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Eventually in formula (49) we obtain

$\Sigma \epsilon ={\frac {k}{I}}\Sigma e{\mathfrak {q}}_{n}\cdot d\sigma$,

where the sum is extended over all ions of sphere I, and

${\overline {\epsilon }}={\frac {k}{I}}\Sigma eq_{n}\cdot d\sigma$,

or by (48)

${\overline {\epsilon }}={\mathfrak {M}}_{n}d\sigma$.

§ 42. Equations ($I_{b}$) — ($VII_{b}$) (§ 20) may form the initial point for the subsequent considerations. At first we notice, that the first of them is equivalent to

$\int {\mathfrak {d}}_{n}d\sigma =E$,

for an arbitrary closed surface (n is to be drawn into the outside), when E is the electric charge enclosed by it. If now in one element $d{\mathfrak {r}}$ of the inner space in equilibrium state, a density $\rho _{0}$ exists, and if (for an element of the surface) $\epsilon$ has the meaning given above, then

$E=\int \rho _{0}d{\mathfrak {r}}-\Sigma \epsilon$,

where the sum is related to all elements $d\sigma$.

By that

$\int {\mathfrak {d}}_{n}d\sigma +\Sigma \epsilon =\int \rho _{0}d{\mathfrak {r}}$.

From the definition of the average we easily find now

$\int {\overline {{\mathfrak {d}}_{n}}}d\sigma +\Sigma {\overline {\epsilon }}=\int {\overline {\rho _{0}}}d{\mathfrak {r}}$.

Since now

${\overline {\rho _{0}}}=0$,

and

${\overline {\epsilon }}={\mathfrak {M}}_{n}d\sigma$.

it is eventually given

$\int \left({\overline {{\mathfrak {d}}_{n}}}+{\mathfrak {M}}_{n}\right)d\sigma =0$.