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Electrostatic energy of a charge distribution.PowerPoint Presentation

Electrostatic energy of a charge distribution.

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Electrostatic energy of a charge distribution.

N.B.: in vacuum – not in a material !!! And, with the boundary condition that Φ(∞) = 0

This one we have already seen! For a system of N (point) charges

Which, for a continuous charge distribution becomes:

The integral can be computed over the whole space, but is only non-zero in places where there is charge (ρ≠0). So, the formula is saying that all the electrostatic energy is concentrated in the region where there are charges, if not directly on the charges themselves. But… we can operate on the equations:

Advanced EM - Master in Physics 2011-2012

In the previous page, the key passage was to apply Gauss’s theorem to convert from a volume integral to a surface integral. Since the volume was the whole space, the surface enclosing it is a spherical surface of radius r with r∞. Then, this surface integral is zero because as r grows the surface grows as r² but the integrand (ΦE·n) decreases as 1/ r³. Where then is the total energy distributed? Only where the charges are?: Or over the whole space, where E≠0?

The two formulas are equivalent. And this leaves us in doubt on where this energy is. But… we are only in the field of electrostatics. We will have to check which formula applies in the general case, i.e. moving charges, and see if it is one of these two or an entirely different one.

In fact, we have used in the derivation the formula:

But we know that this formula is only valid in electrostatics: the general formula is (anticipating a little):

Magnetostatic Energy

(i.e., from a distribution of constant currents)

This situation is similar to one just dealt with in electrostatics. And, we also get similar formulas for the energy distribution, despite the large differences, i.e. E is the gradient of a scalar, while B is the curl of a Vector potential.

Advanced EM - Master in Physics 2011-2012

The two different formulas for the magnetostatic energy are: theorem to convert from a volume integral to a surface integral. Since the volume was the whole space, the surface enclosing it is a spherical surface of radius r with

Again, both integrals are done over the whole space. But the first one is really done only over a limited region, where are contained the currents which generate the fields. The charges and the currents are, of course, all the currents and all the charges.

The charges present under the form of matter have also to be taken into account.

One last remark: in the last form of the energy distribution that we have derived, the fields are sufficient to account for all energy: the charges and the currents are not present in the formulas. This may be a simple consequence of the relations between charges or currents and fields. But it may also be a more fundamental fact.

In any case, the fields,which originally were defined as a simple mathematical artifice to define the forces experimented by a charge as the value of the charge times a field which just lies there,

The fields, again, seem to have more and more physical substance.

Advanced EM - Master in Physics 2011-2012

The electrostatic stress tensor theorem to convert from a volume integral to a surface integral. Since the volume was the whole space, the surface enclosing it is a spherical surface of radius r with

The electric field exerts a force on charges: in fact, it is from this force that the electric field is defined.

Well, how can we calculate the force that a given electric field exerts on a given charge distribution? Keeping in mind, of course, that the two, the charge distribution and the field can not be random – they have to be compatible with each other. Well, given the field E, the force it exerts on an infinitesimal charge ρ dV is:

Where the volume of integration is any volume where charges are contained – well, possibly a volume contained by only one closed surface.

Here too we can, like we did before, take away the charges – i.e. the ρ - from the formula by replacing it with div(E)/4π

And here too we have an integrand as three functions (the components of E) times a derivative of another (E) again, which we can treat as we did before as a derivative of the product of two functions minus the product of the derivatives of the first three functions times the other one (E).

Advanced EM - Master in Physics 2011-2012

Given the fact that the derivatives are in fact a vector operator, we have to be a bit careful in treating them as simple derivatives. This is done by using the standard “Vector Formulas”, as are given, p.ex., in the cover pages of Jackson. Here we shall use the following one:

There is however a complication: the scalar quantity in this formula (ψ) is, in the equation we want to transform, the vector E. That is no problem: E = Exi+Eyj+Ezk. The vector formula above will be applied three times taking as scalar functions the three components of E.

Advanced EM - Master in Physics 2011-2012

The first term (of the last formula) is easy to understand. The second term is less easy: we have used the definition of the quantity

Now, we can use the Vector Formula:

to calculate

The last formula is understood by remembering that the curl of the electric field is zero in electrostatics. We use it by replacing (in the last expression for F)

Advanced EM - Master in Physics 2011-2012

We obtain then: The second term is less easy: we have used the

and are ready to use another Vector Formula:

in order to eventually obtain:

If we write separately the three components of this vector equation, we see that it can be written in the form:

Where the newly introduced quantity TEis the

Electrostatic stress tensor

Advanced EM - Master in Physics 2011-2012

The last formula shows that the force exerted on all the charges contained in a volume V is equal to the integral –on the surface S which encloses the volume V- of the product of the Electrostatic stress tensor with the unitary vector normal to the surface.

Remember at this point that the product of a tensor with a vector is a vector, a vector whose direction depends on n but not in a simple way: it is not necessarily parallel, nor orthogonal ton. What has been obtained now is very similar to what we discussed a few pages ago, about the energy of a system of charges within their fields: it is distributed over the whole space and can be accounted for using only the fields. Here we show that the force exerted over a system of (static) charges and their field can be accounted for with only the knowledge of the field on the enclosing surface of that system.

What said so far for electrostatics is also valid for magnetostatics. The magnetic stress tensor is defined

In presence of charges and currents, there will be both electrical and magnetic forces, and the total force will be determined by the Maxwell stress tensor T

For a physical introduction to tensors in Physics, see Feynman II, ch.31.

Advanced EM - Master in Physics 2011-2012

Magnetostatics charges contained in a volume

Let us start from the Maxwell equations. In the case of statics, electrical and magnetic fields are independent, so only eqs 3 and 4 are relevant.

The B vector field has zero divergence: therefore it is the curl of another vector field, A, so far not yet determined, such that

If we have a field, let’s call it A0, that satisfies this condition, then alsoA1 = A0 + grad Φ, with Φany finite function of space coordinates, satisfies it. This indetermination is called ”the gauge invariance”. Usually the gauge is fixed by determining the div (A).

In the case of statics, the Coulomb gauge is normally chosen:

From previous courses it is known that the general (again…) solution ot Magnetostatic problems is

Biot-Savart’s law, bears some resemblance to the electrostatics law:

Advanced EM - Master in Physics 2011-2012

The two laws compared at the end of previous slide are also somewhat different: one deals with the potential, the other with the field. One is scalar, the other a vector equation, not amenable to three simple ones because the various components are intermixed.

So, if the standard solution for the fields is of difficult handling, let’s try and see if the situation is simpler if –like we did for electrostatics, we try to calculate the potential ,i.e. A first.

Here is how the calculation proceeds, but before the Vector Formula which will be used in the demonstration is shown.

The last equation is really analogue to the electrostatic one.

It consists of three scalar equations, one for each component of the vector potential which is computed by integrating the same component of the current, with exactly the same algorithm as for the scalar potential! Let us write them all down.

Advanced EM - Master in Physics 2011-2012

These four equations are the same, apart from small changes in the constants. And, as an obvious consequence, they admit the same solutions, so that we can apply the “general solution” to determine the vector potential A.

A striking similarity between electricity and magnetism! It can not, however, be carried on too far, since in magnetism we do not have the same boundary conditions as in electrostatics: known potential on some surface.

Advanced EM - Master in Physics 2011-2012

Cylindrical wire of infinite length in the constants. And, as an obvious consequence, they admit the same solutions, so that we can apply the “general solution” to determine the run by a current

We already know that this system generates a magnetic field turning around the wire with modulus B=2I/cr.

z

.

The wire has radius a, then J=(0,0,I/a2π) ; then, A=(0,0,Az)

Ax=0

Ay=0

Az = -(π/c)jr2 inside the wire r≤a

= -(π/c)ja2(1+2ln[r/a]) outside the wire i.e., the same type of solutions than for the electric potential of a cylinder of uniform charge density.

If we now compute the curl of A we find in cylindrical coordinates (r,θ,k)

k

θ

.

y

x

j

.

Advanced EM - Master in Physics 2011-2012

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