Solved thermodynamics exercise

A charged sphere and the question of extensivity

Exercise 16 · Lesson 4 — The First Law of Thermodynamics

Coulomb interaction
electrostatic energy
extensivity
electrical neutrality

Statement

A total charge $Q > 0$ is distributed uniformly throughout the volume of a sphere of radius $R$ , in vacuum. Constraints hold this distribution in place; only its electrostatic energy $U_{el}$ is counted here. This is not a conducting sphere, whose charges would sit on the surface at equilibrium.

The distribution is assembled by bringing charged shells in from infinity. For a charge $Q (r)$ already assembled in a ball of radius $r$ , the potential at the surface is $Φ (r) = k_{e} Q (r) / r$ , where $k_{e} = 1/ (4 π ε_{0})$ . The energy is zero when the charge elements are infinitely far apart.

Draw one stage of the assembly. Express $Q (r)$ and $d Q$ in terms of $Q, R, r$ and $d r$ , then calculate $U_{el}$ .
In a continuum model of $N$ constituents all carrying the same charge $q_{0}$ , set $Q = N q_{0}$ . At fixed particle number density, write $R = r_{0} N^{1/3}$ and determine how the energy depends on $N$ . Compare its sign and its degree of homogeneity with those of gravitation.
At fixed total charge, calculate the change in energy during an expansion from $R_{A}$ to $R_{B} > R_{A}$ . What is the sign of the external work received if this expansion is quasi-static and non-dissipative? The expression for this work represents only the electrical contribution.
For charges of different signs, the microscopic interaction energy is $U_{el} = k_{e} i < j \sum \frac{q _{i} q _{j}}{r _{ij}} .$
Why does the law obtained for a sphere carrying charge of a single sign not prove that all ordinary matter is non-extensive? Is a zero total charge enough to make this sum vanish?

Hint

Use

d U_{el} = Φ (r) d Q

. The shell-by-shell construction counts each pair once. For the last question, consider a single pair of charges

+ q_{0}

and

- q_{0}

a distance

a

apart.

Detailed solution

Solution

Question 1. The diagram shows the already charged ball, of radius

r

, and the shell being brought in, of thickness

d r

. The constant volume charge density is

ρ_{q} = 3 Q / (4 π R^{3})

. Consequently,

Q (r) = ρ_{q} \frac{4 π r ^{3}}{3} = Q \frac{r ^{3}}{R ^{3}}, d Q = 4 π ρ_{q} r^{2} d r = \frac{3 Q}{R ^{3}} r^{2} d r .

The change in energy when the shell is added is

d U_{el} = \frac{k _{e} Q ( r )}{r} d Q = \frac{3 k _{e} Q ^{2}}{R ^{6}} r^{4} d r .

Summing the contributions from the centre out to the surface,

U_{el} = \frac{3 k _{e} Q ^{2}}{R ^{6}} \int_{0}^{R} r^{4} d r = \frac{3 k _{e} Q ^{2}}{R ^{6}} [\frac{r ^{5}}{5}]_{0}^{R} = \frac{3 k _{e} Q ^{2}}{5 R} = \frac{3 Q ^{2}}{20 π ε _{0} R} .

The interaction of each pair of layers is counted only once; no additional factor of $1/2$ is needed. The result is positive: work must be done against the repulsion to assemble the charges. The product $k_{e} Q^{2} / R$ is expressed in joules.

Question 2. At fixed particle number density

ρ_{N} = N / (4 π R^{3} /3)

, we have

r_{0} = (3/ (4 π ρ_{N}))^{1/3}

. Substituting,

U_{el} = \frac{3 k _{e} ( N q _{0} ) ^{2}}{5 r _{0} N ^{1/3}} = \frac{3 k _{e} q _{0}^{2}}{5 r _{0}} N^{5/3} .

This energy is of degree $5/3$ , and therefore not extensive. The energy per particle grows as $N^{2/3}$ . The degree is the same as for the gravitating sphere because both potentials decrease as $1/ r$ ; the sign is opposite, since like charges repel. The continuum model describes the collective contribution, without counting the self-energies of the point charges.

Question 3. At fixed

Q

, the difference is

Δ U_{el} = \frac{3 k _{e} Q ^{2}}{5} (\frac{1}{R _{B}} - \frac{1}{R _{A}}) < 0.

In a quasi-static, non-dissipative process, the electrical contribution to the work received is the change in potential energy. This can also be checked by integration:

W_{ext} = \int_{R_{A}}^{R_{B}} - \frac{3 k _{e} Q ^{2}}{5 R ^{2}} d R = [\frac{3 k _{e} Q ^{2}}{5 R}]_{R_{A}}^{R_{B}} = Δ U_{el} < 0.

The external agent holds back the mutually repelling charges and receives work during the expansion. No other energy storage in the restraining device is included in this electrical balance.

Question 4. For charges of the same sign, all the products

q_{i} q_{j}

are positive: their contributions add up. In electrically neutral matter, by contrast, there are attractive and repulsive contributions that can cancel at large distances. Neutrality and, in media where it occurs, screening thus alter an argument based on adding terms that all have the same sign. Overall neutrality does not, however, imply

U_{el} = 0

. Two charges

+ q_{0}

and

- q_{0}

a distance

a

apart have zero total charge but a nonzero interaction energy

- k_{e} q_{0}^{2} / a

. The arrangement of the charges and the correlations between them must be known to establish a scaling law. The calculation for the uniformly charged sphere is therefore neither a proof that all matter is non-extensive nor a proof that neutral systems are extensive.