The shape doesn’t matter?

Task number: 660

Two metal electrodes with unknown shape are placed into material with conductivity σ (with unknown shape too). Outer surroundings are not conductive.

(a) Show that resistance R between electrodes is determined by capacity C of that arrangement and it is true that

\[R = \frac{\varepsilon_0}{\sigma C}.\]

(b) You can suppose that electrodes are connected to power supply and charged to potential difference U0. Then disconnect the power supply. Show that charge from electrodes is being gradually drained off and for instantaneous voltage u(t) between electrodes is true following equation:

\[u(t) = U_0\,\mathrm{e}^{-t/\tau},\ \]

where

\[\tau = \frac{\varepsilon_0}{\sigma}.\ \]
  • Hint: Part (a)

    You should encircle one of electrodes with (imaginary) closed surface. Calculate total electric current which flows in/out the electrode as a surface integral of current density and use differential form of Ohm’s law.

    Then use Gauss’s law of electrostatics. You gain equation connecting electric current between electrodes with charge of the electrodes.

    At last become conscious of connection between charge of electrodes with capacity and voltage. You gain connection between electric current and voltage, from which you can easily calculate electric resistance of this arrangement.

  • Hint: Part (b)

    Start with equation between instantaneous charge q(t) of electrodes and instantaneous voltage between electrodes u(t)

    q(t) = Cu(t).

    Voltage is not zero and resistance of surroundings between electrodes is not infinite. This means that between electrodes there will flow electric current with instantaneous value i(t) and charge of electrodes and voltage between electrodes will be falling down.

    Formulate instantaneous voltage using instantaneous current and current as a derivation of charge by time. You gain differential equation of charge of electrodes in dependence on time. Calculate this equation. Replace this solution with equation u(t) = q(t)/C and find time dependence of voltage.

  • Analysis

    Capacity of system is constant of proportionality between charge of electrodes and voltage between electrodes. Resistance of system is constant of proportionality between voltage between electrodes and current flowing in the circuit.

    So we need to find relation between charge of electrodes and current in the circuit. For that we can use intensity of electric field \(\vec E\). Charge of electrodes is source of electric field with intensity \(\vec E\) and using differential form of Ohm’s law we know relation between this field and current density. If the voltage is constant, the charge of electrodes is constant too and there flows constant current between electrodes.

    In this situation we can use electrostatics´s laws, because it is stationary case. We need to find total flowing current and we needn’t study its distribution. For its calculation we can integrate current density over any closed surface S which surrounds one of electrodes and if the second electrode is outside the surface. We gain

    \[I = \oint_S \vec j\cdot d\vec S = \sigma\oint_S \vec E\cdot d\vec S.\]

    Thanks to the fact that integral on the right relates with charge of electrodes by Gauss’s law of electrostatics, the needed relation is found.

     

    On condition that changes aren’t so fast we can use results from part (a) to solve part (b).

  • Solution: Part (a)

    We use differential form of Ohm’s law and Gauss’s law of electrostatics (which we can use for stationary current too). It’s true that total current I between electrodes is:

    \[I = \oint_S \vec j\cdot d\vec S,\]

    Where the integral is over any closed surface S which encircles positive electrode (which means electrode with higher potential) and using differential form of Ohm’s law \(\vec j = \sigma\vec E\) we edit right side of the equation to next:

    \[I = \oint_S \vec j\cdot d\vec S = \sigma\oint_S \vec E\cdot d\vec S.\]

    Than we use Gauss’s law of electrostatics:

    \[I = \oint_S \vec j\cdot d\vec S = \sigma\oint_S \vec E\cdot d\vec S = \sigma\frac{Q}{\varepsilon_0},\]

    where Q is total charge of the electrode. Because it is true for any system of electrodes that Q = CU, where C is capacity of the system and U is voltage between electrodes, we gain

    \[I = \sigma\frac{CU}{\varepsilon_0},\]

    and after some editing we gain the missing equation

    \[R = \frac{U}{I} = \frac{\varepsilon_0}{\sigma C}.\]
  • Solution: Part (b)

    When we have potential difference between electrodes (voltage) u = u(t), than according to previous results there flows a current between them: i = i(t) = u(t)/R, impact of is decreasing charge of electrodes. It’s true that

    \[q(t) = C\,u(t) = CR\,i(t) = -CR\frac{dq(t)}{dt},\]

    where the “minus” marks that if the charge of electrode decreases than there is a current between electrodes. From that we gain equation:

    \[\frac{dq(t)}{dt}+\frac{q(t)}{RC} = 0,\]

    which is linear differential equation with constant coefficient. We can write its solution in the form

    \[q(t) = A\, \mathrm{e}^{\alpha t},\]

    where α is a root of characteristic equation

    \[\alpha + \frac{1}{RC} = 0,\]

    From that we immediately have

    \[\alpha = -\frac{1}{RC}.\]

    We set the constant A from starting conditions: in the time t = 0 s is the charge of electrodes q(0) = Q0, thus

    \[Q_0 = q(0) = Ae^{0} = A.\]

    So, we gain equation for charge of electrodes:

    \[q(t) = Q_0\, \mathrm{e}^{-t/{RC}}.\]

    Using the formula for resistance R from part (a) we gain

    \[q(t) = Q_0\, \mathrm{e}^{-t/\tau}, \qquad \textrm{where} \qquad \tau = \frac{\varepsilon_0}{\sigma}.\]

    Constant Q0 marks starting charge of electrodes, for which it is true that: Q0 = CU0, where U0 is voltage of the power supply. So we gain equation for instantaneous voltage

    \[u(t) = \frac{q(t)}{C} = \frac{Q_0}{C}\,\mathrm{e}^{-t/\tau} = U_0\ \mathrm{e}^{-t/\tau}, \]

    Which was the goal to show.

Difficulty level: Level 4 – Undergraduate level
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