V. D. Kuznetsov.

On the Question of the Change in Electrical Conductivity of Liquid Dielectrics Over Time

The causes of changes in the electrical conductivity of liquid dielectrics are very diverse. All kinds of impurities play a major role; some of them are elusive not only quantitatively, but also qualitatively.

Here we shall analyze theoretically the case of a very pure dielectric in which ions of one certain type exist.

If the dielectric is placed in a vessel without metallic electrodes, then the change in the number of ions \(n\) in one \(\mathrm{cm}^3\) occurs according to the equation:

\[ \frac{dn}{dt}=q-\alpha n^2, \tag{1} \]

where \(q\) is the number of ions of one or another sign formed in \(1\ \mathrm{cm}^3\); \(n\) is the number of ions of one sign existing at the given moment in \(1\ \mathrm{cm}^3\); and \(\alpha\) is the recombination coefficient. Equation (1) applies to the case when the numbers of positive and negative ions are equal, i.e. the dielectric is not charged.

Ions in a liquid dielectric poured into a metallic vessel, or into a measuring capacitor with metallic electrodes, are distributed nonuniformly, and their number gradually decreases until a stationary state is established. As a result of the presence of metallic parts, a difference in concentration arises and diffusion of ions takes place. Approaching the electrode, the ion induces a charge in it, is attracted to it and, giving up its charge, turns into a neutral molecule. If the ion consists of a conglomerate or group of molecules gathered around a charged nucleus, then after neutralization of the charge the ion ceases to exist as a separate individual, and the molecules that formed it disperse throughout the entire volume of the liquid. As a result of attraction, the number of ions near the metallic surface rapidly decreases, and at the surface itself the ion concentration is always equal to zero. The distribution of ions in a liquid dielectric between two metallic electrodes is the same as the distribution of droplets of water mist between two surfaces made of a hygroscopic substance, for example sulfuric acid, calcium chloride, phosphoric anhydride, etc.

If the concentration \(n\) of ions is a function of the point \((x,y,z)\), then the change of ions due to diffusion occurs according to the equation:

\[ \frac{\partial n}{\partial t} = D\left( \frac{\partial^2 n}{\partial x^2} + \frac{\partial^2 n}{\partial y^2} + \frac{\partial^2 n}{\partial z^2} \right), \]

where \(D\) is the diffusion coefficient; or according to the equation:

\[ \frac{\partial n}{\partial t} = D\cdot \frac{\partial^2 n}{\partial x^2}, \tag{2} \]

if the concentration depends only on \(x\).

Finally, it remains for us to establish an expression for the change in the number of ions with time for the case when a certain potential difference \(V\) is applied to the electrodes, which creates an electric field inside the dielectric and causes the motion of ions.

Let the electrodes be parallel to one another and be at a distance \(d\). Under the influence of the voltage \(V/d\), a current arises in the dielectric, whose density \(j\) is expressed by

\[ j=\varepsilon\,(K_1+K_2)n\frac{V}{d}, \]

where \(\varepsilon\) is the elementary charge, \(K_1\) and \(K_2\) are the mobilities of the positive and negative ions. The number of ions carried by the current to the electrodes from a column of dielectric of length \(d\) cm and cross-sectional area \(1\ \mathrm{cm}^2\) is equal to

\[ (K_1+K_2)n\frac{V}{d}, \]

and the number of ions removed by the current from a unit volume during one second is equal to:

\[ \frac{\partial n}{\partial t}=(K_1+K_2)n\frac{V}{d^2} \tag{3} \]

Consequently, with the simultaneous action of ionization, recombination, diffusion, and electric current, the equation for the change in the number of ions takes the form:

\[ \frac{\partial n}{\partial t} = q-\alpha n^2+D\frac{\partial^2 n}{\partial x^2} -(K_1+K_2)n\frac{V}{d^2}. \tag{4} \]

Let us consider particular cases of integrating equation (4).

Case I. Before the experiment the given dielectric either was not at all subjected to the action of the electric-field voltage, or the latter was insignificant; then a high potential difference is immediately applied to the electrodes, and it is the principal factor removing ions from the liquid. Neglecting recombination and diffusion of ions, i.e. putting \(\alpha=0\) and \(D=0\), we have:

\[ \frac{dn}{dt} = q-(K_1+K_2)\frac{V}{d^2}n \tag{5} \]

or, denoting

\[ (K_1+K_2)\frac{V}{d^2}=\beta, \]

\[ \frac{dn}{dt}=q-\beta n. \]

The integral of the last equation is written in the form:

\[ n=\frac{q}{\beta}+Ce^{-\beta t}, \]

where \(C\) is an arbitrary constant.

Multiplying both sides of the last expression by \(\varepsilon d\), passing in part to the former notation and introducing the new constant

\[ C_1=C\varepsilon(K_1+K_2)\frac{V}{d}, \]

we obtain

\[ \varepsilon(K_1+K_2)\frac{V}{d}n = \varepsilon qd+C_1e^{-\beta t}. \tag{6} \]

Taking into account that the current density is

\[ j=\varepsilon (K_1+K_2)\frac{V}{d}n \]

and the saturation current density is

\[ J=\varepsilon qd, \]

from (6) we obtain:

\[ j=J+C_1 e^{-\beta t}. \]

If we denote the current density at the initial moment \(t=0\) by \(j_0\), and on this basis determine the constant \(C_1\), then finally we have:

\[ j=J+(j_0-J)e^{-(K_1+K_2)\frac{V\cdot t}{d^2}}, \tag{7} \]

from which it is seen that for \(t=\infty\) the current reaches saturation \((j=J)\).

Consequently, in the case of transition from a low voltage to a sufficiently high one, the electric current, consisting at first of two components, of which one is constant and the other decreases with the passage of time, after an infinitely long interval of time becomes the saturation current.

Case II. The dielectric is for a long time subjected to the action of a high voltage, which then, at the moment \(t=0\), is replaced by so low a voltage that the transport of ions by the current may be neglected in comparison with their destruction as a result of recombination. In addition, we shall neglect the diffusion of ions. In this case equation (4) takes the form:

\[ \frac{dn}{dt}=q-\alpha n^2. \]

The integral of the last equation has the form:

\[ n=\sqrt{\frac{q}{\alpha}\,\frac{1-e^{-2\sqrt{q\alpha}\,t}}{1+e^{-2\sqrt{q\alpha}\,t}}}. \]

The current density at the moment \(t\) is expressed by:

\[ j=j_\infty\frac{1-e^{-2\sqrt{q\alpha}\,t}}{1+e^{-2\sqrt{q\alpha}\,t}}, \]

where

\[ j_\infty=\varepsilon(K_1+K_2)\frac{V}{d}\sqrt{\frac{q}{\alpha}} \tag{8} \]

denotes the current density after an infinitely long interval of time.

Thus, in the transition from a high voltage to a very low one, the current is initially equal to zero, and then, gradually increasing, asymptotically approaches the constant value \(j_\infty\).

Case III. If we neglect only the diffusion of ions and take into account the relation between ionization, recombination, and the transport of ions by the current, then equation (4) takes the form:

\[ \frac{dn}{dt}=q-\alpha n^2-(K_1+K_2)\frac{V}{d^2}n. \]

The solution of this equation reduces to a quadrature and finally takes the form:

\[ \frac{2\alpha n+\beta-\sqrt{\beta^2+4\alpha q}}{2\alpha n+\beta+\sqrt{\beta^2+4\alpha q}} = Ce^{-\sqrt{\beta^2+4\alpha q}\cdot t}. \tag{9} \]

where, in accordance with the previous notation,

\[ \beta=(K_1+K_2)\frac{V}{d^2}. \]

Solving (9) with respect to \(n\), we obtain:

\[ n=\frac{1}{2a}\, \frac{C e^{-\sqrt{\beta^2+4aq}\,t}\left(\sqrt{\beta^2+4aq}+\beta\right)+\left(\sqrt{\beta^2+4aq}-\beta\right)} {1-C e^{-\sqrt{\beta^2+4aq}\,t}} . \tag{10} \]

To determine the arbitrary constant \(C\) in (9), let us assume that at \(t=0\) the number of ions is \(n=n_0\); then

\[ C=\frac{2an_0+\beta-\sqrt{\beta^2+4aq}} {2an_0+\beta+\sqrt{\beta^2+4aq}}. \]

In the particular case where, before the application of a potential difference to the electrodes, a stationary state had been established between ionization and recombination, then

\[ n_0=\sqrt{\frac{q}{a}}. \]

From (9) we have, for \(t=\infty\),

\[ n_{\infty}=\frac{1}{2a}\left(\sqrt{\beta^2+4aq}-\beta\right). \]

Taking into account this latter value \(n_{\infty}\), we rewrite formula (10) in the form:

\[ n=n_{\infty}\, \frac{ 1+ \frac{\sqrt{\beta^2+4aq}+\beta}{\sqrt{\beta^2+4aq}-\beta}\, C e^{-\sqrt{\beta^2+4aq}\,t} } { 1-Ce^{-\sqrt{\beta^2+4aq}\,t} }. \tag{11} \]

Using the expression for the current density,

\[ j=en(K_1+K_2)\frac{V}{d}, \]

from (11) we find:

\[ j=j_{\infty}\, \frac{ 1+C_1 e^{-\sqrt{\beta^2+4aq}\,t} } { 1-Ce^{-\sqrt{\beta^2+4aq}\,t} }, \tag{12} \]

where

\[ C_1=\frac{\sqrt{\beta^2+4aq}+\beta} {\sqrt{\beta^2+4aq}-\beta}\,C. \]

Formula (12) shows that in this case as well the current density, gradually decreasing, assumes at \(t=\infty\) a constant value,

\[ j_{\infty}=en_{\infty}(K_1+K_2)\frac{V}{d}, \]

which, in the case of high voltage, will be the saturation current density.