Правка разделов «Solution», «1. Potential of a point current source i», «2. Effect of the sea surface»

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### Solution
−![For problem $8.2.18$ |2170x3140, 31%](../../img/8.2.18/scan-0.png)
+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 <b>same sign</b> 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 <b>one</b> 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 <b>between two</b> 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.
+
#### Answer
−R = 0.13 Ω
+$$
+\boxed{R\approx0.27\,\Omega}
+$$