Solution

Let the radius of each metallic sphere be denoted by $a$:

$$a=\frac{30\,\text{cm}}{2}=0.15\,\text{m}.$$

The electrodes are located at a depth

$$h=60\,\text{m},$$

their center-to-center separation is

$$L=300\,\text{m},$$

and the electrical conductivity of seawater is

$$\sigma=4\,\text{S/m}.$$

We need to determine the electrical resistance of the seawater between the two spherical electrodes.

Since the size of each electrode is much smaller than both the depth and the separation between the electrodes,

$$a\ll h,\qquad a\ll L,$$

each electrode may be treated, to a good approximation, as a point current source or sink, while its finite radius $a$ is retained when calculating its self-potential.

1. Potential of a point current source in a conducting medium

Suppose a current $I$ flows out of a small spherical electrode into an infinite homogeneous conducting medium.

At a distance $r$ from the center, the current is distributed approximately spherically. Therefore, the current density is

$$j(r)=\frac{I}{4\pi r^2}.$$

The differential form of Ohm's law is

$$\mathbf j=\sigma\mathbf E.$$

Hence,

$$E(r)=\frac{j(r)}{\sigma}.$$

Substituting the expression for the current density,

$$E(r)=\frac{I}{4\pi\sigma r^2}.$$

The electric potential relative to infinity is

$$\varphi(r)=\int_r^\infty E(r')\,dr'.$$

Therefore,

$$\varphi(r)=\frac{I}{4\pi\sigma}\int_r^\infty\frac{dr'}{r'^2}.$$

Thus,

$$\boxed{\varphi(r)=\frac{I}{4\pi\sigma r}}.$$

For a spherical electrode of radius $a$, its self-contribution to the potential is

$$\boxed{\varphi_{\text{self}}=\frac{I}{4\pi\sigma a}}.$$

2. Effect of the sea surface

Seawater occupies only the lower half-space, while the air above it has negligible conductivity compared with seawater.

Therefore, electric current cannot flow through the water-air boundary. The normal component of the current density at the surface must vanish:

$$j_n=0.$$

Since

$$\mathbf j=\sigma\mathbf E=-\sigma\nabla\varphi,$$

this boundary condition is equivalent to

$$\frac{\partial\varphi}{\partial n}=0.$$

This condition can be satisfied using the method of images.

For a point current source located at depth $h$, one introduces a mirror source of the same sign at the symmetric point a distance $h$ above the surface.

Thus, each real electrode has an image electrode of the same sign.

3. Potential of the first electrode

Let the first electrode act as a current source $+I$, while the second electrode acts as a current sink $-I$.

The potential of the first electrode is the sum of four contributions:

1. its own field;
2. the field of its own image;
3. the field of the second electrode;
4. the field of the image of the second electrode.

The self-potential contribution is

$$\varphi_{11}=\frac{I}{4\pi\sigma a}.$$

The distance from the first electrode to its own image is

$$2h,$$

so the corresponding contribution is

$$\varphi_{1\,\text{im1}}=\frac{I}{4\pi\sigma\,2h}.$$

The second electrode is at distance $L$ and has opposite current sign, therefore

$$\varphi_{12}=-\frac{I}{4\pi\sigma L}.$$

The distance from the first electrode to the image of the second electrode is

$$\sqrt{L^2+(2h)^2}.$$

Hence,

$$\varphi_{1\,\text{im2}}=-\frac{I}{4\pi\sigma\sqrt{L^2+4h^2}}.$$

Therefore, the total potential of the first electrode is

$$\varphi_1=\frac{I}{4\pi\sigma}\left(\frac1a+\frac1{2h}-\frac1L-\frac1{\sqrt{L^2+4h^2}}\right).$$

4. Potential of the second electrode

By symmetry, the second electrode has the same potential magnitude but opposite sign:

$$\varphi_2=-\varphi_1.$$

Therefore, the potential difference between the two electrodes is

$$U=\varphi_1-\varphi_2=2\varphi_1.$$

Thus,

$$U=\frac{I}{2\pi\sigma}\left(\frac1a+\frac1{2h}-\frac1L-\frac1{\sqrt{L^2+4h^2}}\right).$$

By definition,

$$R=\frac{U}{I}.$$

Hence,

$$\boxed{R=\frac{1}{2\pi\sigma}\left(\frac1a+\frac1{2h}-\frac1L-\frac1{\sqrt{L^2+4h^2}}\right)}.$$

This expression takes into account both the finite separation between the electrodes and the presence of the sea surface.

5. Numerical evaluation

Substitute

$$a=0.15\,\text{m},$$

$$h=60\,\text{m},$$

$$L=300\,\text{m},$$

$$\sigma=4\,\text{S/m}.$$

Then

$$R=\frac{1}{2\pi\cdot4}\left(\frac1{0.15}+\frac1{120}-\frac1{300}-\frac1{\sqrt{300^2+120^2}}\right).$$

The individual terms are

$$\frac1{0.15}\approx6.6667\,\text{m}^{-1},$$

$$\frac1{120}\approx0.00833\,\text{m}^{-1},$$

$$\frac1{300}\approx0.00333\,\text{m}^{-1},$$

$$\frac1{\sqrt{300^2+120^2}}\approx0.00309\,\text{m}^{-1}.$$

Therefore,

$$R\approx\frac{6.6667+0.00833-0.00333-0.00309}{8\pi}.$$

Hence,

$$\boxed{R\approx0.265\,\Omega}.$$

To the appropriate accuracy,

$$\boxed{R\approx0.27\,\Omega}.$$

6. Why the depth has only a very small effect

The dominant contribution to the resistance comes from the region close to the surface of each electrode.

Indeed,

$$\frac1a\approx6.67\,\text{m}^{-1},$$

whereas the corrections associated with the depth and the finite electrode separation are only of order

$$10^{-3}-10^{-2}\,\text{m}^{-1}.$$

Therefore, even the simpler estimate

$$R\approx\frac{1}{2\pi\sigma a}$$

gives

$$R\approx\frac{1}{2\pi\cdot4\cdot0.15}\approx0.265\,\Omega,$$

which is practically the same result.

The depth $60\,\text{m}$ is useful for checking that the electrodes are effectively embedded in a nearly infinite conducting medium and for estimating the correction caused by the water-air interface.

Note

The spreading resistance of one spherical electrode relative to a distant region of the conducting medium is

$$R_1=\frac{1}{4\pi\sigma a}.$$

For the present data,

$$R_1\approx0.133\,\Omega.$$

This is the value obtained if only one electrode is accounted for.

However, the problem asks for the resistance between two widely separated electrodes. To first approximation, the spreading resistances of the two electrodes add:

$$R\approx2R_1.$$

Therefore,

$$R\approx0.265\,\Omega.$$

Thus, a value of approximately $0.13$–$0.14\,\Omega$ corresponds to the spreading resistance of a single sphere, rather than to the total resistance of the seawater between the two electrodes.