Measuring of the electrical conductivity of sea water

Task number: 1040

Two concentric spherical metal conductors have radiuses a and b. They are separated by a slightly conductive material with electrical conductivity σ.

a) Determine the electrical resistance between the conductors. What is the value of the electric current that flows between them, if the potential difference between the conductors has a constant value U?

b) Use the previous results and determine the electric current flows between two spherical electrodes, each with a radius a. Both electrodes are dipped deep in the sea and they are located in a considerable distance. The voltage between the electrodes has a constant value U.

Note: The part a) of this task can be solved by three different methods:

I. By straight integration of individual parts of the conductor,
II. by using the Ohm’s law written in a differential form,
III. by formal analogy with electrostatics laws.

  • I. Hints

  • I. Analysis

    The space between conductors can be divided into thin spherical shells with radius r (a < r < b).

    The space between the electrodes divided into spherical surfaces

    The whole conductor can be now understood as a series connection of these shells. The total electrical resistance is equal to a summ (integral) of resistances of each individual shell.

    TThe electrical resistance of a spherical shell is directly related to its “length” and inversely related to the area of the shell and its electrical conductivity.

    The electric current flowing between the electrodes is equal to the ratio of voltage and electrical resistance between these electrodes.

  • I. Solution

    The space between the electrodes can be divided into thin shells with the radiuses r (a < r < b) and the “thickness” dr.

    The space between the electrodes divided into spherical surfaces

    The electrical resistance of the spherical shell can be calculated from the equation:

    \[R\,=\,\frac{1}{\sigma}\frac{l}{S}\,,\]

    where σ is electrical conductivity of the material, l is the length of the conductor and S is the area of the conductor cross-section.

    In this case, the “area” of the spherical shell is equal to the area of the spherical surface, i.e. 4πr2. Let’s mark the “length” of the shell dr. The electrical resistance dR of the shell with the radius r can be expressed by the formula:

    \[\mathrm{d}R\,=\,\frac{1}{\sigma}\frac{\mathrm{d}r}{4\pi r^2}\,.\]

    The individual shells are connected in series. The total electrical resistance between the spherical electrodes is equal to the sum (integral) of the electrical resistance dR of all shells. Limits of the integration are the radiuses a and b.

    \[ R\,=\,\int_a^b\,\frac{1}{\sigma}\,\frac{\mathrm{d}r}{4\pi r^2}\,=\, \frac{1}{4\pi \sigma}\int_a^b\frac{\mathrm{d}r}{r^2}\,=\,\frac{1}{4\pi \sigma}\,\left[-\frac{1}{r}\right]_a^b \,. \]

    The total electrical resistance between the spherical electrodes can be calculate as:

    \[R\,=\,\frac{1}{4\pi \sigma}\left(\frac{1}{a}\,-\,\frac{1}{b}\right)\] \[R\,=\, \frac{1}{4\pi \sigma}\,\frac{b\,-\,a}{ab}\] \[R\,=\,\frac{b\,-\,a}{4\pi \sigma ab}\,.\]

    The electric current flowing between the electrodes is equal to the ratio of the voltage U and the electrical resistance R between the electrodes:

    \[I\,=\,\frac{U}{R}\,=\,\frac{4\pi \sigma ab}{ b\,-\,a}\,.\]
  • II. Hints

  • II. Analysis

    The space between spherical conductors can be divided into thin spherical shells with the radiuses r, where a < r < b

    Space between electrodes divided into spherical peels

    It follows from the spherical symmetry that the electric density is equivalent in all points of such a formed shell. The electric density can be expressed using the total current and the surface of the shell. The total current between electrodes is constant and independent of the radius r.

    The Ohm’s law in the differential form for homogenous conductors says that the density of current flowing through the conductor is equal to the product of electrical conductivity and the electric field of the conductor. We can express the electric field from this law

    We gain the electrical voltage between the electrodes by the integration of the electric field. The total electrical resistance of the material can be determined as a ratio of the electrical voltage and the electric current flowing between the electrodes.

  • II. Solution

    The space between spherical conductors can be divided into thin spherical shells with the radiuses r, where a < r < b. From the spherical symmetry it follows that the current density j is equivalent in all points of such a shell. The total current flowing through the shell can be calculated as:

    \[ I=S\,j\left(r\right)=4\pi r^2\,j\left(r\right)\,. \]

    The total electric current is constant and therefore it is independent of the radius r.

    The differential form of Ohm’s law for a homogenous conductor says:

    \[ \vec{j\,}=\,\sigma\vec{E}, \]

    where σ means an electrical conductivity and \(\vec{E}\) is an electric field. Now we can determine the size of the electric field in the spherical shell with the radius r:

    \[ E\left(r\right)\,=\,\frac{j\left(r\right)}{\sigma} \]

    Just as the current density j(r) so the magnitude of the electric field E(r) depends on the radius of the shell r. The current density j(r) in a spherical conductor with the radius r can be calculated from the equation for the total current.

    \[ E\left(r\right)\,=\,\frac{I}{4\pi r^2\sigma} \]

    The electrical voltage U between the electrodes can be determined by integration of the electric field:

    \[ U\,=\,\int_a^b \vec{E}\left(r\right)\cdot \mathrm{d}\vec{r\,} \]

    Thanks to the spherical symmetry, the electric field will point in the radial direction at each point of the conductor. Therefore, the vectors \(\vec{E}\) and d\(\vec{r}\) are parallel and their scalar product is equal to the product of their amplitudes:

    \[ U\,=\,\int_a^b E\left(r\right)\,\mathrm{d}r\,=\,\frac{I}{4\pi \sigma}\int_a^b \frac{\mathrm{d}r}{r^2}\,=\,\frac{I}{4\pi \sigma}\,\left[\frac{-1}{r}\right]_a^b\,=\,\frac{I}{4\pi \sigma}\,\left(\frac{1}{a}\,-\,\frac{1}{b}\right) \]

    The electrical resistance can be calculated as a ratio of the electrical voltage and the electric current flowing between the electrodes:

    \[ R\,=\,\frac{U}{I}\,=\,\frac{\frac{I}{4\pi \sigma}\,\left(\frac{1}{a}\,-\,\frac{1}{b}\right)}{I} \]

    The total electrical resistance between the two electrodes can be determined from the equation:

    \[ R\,=\,\frac{1}{4\pi\sigma}\left(\frac{1}{a}\,-\,\frac{1}{b}\right) \] \[ R\,=\,\frac{1}{4\pi\sigma}\,\frac{b-a}{ab} \] \[ R\,=\,\frac{b-a}{4\pi\sigma ab} \]

    The electric current flowing between the electrodes is equal to the ratio of the electrical voltage and the electrical resistance between the electrodes:

    \[ I\,=\,\frac{U}{R}\,=\,\frac{4\pi\sigma ab}{b-a}\,U \]
  • III. Hints

  • III. Analysis

    The electrical resistance and the electric current flowing between two spherical electrodes can be calculated using the formal analogy with the electrostatics field.

    Consider similar electrodes submerged into a dielectric with a definite electric permittivity. The electric charges of the same size but opposite signs are brought to the electrodes. These electrodes keep the original values of potentials. Therefore, the equal course of the electric field will be between the electrodes. The Gauss’s law in the integral form can be used for expressing the electric field.

    The Ohm’s law for a homogenous conductor in differential form can be substituted for the equation of the electric field. This way, we gain the electric current flowing between the electrodes depending on the size of the electric charge of the spherical electrodes. Therefore, the electric charge can be expressed using the capacity of a spherical capacitor.

    According to the Ohm’s law, the electrical resistance between the spherical electrodes is equal to the ratio of voltage between the electrodes and the electric current flowing between them.

  • III. Solution

    Let’s imagine that a dielectric with a permittivity ε is put between the electrodes, and charges with the same quantity Q but opposite signs are brought to the electrodes. The charges keep original value of the electric potentials. To solve the task, we can use the fact that the course of the electric field is the same as in this stationary case, when a steady electric current flows between the electrodes.

    At first we write two equations describing the spherical capacitor that we will use later. The interrelation between the size of a charge Q, a potential difference (i.e. voltage U), and a capacity C can be expressed by an equation:

    \[ Q\,=\,UC \]

    The capacity of the spherical capacitor is:

    \[ C\,=\,\frac{4\pi\epsilon}{\frac{1}{a}\,-\,\frac{1}{b}}\,=\,\frac{4\pi\epsilon ab}{b-a}\,, \tag{*}\]

    where ε means a permittivity of the environment, and a, b (a < b) are radiuses of the spherical electrodes creating the capacitor.

    Let’s assail the very solution of the task. We apply the integral form of the Gauss’s law:

    \[ \oint \vec{E}\cdot\mathrm{d}\vec{S}\,=\,\frac{Q}{\epsilon}\,, \]

    where \(\vec{E}\) means an electric field , Q is a charge brought to the spherical electrodes with a surface S, and ε means a permittivity of the stated material.

    The formula for the Ohm’s law for a homogeneous conductor in the differential form is:

    \[ \vec{j}\,=\,\sigma \vec{E} \]

    where \(\vec{j}\) is a current density, and σ is a material dependent parameter called conductivity.

    The total electric current I flowing through the electrodes can be expressed by using the current density  \(\vec{j}\):

    \[ I\,=\,\oint \vec{j\,}\cdot\mathrm{d}\vec{S}, \]

    where the current density \(\vec{j\,}\) can be substituted from the Ohm’s law. Further, we use the Gauss’s law for the modification of the equation:

    \[ I\,=\,\oint \sigma \vec{E}\cdot\mathrm{d}\vec{S}\,=\,\sigma\oint\vec{E}\cdot\mathrm{d}\vec{S}\,=\,\sigma\frac{Q}{\epsilon} \]

    The electric charge Q placed on one electrode can be expressed by using the capacity: \(Q\,=\,CU\); i.e.:

    \[ I\,=\,\sigma\frac{CU}{\epsilon} \]

    We gain an equation for computation of the total electric current flowing through the electrodes thanks to the substitution of the capacity of the spherical capacitor (*) for C.

    \[ I\,=\,\sigma\frac{\frac{4\pi\epsilon ab U}{b-a}}{\epsilon} \] \[ I\,=\,\frac{4\pi\sigma ab}{b-a}\,U \]

    The electrical resistance between the spherical electrodes can be calculated as a ratio of the voltage U between the electrodes and the electric current I flowing between the electrodes:

    \[ R\,=\,\frac{U}{I}\,=\,\frac{U}{\frac{4\pi\sigma ab}{b-a}\,U} \] \[ R\,=\,\frac{b-a}{4\pi\sigma ab} \]
  • b) Hints

  • b) Analysis

    Let’s suppose we have two concentric spherical metal conductors with radiuses a and b (a << b). Thanks to this condition, we gain an equation for the electrical resistance of the surroundings of the spherical electrode with the radius a from the previous results.

    Therefore, the electric current flowing between two spherical electrodes that are submerged deep in the sea and are located in a considerable distance will be equal to the electric current flowing through two resistors connected in a series with resistances corresponding to the resistances of the surroundings of both electrodes.

  • b) Solution

    We have calculated the electrical resistance between the two concentric spherical electrodes with the radiuses a and b (a < b) in the previous parts of the solution:

    \[ R\,=\,\frac{b-a}{4\pi\sigma ab}\,=\,\frac{1}{4\pi\sigma}\left(\frac{1}{a}\,-\,\frac{1}{b}\right), \tag{*}\]

    where σ is the electrical conductivity of the material, i.e. of the sea water in this case.

    If b >> a, then \[\frac{1}{b}\,<<\,\frac{1}{a}\] . Which means that the term \[\frac{1}{b}\] in the equation (*) can be neglected. Therefore, the equation for the electric resistivity of the environment is:

    \[ R\,\approx\,\frac{1}{4\pi\sigma a} \]

    The resistance of the environment is concentrated mostly near the surface of the sphere. Therefore, this equation can be considered in first approximation to give the resistance of the surroundings of the spherical electrode with the radius a.

    The electric current flowing between two spherical electrodes that are submerged deep in the sea and are located in a considerable distance will be equal to the electric current flowing through two resistors connected in a series with resistances corresponding to the resistances of the surroundings of stated electrodes. The total resistance of this series connection will be equal to the sum of these resistances:

    \[ R_{total}\,=\,2R\,\approx\,2\,\frac{1}{4\pi\sigma a}\,=\,\frac{1}{2\pi\sigma a} \]

    The electric current flowing between the spherical electrodes can be calculated as a ratio of the voltage U between the electrodes and the total resistance of the electrodes Rtotal; i.e.:

    \[ I\,\approx\,\frac{U}{R_{total}}\,=\,\frac{U}{\frac{1}{2\pi\sigma a}} \] \[ I\,\approx\,2\pi\sigma aU \]
  • Answer

    a) The electrical resistance between the two concentric spherical electrodes with the radiuses a and b can be calculated from the equation:

    \[ R\,=\,\frac{b-a}{4\pi\sigma ab}. \]

    The electric current flowing between the electrodes is equal to:

    \[ I\,=\,\frac{4\pi\sigma ab}{b-a}\,U, \]

    where σ is the electrical conductivity of the material and U means the voltage between the electrodes.

    Note: All the three methods led to the same result.

    b) The electric current flowing between two spherical electrodes with the radius a that are dipped deep in the sea is:

    \[ I\,=\,2\pi\sigma aU. \]
Difficulty level: Level 4 – Undergraduate level
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