Coulomb Energy of a Uniformly Charged Nucleus

Derivation

Coulomb Energy of a Uniformly Charged Nucleus

D-266 Home PU-304 Threads energy · fields · force Depends on electrostatic-energy-uniform-sphere

Statement

For a nucleus modelled as a total charge \(Q = Ze\) distributed uniformly throughout a sphere of radius \(R\), the electrostatic self-energy — the work required to assemble the charge against its own field — is \(\displaystyle U = \frac{3}{5}\frac{Z^2 e^2}{4\pi\varepsilon_0 R}\). This is the origin of the Coulomb (electrostatic) term \(a_C\, Z^2 A^{-1/3}\) in the semi-empirical mass formula.

Why it matters

The Coulomb term is the only long-range contribution in the liquid-drop model, and it is what makes heavy nuclei unstable: the strong force saturates and scales like the volume, but the electrostatic energy grows like \(Z^2/R \sim Z^2 A^{-1/3}\), so beyond a critical charge the drop prefers to split. This single term drives the systematics of fission, the location of the valley of stability, and the \(Z^2/A\) fissility parameter.

It is also a clean, exactly solvable electrostatics problem whose answer — the numerical factor \(3/5\) — is quoted constantly. Getting it right requires being careful about the difference between the field energy of a full sphere and the assembly energy of shells, and about the factor-of-two conventions that trip students up.

Assumptions

The charge density is uniform, \(\rho = Q/\left(\tfrac{4}{3}\pi R^3\right)\) constant inside \(r<R\) and zero outside.If the density varies (a realistic Fermi/Woods–Saxon profile with a diffuse surface), the coefficient \(3/5\) shifts and the surface region contributes an additional correction; the clean power law survives but the prefactor is model-dependent.

The charge distribution is rigid and classical, a continuous fluid rather than \(Z\) discrete protons.If treated as \(Z\) point charges one must remove the divergent proton self-energies, replacing \(Z^2\) by \(Z(Z-1)\); the exchange (Fock) term is also absent classically. Dropping this gives the wrong small-\(Z\) behaviour.

Electrostatics is the only interaction counted, and the assembly is quasi-static.If magnetic, retardation, or vacuum-polarization effects were included the energy would acquire relativistic corrections of order \((v/c)^2\); for nuclear charges these are small but nonzero, so the result is the leading static term only.

Derivation

\[ \rho = \frac{Q}{\tfrac{4}{3}\pi R^3} = \frac{3Q}{4\pi R^3} \]

Definition of uniform volume charge density: total charge divided by total volume. A

\[ q(r) = \rho\cdot \frac{4}{3}\pi r^3 = Q\,\frac{r^3}{R^3} \]

Charge enclosed within radius \(r<R\); by Gauss's law only this interior charge sources the field at \(r\). A

\[ E(r) = \frac{1}{4\pi\varepsilon_0}\frac{q(r)}{r^2} = \frac{1}{4\pi\varepsilon_0}\frac{Q\,r}{R^3},\qquad r<R \]

Gauss's law with spherical symmetry: \(\oint \vec E\cdot d\vec A = q_{\text{enc}}/\varepsilon_0\) gives \(E\cdot 4\pi r^2 = q(r)/\varepsilon_0\). A

\[ E(r) = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2},\qquad r>R \]

Outside the sphere all charge is enclosed, so the field is that of a point charge \(Q\) at the centre (shell theorem). A

\[ U = \frac{\varepsilon_0}{2}\int_{\text{all space}} E^2 \, dV \]

The energy stored in an electrostatic field is \(u = \tfrac{1}{2}\varepsilon_0 E^2\) per unit volume; integrating the field-energy density gives the total assembly energy. This is the assumed prior result B

\[ U = \frac{\varepsilon_0}{2}\int_0^R E_{\text{in}}^2\,4\pi r^2\,dr \;+\; \frac{\varepsilon_0}{2}\int_R^{\infty} E_{\text{out}}^2\,4\pi r^2\,dr \]

Split the all-space integral at \(r=R\) using \(dV = 4\pi r^2\,dr\) for spherical symmetry; the two fields differ, so they must be integrated separately. B

\[ U_{\text{in}} = \frac{\varepsilon_0}{2}\int_0^R \left(\frac{Q\,r}{4\pi\varepsilon_0 R^3}\right)^2 4\pi r^2\,dr = \frac{Q^2}{8\pi\varepsilon_0 R^6}\int_0^R r^4\,dr \]

Substitute \(E_{\text{in}}\) from step 3 and collect constants; \(4\pi\) partially cancels the \((4\pi)^2\) in the denominator. A

\[ \int_0^R r^4\,dr = \frac{R^5}{5} \quad\Longrightarrow\quad U_{\text{in}} = \frac{Q^2}{8\pi\varepsilon_0 R^6}\cdot\frac{R^5}{5} = \frac{Q^2}{40\pi\varepsilon_0 R} \]

Elementary power-law integral, then simplify. This is the interior field-energy contribution. A

\[ U_{\text{out}} = \frac{\varepsilon_0}{2}\int_R^{\infty}\left(\frac{Q}{4\pi\varepsilon_0 r^2}\right)^2 4\pi r^2\,dr = \frac{Q^2}{8\pi\varepsilon_0}\int_R^{\infty}\frac{dr}{r^2} \]

Substitute \(E_{\text{out}}\) from step 4; the \(r^2\) from the volume element cancels two powers of \(r^{-2}\) from \(E^2\), leaving \(r^{-2}\). A

\[ \int_R^{\infty}\frac{dr}{r^2} = \frac{1}{R} \quad\Longrightarrow\quad U_{\text{out}} = \frac{Q^2}{8\pi\varepsilon_0 R} = \frac{5Q^2}{40\pi\varepsilon_0 R} \]

Convergent improper integral (the exterior energy of a point charge is finite because it starts at \(r=R\), not \(0\)). Rewritten over \(40\) to add to step 8. A

\[ U = U_{\text{in}} + U_{\text{out}} = \frac{Q^2}{40\pi\varepsilon_0 R} + \frac{5Q^2}{40\pi\varepsilon_0 R} = \frac{6Q^2}{40\pi\varepsilon_0 R} = \frac{3}{5}\frac{Q^2}{4\pi\varepsilon_0 R} \]

Add the two contributions; \(6/40 = 3/20 = (3/5)(1/4)\), which factors cleanly into the standard form. A

\[ U = \frac{3}{5}\frac{(Ze)^2}{4\pi\varepsilon_0 R} = \frac{3}{5}\frac{Z^2 e^2}{4\pi\varepsilon_0 R} \]

Insert the nuclear charge \(Q = Ze\). With \(R = r_0 A^{1/3}\) this is exactly \(a_C Z^2 A^{-1/3}\), the Coulomb term of the mass formula. B

Result

\[ U = \frac{3}{5}\frac{Z^2 e^2}{4\pi\varepsilon_0 R} \]

Reading. The electrostatic self-energy of a uniformly charged sphere is \(3/5\) of the energy \(Q^2/(4\pi\varepsilon_0 R)\) you would (naively) assign to charge \(Q\) at separation \(R\). The factor \(3/5\) reflects that most charge sits at radii smaller than \(R\), so the effective mutual separation is less than \(R\), raising the energy above that of a single thin shell (which would give a factor \(1/2\)). Substituting \(R = r_0 A^{1/3}\) reproduces the mass-formula scaling \(a_C\, Z^2 A^{-1/3}\), with \(a_C = \tfrac{3}{5}\,e^2/(4\pi\varepsilon_0 r_0) \approx 0.72\text{–}0.77\ \text{MeV}\) once the point-charge correction \(Z^2 \to Z(Z-1)\) and diffuseness are folded in empirically.

Units check. \(\dfrac{e^2}{4\pi\varepsilon_0 R}\) has units \(\dfrac{\text{C}^2}{(\text{C}^2\,\text{N}^{-1}\text{m}^{-2})\cdot\text{m}} = \text{N}\cdot\text{m} = \text{J}\). The prefactor \(3/5\) and \(Z^2\) are dimensionless, so \(U\) is an energy, as required.

Limiting cases

Thin shell limit: if instead all charge sat on the surface \(r=R\), the interior field vanishes and only \(U_{\text{out}}\) survives, giving \(U = \tfrac{1}{2}Q^2/(4\pi\varepsilon_0 R)\) — factor \(1/2\), not \(3/5\). The uniform sphere is more tightly bound because charge overlaps.
Point charge \(R\to 0\): \(U \to \infty\). The classical self-energy diverges as \(1/R\); a finite nuclear radius is what keeps it finite, which is precisely why the term depends on \(A^{1/3}\).
Large \(R\) (dilute): \(U\to 0\) as \(1/R\); spreading the same charge over a larger volume always lowers the electrostatic energy.
\(Z=1\): the formula gives a nonzero \(U\), but physically a single proton has no proton–proton repulsion; this signals the need for the \(Z(Z-1)\) correction (see Assumptions).

Breaks when

Few protons / small \(Z\): the continuum \(Z^2\) overcounts by including each proton's self-interaction. The discrete result is \(\propto Z(Z-1)\), so for light nuclei (e.g. \(Z=1,2\)) the uniform-sphere expression is quantitatively wrong and must be corrected.
Diffuse or deformed surface: real nuclei have a Woods–Saxon density with surface thickness \(\sim 2.4\ \text{fm}\), and near fission they deform. Both change the geometric coefficient away from \(3/5\); the sharp-sphere assumption fails and one needs a surface-diffuseness correction and, for large deformation, a shape-dependent Coulomb energy.
Strong-field / relativistic regime (superheavy, \(Z\alpha \sim 1\)): for very large \(Z\) the electric field near the surface becomes strong enough that vacuum polarization and relativistic corrections matter, and the simple static electrostatic self-energy is no longer the whole story.

Failure modes

Forgetting the interior field. Integrating only \(U_{\text{out}}\) gives \(1/2\) instead of \(3/5\); the inside of the sphere stores real field energy and must be included.
Double-counting with a factor \(\tfrac12\). Using both the energy density \(\tfrac12\varepsilon_0 E^2\) and a separate \(\tfrac12\sum q_i V_i\) sum, or inserting an extra \(\tfrac12\) into the field integral, halves the answer. The \(\tfrac12\) is already in \(u=\tfrac12\varepsilon_0E^2\).
Using \(E = Q/(4\pi\varepsilon_0 r^2)\) inside the sphere. The point-charge field is wrong for \(r<R\); inside, \(E \propto r\) (only enclosed charge counts).
Wrong volume element. Writing \(dV = dr\) or \(4\pi r\,dr\) instead of \(4\pi r^2\,dr\) breaks every power of \(R\).
Keeping \(Z^2\) for light nuclei. Reporting a proton–proton repulsion for \(Z=1\) — physically impossible — because the \(Z(Z-1)\) correction was ignored.
Dropping \(4\pi\varepsilon_0\) (Gaussian confusion). Mixing SI and Gaussian conventions loses the \(4\pi\varepsilon_0\), giving numbers off by that factor.

Discussion

The number \(3/5\) is purely geometric: it encodes how a uniform density weights the mutual Coulomb interaction across all pairs of charge elements. An equivalent derivation builds the sphere shell by shell, \(dU = V(r)\,dq\) with \(V(r)=q(r)/(4\pi\varepsilon_0 r)\) the potential at the growing surface; integrating \(dU\) from \(0\) to \(R\) gives the same \(3/5\). That "assembly" picture and the "field-energy" picture used above are two faces of the same conservation statement — the work done to assemble the charge equals the energy stored in the resulting field — and it is instructive that they agree exactly.

In the semi-empirical mass formula the Coulomb term enters with a minus sign in the binding energy, \(B = a_V A - a_S A^{2/3} - a_C Z^2 A^{-1/3} - \dots\), because electrostatic self-energy destabilises the nucleus. Its competition with the surface term \(a_S A^{2/3}\) under deformation is exactly the Bohr–Wheeler analysis of fission: a small quadrupole distortion lowers the Coulomb energy but raises the surface energy, and the sign of the net change defines the fissility parameter \(x = E_C/(2E_S) \propto Z^2/A\). When \(x\gtrsim 1\) the nucleus is unstable to spontaneous fission.

Empirically \(a_C \approx 0.71\ \text{MeV}\) implies \(r_0 \approx 1.2\ \text{fm}\) via \(a_C = \tfrac35 e^2/(4\pi\varepsilon_0 r_0)\) — remarkably, an electrostatics coefficient fitted to nuclear masses recovers the nuclear-radius constant independently measured by electron scattering. This consistency is one of the quiet triumphs of the liquid-drop picture.

At a deeper level the classical \(Z^2\) should be \(Z(Z-1)\) plus a quantum-mechanical exchange (Fock) correction. Treating the protons as a degenerate Fermi gas, the exchange energy scales as \(-Z^{4/3}\), giving the small \(a_C^{\text{exch}}\) correction sometimes written \(-\tfrac34(3/2\pi)^{2/3} e^2/(4\pi\varepsilon_0 r_0)\,Z^{4/3}A^{-1/3}\). The uniform-sphere result is thus the leading direct (Hartree) term; the full nuclear Coulomb energy is direct plus exchange plus surface-diffuseness corrections, all built on this backbone.

Common misconceptions. The \(3/5\) is not a probability or an averaging artefact and has nothing to do with three dimensions "over five" of anything; it is \(\int_0^R r^4\,dr\)-type geometry. And the self-energy is not "the energy to bring \(Q\) from infinity to a point at distance \(R\)" — that would be \(Q^2/(4\pi\varepsilon_0 R)\); the \(3/5\) specifically accounts for assembling the charge throughout the volume.

Worked examples

Coulomb self-energy of \(^{16}\)O (\(Z=8\), \(A=16\)).

\[ R = r_0 A^{1/3},\qquad U = \frac{3}{5}\frac{Z^2 e^2}{4\pi\varepsilon_0 R} \]

Take \(r_0 = 1.2\ \text{fm}\), so \(R = 1.2\times 16^{1/3} = 1.2\times 2.520 = 3.02\ \text{fm}\). A

\[ \frac{e^2}{4\pi\varepsilon_0} = 1.44\ \text{MeV·fm} \]

Standard nuclear-physics shortcut for the Coulomb constant. A

\[ U = \frac{3}{5}\cdot\frac{8^2\times 1.44\ \text{MeV·fm}}{3.02\ \text{fm}} = \frac{3}{5}\cdot\frac{64\times 1.44}{3.02}\ \text{MeV} \]

Insert numbers; the femtometres cancel, leaving MeV. A

\[ U = 0.6\times \frac{92.16}{3.02}\ \text{MeV} = 0.6\times 30.5\ \text{MeV} = 18.3\ \text{MeV} \]

Arithmetic. Using the point-charge correction \(Z(Z-1)=56\) instead of \(64\) gives \(16.0\ \text{MeV}\), closer to experiment. B

\[ U \approx 18\ \text{MeV}\quad(\text{uniform sphere}),\ \ \approx 16\ \text{MeV}\ \text{with }Z(Z-1) \]

Reading. The mutual electrostatic repulsion of the eight protons in oxygen-16 amounts to roughly 16–18 MeV — comparable to the total binding energy scale, showing why Coulomb energy is never negligible.

Units check. \(\text{MeV·fm}/\text{fm}=\text{MeV}\); dimensionless prefactors, so the result is an energy.

Coulomb energy difference of the mirror pair \(^{15}\)O (\(Z=8\)) and \(^{15}\)N (\(Z=7\)).

\[ \Delta U = \frac{3}{5}\frac{e^2}{4\pi\varepsilon_0 R}\left(Z_1^2 - Z_2^2\right) \]

Both share \(A=15\) so the same radius \(R\); only \(Z\) differs. Mirror nuclei differ by one proton↔neutron swap, so the mass difference is (almost) pure Coulomb. B

\[ R = 1.2\times 15^{1/3} = 1.2\times 2.466 = 2.96\ \text{fm} \]

Same \(r_0\). A

\[ Z_1^2 - Z_2^2 = 8^2 - 7^2 = 64-49 = 15 \]

Note \(Z_1^2 - Z_2^2 = (Z_1+Z_2)(Z_1-Z_2) = 15\times 1\); equivalently \(2Z-1\) for adjacent \(Z\). A

\[ \Delta U = 0.6\times\frac{1.44\ \text{MeV·fm}}{2.96\ \text{fm}}\times 15 = 0.6\times 0.486\times 15\ \text{MeV} \]

Insert numbers. A

\[ \Delta U \approx 4.4\ \text{MeV} \]

Reading. The measured Coulomb energy difference of the \(^{15}\)O–\(^{15}\)N mirror pair is \(\approx 3.5\ \text{MeV}\); the uniform-sphere estimate \(4.4\ \text{MeV}\) is the right order and slightly high, as expected before diffuseness and \(Z(Z-1)\) corrections. Such differences are how nuclear radii are extracted from mirror masses.

Units check. \(\text{MeV·fm}/\text{fm}\times(\text{dimensionless})=\text{MeV}\).

Problems

Show, by the shell-assembly method \(dU = \dfrac{q(r)}{4\pi\varepsilon_0 r}\,dq\) with \(q(r)=Q r^3/R^3\), that the self-energy is \(\tfrac35 Q^2/(4\pi\varepsilon_0 R)\), confirming the field-energy derivation.
Solution
With \(q(r)=Q r^3/R^3\), \(dq = \rho\,4\pi r^2\,dr = \dfrac{3Q}{R^3}r^2\,dr\). Then \(dU = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q(r)\,dq}{r} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{Q r^3/R^3}{r}\cdot\dfrac{3Q}{R^3}r^2\,dr = \dfrac{3Q^2}{4\pi\varepsilon_0 R^6}r^4\,dr\). Integrate: \(U = \dfrac{3Q^2}{4\pi\varepsilon_0 R^6}\cdot\dfrac{R^5}{5} = \dfrac{3}{5}\dfrac{Q^2}{4\pi\varepsilon_0 R}\). Same result.
A uniformly charged sphere has \(Q\) and \(R\). What fraction of its total electrostatic energy is stored inside the sphere?
Solution
From the derivation \(U_{\text{in}} = \dfrac{Q^2}{40\pi\varepsilon_0 R}\) and \(U_{\text{tot}} = \dfrac{6Q^2}{40\pi\varepsilon_0 R}\). Fraction inside \(= U_{\text{in}}/U_{\text{tot}} = 1/6 \approx 16.7\%\). The remaining \(5/6\) is stored in the external field.
Estimate the Coulomb self-energy of \(^{208}\)Pb (\(Z=82\), \(A=208\)) using \(r_0=1.2\ \text{fm}\) and the uniform-sphere formula. Then recompute with the \(Z(Z-1)\) correction.
Solution
\(R = 1.2\times 208^{1/3} = 1.2\times 5.925 = 7.11\ \text{fm}\). \(U = 0.6\times\dfrac{82^2\times 1.44}{7.11} = 0.6\times\dfrac{6724\times1.44}{7.11} = 0.6\times\dfrac{9683}{7.11} = 0.6\times 1362 = 817\ \text{MeV}\). With \(Z(Z-1)=82\times81=6642\): \(U = 0.6\times\dfrac{6642\times1.44}{7.11}=0.6\times 1345 = 807\ \text{MeV}\). The correction is small (\(\sim 1\%\)) for heavy nuclei since \(Z(Z-1)/Z^2 = 1-1/Z\) is near unity.
Using the empirical Coulomb coefficient \(a_C = 0.711\ \text{MeV}\), infer the radius constant \(r_0\) from \(a_C = \tfrac35\,e^2/(4\pi\varepsilon_0 r_0)\).
Solution
\(r_0 = \dfrac{3}{5}\dfrac{e^2/(4\pi\varepsilon_0)}{a_C} = \dfrac{3}{5}\dfrac{1.44\ \text{MeV·fm}}{0.711\ \text{MeV}} = 0.6\times 2.025\ \text{fm} = 1.22\ \text{fm}\). This matches the electron-scattering value \(r_0\approx 1.2\ \text{fm}\) to within a few percent, an independent confirmation of the nuclear radius.
For the mirror pair \(^{27}\)Si (\(Z=14\)) and \(^{27}\)Al (\(Z=13\)), predict the Coulomb energy difference with \(r_0=1.2\ \text{fm}\), and compare with the measured \(\approx 5.6\ \text{MeV}\).
Solution
\(R = 1.2\times 27^{1/3} = 1.2\times 3 = 3.60\ \text{fm}\). \(Z_1^2 - Z_2^2 = 14^2-13^2 = 196-169 = 27\) (indeed \(2Z-1 = 27\)). \(\Delta U = 0.6\times\dfrac{1.44}{3.60}\times 27 = 0.6\times 0.400\times 27 = 6.48\ \text{MeV}\). Predicted \(6.5\ \text{MeV}\) vs measured \(5.6\ \text{MeV}\); the sharp-sphere model overestimates by \(\sim 15\%\), the discrepancy absorbed by surface diffuseness.