potential that a point charge, 𝑄𝑖 , (situated at 𝐫𝑖 ) produces at a
point r.
𝑄𝑖
So
𝜓𝑖 (𝐫) =
eq 1
4𝜋𝜀0 |𝐫 − 𝐫𝒊 |
Suppose the charges to be brought from a very long way away to their final places, {𝐫𝑖 }, one
by one... No work is needed to bring the first, 𝑄1, to 𝐫1. But 𝑄2 now needs work 𝑄2 𝜓1 (𝐫𝟐 ) to
bring it to 𝐫2 . As eq 1 shows, we could equally well have written the work as 𝑄1 𝜓2 (𝐫𝟏 ), as if
we had installed 𝑄2 before 𝑄1. To emphasise the symmetry we shall write the work as
1
2
{𝑄1 𝜓2 (𝐫𝟏 ) + 𝑄2 𝜓1 (𝐫𝟐 )}.
Now let us bring on the third charge, into the fields of 𝑄1 and 𝑄2 . This will require extra
work of
1
{𝑄1 𝜓3 (𝐫𝟏 ) + 𝑄3 𝜓1 (𝐫𝟑 ) + 𝑄2 𝜓3 (𝐫𝟐 ) + 𝑄3 𝜓2 (𝐫𝟑 )}.
2
The total work, that is the total mutual energy of all three charges, is therefore
1
2
{𝑄1 [𝜓2 (𝐫𝟏 ) + 𝜓3 (𝐫𝟏 )] + 𝑄2 [𝜓1 (𝐫𝟐 ) + 𝜓3 (𝐫𝟐 )] + 𝑄3 [𝜓1 (𝐫𝟑 ) + 𝜓2 (𝐫𝟑 )]}.
Each further charge, 𝑄𝑖 , that we bring up from infinity will add a 𝜓𝑖 term to each existing
square bracket, and another whole work term, 12𝑄𝑖 [𝜓1 (𝐫𝒊 ) + 𝜓2 (𝐫𝒊 ) . . . +𝜓𝑖−1 (𝐫𝒊 )], to the
progressive energy total.
We now define 𝜙𝑖 (𝐫) to mean the potential at r due to all the n charges except 𝑄𝑖 , that is the
sum of all the potentials in the i’th square bracket.
Thus
𝜙𝑖 (𝐫) = 𝜓1 (𝐫) + 𝜓2 (𝐫) . . . +𝜓𝑖−1 (𝐫) + 𝜓𝑖+1 (𝐫) . . . +𝜓𝑛 (𝐫).
eq 2
This enables us to write the mutual energy, 𝑈𝑚𝑢𝑡 , of the system as
1
𝑈𝑚𝑢𝑡 = 2 ∑
𝑛
𝑄𝑖 𝜙𝑖 (𝐫𝒊 )
𝑖=1
eq 3
The energy in an electric field
Where does the mutual energy of the charges reside? Eq. 3 possibly suggests that we should
associate it with the charges themselves. But, rather marvellously, the mutual energy can be
expressed in terms of electric field strengths in the space surrounding the charges. We shall
do this for a group of point charges in the section after this one. In fact, for point charges, the
process is a bit fiddly, because in order to avoid infinities we need to exclude the charges
themselves from our treatment of the surrounding space. We shall first consider the easier
case of charge distributed with a finite charge density (charge per unit volume) 𝜌(𝐫).
To this end we modify eq 3 by writing 𝑄𝑖 = 𝜌d𝑉, in which d𝑉 is a volume element over which
𝜌 is the charge density. Replacing the summation by an integration over a volume V including
all the charge we’re interested in, eq 3 becomes
1
𝑈 = 2 ∭ 𝜙(𝐫) 𝜌(𝐫)d𝑉
1
or just
𝑈 = 2 ∭ 𝜙 𝜌 d𝑉
Recall that 𝜙𝑖 (𝐫), as defined by eq 2, specifically excludes the potential due to 𝑄𝑖 . With a finite
charge density, the contribution to 𝜙(𝐫)of local charge (charge in a region dV centred on r) is
zero. This is easily demonstrated by calculating the pd between the outside and the centre of
a uniformly charged sphere: as its radius goes to zero, so does the pd. This implies that 𝜙(𝐫)
may be taken as the potential due to all charge in V, with no exclusions. Incidentally, that’s
why we’ve dropped the mut subscript on U: mutuality no longer needs to be emphasised.
We now use Gauss’s law to express 𝜌(𝐫) in terms of the local electric field strength, −∇𝜙(𝐫).
𝜌(𝐫) = 𝜀0 𝛁. [−𝛁𝜙(𝐫)]
𝜌(𝐫) = −𝜀0 ∇2 𝜙(𝐫).
that is
𝜀
𝑈 = − 20 ∭ 𝜙 ∇2 𝜙 d𝑉.
So
Now for any vector field A and scalar field s we have the identity 𝛁. (𝑠𝐀) = 𝑠𝛁. 𝐀 + 𝛁𝑠. 𝐀.
Putting 𝑠 = 𝜙,
Thus
𝐀 = 𝛁𝜙,
𝛁. (𝜙𝛁𝜙) = 𝜙∇2 𝜙 + 𝛁𝜙. 𝛁𝜙.
we have
𝜀
𝑈 = − 20 ∭ 𝛁. (𝜙𝛁𝜙)d𝑉 +
𝜀0
2
∭ 𝛁𝜙. 𝛁𝜙 d𝑉.
Using the divergence theorem on the left hand integral, and remembering that −𝛁𝜙 = 𝐄
𝜀
𝑈 = − 20 ∬ 𝜙𝛁𝜙. 𝐧 d𝑆 +
𝑆
𝜀0
2
∭ 𝐸 2 d𝑉.
Here, n, which we could write as n(r), is the unit normal vector pointing from V outward
through the various patches, dS of its boundary, S. The surface ‘outflow’ integral can be
made to vanish by pushing S far out beyond the region containing charge. We can see this
by making S a sphere of increasing radius r with the charges near its centre.
Then,
1
𝜙(𝐫) ~ 𝑟
and
1
𝛁𝜙(𝐫) ~ 𝑟2
whilst 𝑆 ~ 𝑟 2 .
We are left with the very simple equation
𝑈=
𝜀0
2
∭ 𝐸 2 d𝑉.
eq 4
This is consistent with an energy density (energy per unit volume of space), u, given by
𝑢=
𝜀0
2
𝐸2.
An example which, nicely, involves only a finite volume of space, is the electric field between
the plates (area A, separation s) of a parallel plate, air-spaced capacitor. By considering the
charging process we find that, for a pd ∆𝜙 between the plates, the stored energy is
𝑈 = 12𝐶(∆𝜙)2 = 12
𝜀0 𝐴
∆𝑠
(𝐸∆𝑠)2 = 𝜀20 𝐸 2 (𝐴∆𝑠).
eq 5
The mutual energy of a group of n stationary point charges re-expressed
The mutual energy can be expressed in terms of electric field strengths in the space
surrounding the group. To do this we start by using Gauss’s law to express a charge 𝑄𝑖 in
terms of the flux it sets up through a surface 𝑆𝑖 enclosing it (but no other charges).
𝑄𝑖 = 𝜀0 ∬ E𝒊 (𝐫).m dS = −𝜀0 ∬ 𝛁𝜓𝑖 (𝐫).m dS
𝑆𝑖
𝑆𝑖
in which E𝒊 (𝐫) is the electric field strength due to 𝑄𝑖 , expressible as minus the grad of the
potential due to 𝑄𝑖 , and m is the outward unit vector normal to a patch dS of S𝑖 .
From eq 3
𝑈𝑚𝑢𝑡 =
1
∑
2
𝑛
𝜀
𝑄𝑖 𝜙𝑖 (𝐫𝒊 ) = − 20 ∑
𝑖=1
𝑛
𝑖=1
∬ 𝜙𝑖 (𝐫𝒊 )𝛁𝜓𝑖 (𝐫).m dS
𝑆𝑖
It was perfectly permissible to take 𝜙𝑖 (𝐫𝒊 ) inside the integral, as it is a constant.
Next we shrink each S𝑖 so closely around its charge, 𝑄𝑖 , that we can replace 𝜙𝑖 (𝐫𝒊 ) by 𝜙𝑖 (𝐫).
This is because the value of 𝜙𝑖 (𝐫) at any point on the surface S𝑖 is now indistinguishable
from 𝜙𝑖 (𝐫𝒊 ). Remember that 𝜙𝑖 (𝐫) arises from ‘distant’ charges, that is charges other than
𝑄𝑖 , and so varies smoothly and slowly with r in the vicinity of 𝑄𝑖 . So we have...
𝜀
𝑛
𝑈𝑚𝑢𝑡 = − 20 ∑
𝜀
𝑖=1
∬ 𝜙𝑖 (𝐫)𝛁𝜓𝑖 (𝐫).m dS or just 𝑈𝑚𝑢𝑡 = − 20 ∑
𝑆𝑖
𝑛
𝑖=1
∬ 𝜙𝑖 𝛁𝜓𝑖 .m dS
eq 6
𝑆𝑖
We now switch our attention to the space V that
surrounds the enclosures {S𝑖 }, rather as cheese
surrounds the holes in a Swiss cheese. We give V an
outer boundary, Sout , enclosing all charges. The total
outflow of the quantity 𝜙𝑖 𝛁𝜓𝑖 through all V’s
boundaries is
∯ 𝜙𝑖 𝛁𝜓𝑖 .n dS = ∬
𝜙𝑖 𝛁𝜓𝑖 .n dS
𝑆𝑜𝑢𝑡
+ ∬ 𝜙𝑖 𝛁𝜓𝑖 .n dS + ∬ 𝜙𝑖 𝛁𝜓𝑖 .n dS . . . + ∬ 𝜙𝑖 𝛁𝜓𝑖 .n dS . . . + ∬ 𝜙𝑖 𝛁𝜓𝑖 .n dS
𝑆1
𝑆2
𝑆𝑖
𝑆𝑛
Here, n, which we could write as n(r), is the unit normal vector pointing from V outward
through the various patches, dS of its boundaries. Thus for S1, S1, . . . S𝑖 , . . . Sn , 𝐧 = −𝐦.
We proceed to examine the integrals on the right hand side. The first can be made to vanish
by pushing Sout far out beyond the group of charges. We can see that the outflow integral
goes to zero by making Sout a sphere of increasing radius r with the charges near its centre.
1
𝜙𝑖 (𝐫) ~ 𝑟
Then,
and
1
whilst Sout ~ 𝑟 2 .
𝛁𝜓𝑖 (𝐫) ~ 𝑟 2
Now we deal with ∬𝑆 𝜙𝑖 𝛁𝜓𝑖 .n dS when 𝑗 ≠ 𝑖. Replacing 𝜙𝑖 by 𝜓𝑗 , the only term in 𝜙𝑖 (see
𝑗
eq 2) which is rapidly varying around 𝑄𝑗 , and taking S𝑗 as a sphere centred on Q𝑗 , of radius
approaching zero,
∬𝑆 𝜙𝑖 𝛁𝜓𝑖 .n dS = ∬𝑆 𝜓𝑗 𝛁𝜓𝑖 .n dS ~
𝑗
We are left with
𝑗
1
|r−r𝑖 |
× |𝒓 − 𝒓𝑖 |2 → 0
∯ 𝜙𝑖 𝛁𝜓𝑖 .n dS = ∬ 𝜙𝑖 𝛁𝜓𝑖 .n dS
𝑆𝑖
The outflow of 𝜙𝑖 𝛁𝜓𝑖 from V is therefore simply the inflow to S𝑖 . Remembering that n = –m,
eq 6 can be written as
𝑈𝑚𝑢𝑡 =
Applying the divergence theorem: 𝑈𝑚𝑢𝑡 =
But, purely mathematically,
𝑛
𝜀0
∑
∯ 𝜙𝑖 𝛁𝜓𝑖 .n dS
2
𝑖=1
𝜀0
∑
2
eq 7
𝑛
𝑖=1
∭ 𝛁. [𝜙𝑖 𝛁𝜓𝑖 ] d 𝑉
𝛁. [𝜙𝑖 𝛁𝜓𝑖 ] = 𝛁𝜙𝑖 . 𝛁𝜓𝑖 + 𝜙𝑖 𝛁. 𝛁𝜓𝑖
in which 𝛁𝜓𝑖 = −𝐄𝑖 , so 𝛁. 𝛁𝜓𝑖 = −𝛁. 𝐄𝑖 = 0 since there are no charges in 𝑉 itself.
and 𝛁𝜙𝑖 = 𝛁(𝜓1 + 𝜓2 . . . +𝜓𝑖−1 + 𝜓𝑖+1 . . . +𝜓𝑛 ) = −(𝐄1 + 𝐄2 . . . +𝐄𝑖−1 + 𝐄𝑖+1 . . . +𝐄𝑛 )
Thus
That is
That is
In other words:
in which
𝑛
𝑈𝑚𝑢𝑡 =
𝜀0
∭∑
2
𝑈𝑚𝑢𝑡 =
𝜀0
∭∑
2
𝑈𝑚𝑢𝑡 =
𝜀0
∭ {𝐸 2
2
𝐄𝑖 . (𝐄1 + 𝐄2 . . . +𝐄𝑖−1 + 𝐄𝑖+1 . . . +𝐄𝑛 ) d 𝑉
𝑖=1
𝑛
𝑖=1
𝑛
𝐄𝑖 . {[∑
𝑛
−∑
𝐄𝑖 ] − 𝐄𝑖 } d 𝑉
𝑖=1
𝑛
𝐄𝑖 2 } d𝑉
in which
𝑖=1
𝐄= ∑
𝐄𝑖
𝑖=1
𝑈𝑚𝑢𝑡 = 𝑈𝑡𝑜𝑡 − 𝑈𝑠𝑒𝑙𝑓
𝑈𝑡𝑜𝑡 =
𝜀0
∭ 𝐸 2 d𝑉 ,
2
𝑛
𝑈𝑠𝑒𝑙𝑓 = ∑
𝑖=1
𝜀0
∭ 𝐄𝑖 2 d𝑉
2
The interpretation of this equation is straightforward ... 𝑈𝑡𝑜𝑡 = 𝜀20 ∭ 𝐸 2 d𝑉 is the field
energy, as derived earlier (eq 4). 𝑈𝑠𝑒𝑙𝑓 = ∑𝑛𝑖=1 𝜀20 ∭ 𝐄𝑖 2 d𝑉 is independent of the relative
positions of the charges. It is equal to the sum of the energies of the fields of the charges if
they were a very long way away from each other – effectively by themselves. We can’t
evaluate it; for example, if we treat the charges as spheres and shrink their radii to zero we
get infinite field energies. Clearly 𝑈𝑡𝑜𝑡 contains 𝑈𝑠𝑒𝑙𝑓 and must also be impossible to
calculate. But 𝑈𝑡𝑜𝑡 − 𝑈𝑠𝑒𝑙𝑓 is finite and, as we’ve shown, gives the energy arising from the
relative positions of the particles.