Rudolf Ortvay 2015 Problem 13

Physicsguy51 · Messaggio da **Physicsguy51** » 26 ago 2023, 5:36

Hello!
I am new to this forum. I am not from Italy. I was introduced to this forum by a friend of mine who is preparing for IPhO 2024. I was recently trying to solve this problem from Rudolf Ortvay 2015 but I could not understand it. Could anyone please help me solve this problem and understand the techniques involved in this?
Problem: A 2 × 2 × 2 Rubik's cube consists of eight solid, uniform, metallic cubes of side-length $a$ . Four cubes out of the eight have conductivity $2\sigma$ , while the conductivity of the other four cubes is $\sigma$ . The cubes with different conductivities have no common faces (see the figure). Two metallic plates of very good conductivity are attached to the two opposite sides of the Rubik's cube (not shown). Find the total current flowing between the two plates if a battery of voltage $U_0$ is connected to them.
Figure: https://ibb.co/G2Km8YY

Tarapìa Tapioco · Messaggio da **Tarapìa Tapioco** » 27 ago 2023, 1:08

Physicsguy51 ha scritto: ↑26 ago 2023, 5:36 Hello!
I am new to this forum. I am not from Italy. I was introduced to this forum by a friend of mine who is preparing for IPhO 2024. I was recently trying to solve this problem from Rudolf Ortvay 2015 but I could not understand it. Could anyone please help me solve this problem and understand the techniques involved in this?

Hi, Physicsguy51. As a matter of fact, this is an interesting and quite difficult problem, but it will be my pleasure to help you.

Physicsguy51 ha scritto: ↑26 ago 2023, 5:36 Problem: A 2 × 2 × 2 Rubik's cube consists of eight solid, uniform, metallic cubes of side-length $a$ . Four cubes out of the eight have conductivity $2\sigma$ , while the conductivity of the other four cubes is $\sigma$ . The cubes with different conductivities have no common faces (see the figure). Two metallic plates of very good conductivity are attached to the two opposite sides of the Rubik's cube (not shown). Find the total current flowing between the two plates if a battery of voltage $U_0$ is connected to them.
Figure: https://ibb.co/G2Km8YY

First, analyze the physics model. Since four out of eight cubes have conductivity $2\sigma$ , while the conductivity of the other four is $\sigma$ , assume - from the figure under consideration - that the conductivity of the light squares is $\sigma_1 = \sigma$ , while that of the dark squares is $\sigma_2 = 2 \sigma$ . Assuming that a steady voltage $U_0$ is applied to the terminals and neglecting any interface resistances between the squares, it can be seen that the current distribution and the electric field, respectively, can be characterised by current-streamlines and equipotential lines, which for all practical purposes lie in one plane: these form arrays of mutually orthogonal curves; rotating the cube through $90^\circ$ reverses the two arrays of curves, but its equivalent resistance remains unaltered. It can be seen that the common thickness $d$ of the metal cubes is not - as is the case, on the other hand, in a chessboard composed of plates - very much less (thus, almost negligible) than the length $L$ of the panel: since the cubes form a $2 \times 2 \times 2$ Rubik's cube, the thickness $d$ of the cubes is equal to the length $L$ of the face of the larger cube, which, being the sum of the length $a$ of the faces of the two cubes from which it is composed, is equal to $2a$ . Thus: $d = L = 2a$ .

If all the squares on the cube were made from a homogeneous metal plate with conductivity $\sigma_1$ , then the magnitudes of the electric field strength and current density would be $E = \frac{U_0}{L}$ and $j_1 = \sigma_1 E$ , respectively, and the total current flowing $I_1$ through the board would be:

$I_1 = j_1S$ , with $S = Ld$ . Therefore: $I_1 = j_1 Ld$

Substituting $j_1$ and $E$ into $I_1$ , we obtain:

$I_1 = j_1 Ld$ $\Rightarrow$ $I_1 =\sigma_1 ELd$ $= \frac{U_0}{L}\sigma_1 Ld$ $\Rightarrow$ $I_1 = U_0\sigma_1d$ .

Similarly, in another homogeneous metal plate with conductivity $\sigma_2$ , under the same voltage $U_0$ , the total current $I_2$ would be:

$I_2 = j_2S$ , with $S = Ld$ . Therefore: $I_2 = j_2 Ld$ .

Substituting $j_2 = \sigma_2 E$ and $E = \frac{U_0}{L}$ into $I_2$ , we obtain:

$I_2 = j_2 Ld$ $\Rightarrow$ $I_2 =\sigma_2 ELd$ $= \frac{U_0}{L}\sigma_2 Ld$ $\Rightarrow$ $I_2 = U_0\sigma_2d$ .
We note in passing that the current does not depend on the length $L$ of the board.

Now, consider first the electric field formed in the board, and the currents that flow through the squares of the plate. A conservative electric field can be fully characterised by the electric potential, denoted by $\Phi$ , and represented graphically by equipotentials, drawn with dashed lines in the figure below. The "bottom" edge of the square plate is chosen as zero potential, and so the potential of the "top" of it is equal to the voltage $U_0$ of the battery.

Immagine

The electric field, and consequently the current density vector, which is proportional to it, are both perpendicular to the equipotential curves. It follows that the current-streamlines, drawn in the figure with continuous lines, are also perpendicular to the same lines. Let us associate with each current-streamline a number $\Psi$ that is equal to the total current that flows between it and the left-hand edge of the plate.
It is obvious that the left-hand edge of the plate is a streamline, and corresponds to $\Psi = 0$ , and that the right-hand edge is $\Psi = I$ . From here on, let us call $\Phi$ the voltage-potential and $\Psi$ the current-potential. If, in the vicinity of an arbitrary point $A$ of the plate, a distance $\Delta \ell$ between two equipotential lines corresponds to a small potential difference $\Delta \Phi$ between them, then the magnitude of the electric field at $A$ is

$E_A = \frac{\Delta \Phi}{\Delta \ell}$ ,

and the local electric current density is

$j_A = \sigma_A E_A$ $\Rightarrow$ $j_A = \sigma_A\frac{\Delta \Phi}{\Delta \ell}$ .

Here, $\sigma_A$ denotes the conductivity around point $A$ , which could be $\sigma_1$ or $\sigma_2$ , depending on the "colour" of the particular square in which $A$ is located. Now, from the definition of the current-potential, the current element $j_A d \Delta s$ that flows through a cross-sectional area $d \Delta s$ of the plate must be equal to the change $\Delta \Psi$ in the current-potential over the small distance $\Delta s$ :

$j_A d\Delta s = \sigma_A\frac{\Delta \Phi}{\Delta \ell} d \Delta s$ $= \Delta \Psi \Rightarrow \sigma_A\frac{\Delta \Phi}{\Delta \ell} d \Delta s = \Delta \Psi$ ,

from which we get

$d\sigma_A\frac{\Delta \Phi}{\Delta \ell}$ $= \frac{\Delta \Psi}{\Delta s}$ . $\space$ $\space$ $(1)$

Next, consider rotating the first figure anticlockwise by $90^\circ$ in the plane of the plate, and obtaining the arrangement shown in the figure below.

Immagine

The question now arises whether this arrangement, after interchanging the roles of the voltage- and current-potentials, and multiplying them by appropriately chosen factors, is a suitable description of the currents flowing through the original plate.
Let the renaming and scaling be:

$\Psi '= \frac{I}{U_0}\Phi$ $\space$ $\space$ $\space$ and $\space$ $\space$ $\space$ $\Phi '= \frac{U_0}{I}\Psi$

and denote the separations of the neighbouring pairs of curves by

$\Delta \ell' = \Delta s$ $\space$ $\space$ $\space$ and $\space$ $\space$ $\space$ $\Delta s' = \Delta \ell$ .

The scaling factors are chosen so that the maximal value for $\Phi '$ (the new voltage-potential) is $U_0$ , and that for $\Psi '$ (the new current-potential) is $I$ . If these new (primed) variables are used to label the original (unrotated) plate, then we get the arrangement shown in the figure below.

Immagine

Point $B$ , shown in the figure above, is the place to which point $A$ has been moved by the rotation. As noted earlier, $B$ is bound to be on a different coloured square than $A$ was, and the product of the corresponding conductivities is:

$\sigma_A \sigma_B = \sigma_1 \sigma_2$ . $\space$ $\space$ $(2)$

This always has the same value, wherever $A$ , and hence $B$ , are situated on the cube.

Ohm’s law will be obeyed (as it needs to be) by the new voltage- and current-potentials, if an equation corresponding to $(1)$ is satisfied, that is, if

$d\sigma_B\frac{\Delta \Phi'}{\Delta \ell'}$ $= \frac{\Delta \Psi'}{\Delta s'}$ .

Using the connections between the primed and unprimed quantities, this requirement can be written in the form:

$d\sigma_B\frac{U_0}{I}\frac{\Delta \Psi}{\Delta s}$ $= \frac{I}{U_0}\frac{\Delta \Phi}{\Delta\ell}$ .

Substituting the equation $(1)$ into the previous:

$d\sigma_B\frac{U_0}{I}d\sigma_A\frac{\Delta \Phi}{\Delta \ell} = \frac{I}{U_0}\frac{\Delta \Phi}{\Delta\ell}$ ,

from which we obtain:

$d\sigma_B\frac{U_0}{I}d\sigma_A\frac{\Delta \Phi}{\Delta \ell} = \frac{I}{U_0}\frac{\Delta \Phi}{\Delta\ell} \Rightarrow d^2\sigma_A\sigma_B\frac{U_0}{I}$ $= \frac{I}{U_0}$ $\Rightarrow$

$I^2 = U_0^2d^2\sigma_A\sigma_B$ $\Rightarrow$ $I = U_0d\sqrt{\sigma_A\sigma_B}$ .

Using $(2)$ , this can be put in the form:

$I = U_0d\sqrt{\sigma_A\sigma_B}$ $\Rightarrow$ $\boxed{I = U_0d\sqrt{\sigma_1\sigma_2}}$ .

Substituting the specific values of $d = L = 2a$ , $\sigma_1 = \sigma$ and $\sigma_2 = 2\sigma$ initially found, we have:

$I = U_0 \cdot 2a \sqrt{\sigma \cdot 2 \sigma}$ $\Rightarrow$ $I = 2\sqrt{2}\cdot U_0a \sqrt{\sigma^2}$ $\Rightarrow$ $\color{Red} \boxed{\color{Black} I = 2 \sqrt{2}\cdot U_0 \cdot a \cdot \sigma}$ .

We have just proved that the current flowing through the inhomogeneous cube is the geometric mean of $I_1$ and $I_2$ . This statement (about the geometric mean giving the required current) is valid for any set of square plates of constant thickness that satisfy the requirement according to which for a pair of points, $A$ and $B$ , which can be transformed into each other by rotating the square through $90^\circ$ , the product of the conductivities at those two points must have a value that is independent of how the point pair is chosen.
It is clear that, in our case, this condition will be satisfied, because a rotation by $90^\circ$ will always transform a point on a light square into one on a dark square, and viceversa; i.e. points $A$ and $B$ will always be on different coloured squares and $\sigma_A\sigma_B$ will always have the value $\sigma_1\sigma_2$ .

Notes. $1)$ All of the arguments used in the solution are valid not only for the $2 \times 2$ board: more generally, it can be stated that the final result is true for all boards with $n \times n$ squares if $n$ is even, but not true if $n$ is an odd number. In the latter case, a rotation by $90^\circ$ does not change the "colour" of the square on which any particular point is located, and further, since the numbers of light and dark squares are different, the result cannot be symmetric in $\sigma_1$ and $\sigma_2$ .
$2)$ Using numerical methods, the current-streamlines and the equipotential lines for any particular cube can be found and plotted. Such a computer-generated map, for a $2 \times 2$ "cube", is shown in figure below; in it, the conductivity $\sigma_2$ of the dark squares is twice that, $\sigma_1$ , of the light ones. Because of the boundary conditions, both the current-streamlines and the equipotential lines change direction at the interfaces – in much the same way that light is refracted at the boundary between two different media.

Immagine

Physicsguy51 · Messaggio da **Physicsguy51** » 27 ago 2023, 2:54

Tarapìa Tapioco ha scritto: ↑27 ago 2023, 1:08 Hi, Physicsguy51. As a matter of fact, this is an interesting and quite difficult problem, but it will be my pleasure to help you.

First, analyze the physics model. Since four out of eight cubes have conductivity $2\sigma$ , while the conductivity of the other four is $\sigma$ , assume - from the figure under consideration - that the conductivity of the light squares is $\sigma_1 = \sigma$ , while that of the dark squares is $\sigma_2 = 2 \sigma$ . Assuming that a steady voltage $U_0$ is applied to the terminals and neglecting any interface resistances between the squares, it can be seen that the current distribution and the electric field, respectively, can be characterised by current-streamlines and equipotential lines, which for all practical purposes lie in one plane: these form arrays of mutually orthogonal curves; rotating the cube through 90° reverses the two arrays of curves, but its equivalent resistance remains unaltered. It can be seen that the common thickness $d$ of the metal cubes is not - as is the case, on the other hand, in a chessboard composed of plates - very much less (thus, almost negligible) than the length $L$ of the panel: since the cubes form a $2 \times 2 \times 2$ Rubik's cube, the thickness $d$ of the cubes is equal to the length $L$ of the face of the larger cube, which, being the sum of the length $a$ of the faces of the two cubes from which it is composed, is equal to $2a$ . Thus: $d = L = 2a$ .

If all the squares on the cube were made from a homogeneous metal plate with conductivity $\sigma_1$ , then the magnitudes of the electric field strength and current density would be $E = \frac{U_0}{L}$ and $j_1 = \sigma_1 E$ , respectively, and the total current flowing $I_1$ through the board would be:

$I_1 = j_1S$ , with $S = Ld$ . Therefore: $I_1 = j_1 Ld$

Substituting $j_1$ and $E$ into $I_1$ , we obtain:

$I_1 = j_1 Ld$ $\Rightarrow$ $I_1 =\sigma_1 ELd$ $= \frac{U_0}{L}\sigma_1 Ld$ $\Rightarrow$ $I_1 = U_0\sigma_1d$

Similarly, in another homogeneous metal plate with conductivity $\sigma_2$ , under the same voltage $U_0$ , the total current $I_2$ would be:

$I_2 = j_2S$ , with $S = Ld$ . Therefore: $I_2 = j_2 Ld$

Substituting $j_2 = \sigma_2 E$ and $E = \frac{U_0}{L}$ into $I_2$ , we obtain:

$I_2 = j_2 Ld$ $\Rightarrow$ $I_2 =\sigma_2 ELd$ $= \frac{U_0}{L}\sigma_2 Ld$ $\Rightarrow$ $I_2 = U_0\sigma_2d$
We note in passing that the current does not depend on the length $L$ of the board.

Now, consider first the electric field formed in the board, and the currents that flow through the squares of the plate. A conservative electric field can be fully characterised by the electric potential, denoted by $\Phi$ , and represented graphically by equipotentials, drawn with dashed lines in the figure below. The "bottom" edge of the square plate is chosen as zero potential, and so the potential of the "top" of it is equal to the voltage $U_0$ of the battery.

The electric field, and consequently the current density vector, which is proportional to it, are both perpendicular to the equipotential curves. It follows that the current-streamlines, drawn in the figure with continuous lines, are also perpendicular to the same lines. Let us associate with each current-streamline a number $\Psi$ that is equal to the total current that flows between it and the left-hand edge of the plate.
It is obvious that the left-hand edge of the plate is a streamline, and corresponds to $\Psi = 0$ , and that the right-hand edge is $\Psi = I$ . From here on, let us call $\Phi$ the voltage-potential and $\Psi$ the current-potential. If, in the vicinity of an arbitrary point $A$ of the plate, a distance $\Delta \ell$ between two equipotential lines corresponds to a small potential difference $\Delta \Phi$ between them, then the magnitude of the electric field at $A$ is

$E_A = \frac{\Delta \Phi}{\Delta \ell}$ ,

and the local electric current density is

$j_A = \sigma_A E_A$ $\Rightarrow$ $j_A = \sigma_A\frac{\Delta \Phi}{\Delta \ell}$

Here, $\sigma_A$ denotes the conductivity around point $A$ , which could be $\sigma_1$ or $\sigma_2$ , depending on the "colour" of the particular square in which $A$ is located. Now, from the definition of the current-potential, the current element $j_A d \Delta s$ that flows through a cross-sectional area $d \Delta s$ of the plate must be equal to the change $\Delta \Psi$ in the current-potential over the small distance $\Delta s$ :

$j_A d\Delta s = \sigma_A\frac{\Delta \Phi}{\Delta \ell} d \Delta s$ $= \Delta \Psi \Rightarrow \sigma_A\frac{\Delta \Phi}{\Delta \ell} d \Delta s = \Delta \Psi$ ,

from which we get

$d\sigma_A\frac{\Delta \Phi}{\Delta \ell}$ $= \frac{\Delta \Psi}{\Delta s}$ $\space$ $\space$ $(1)$

Next, consider rotating the first figure anticlockwise by 90°in the plane of the plate, and obtaining the arrangement shown in the figure below.

The question now arises whether this arrangement, after interchanging the roles of the voltage- and current-potentials, and multiplying them by appropriately chosen factors, is a suitable description of the currents flowing through the original plate.
Let the renaming and scaling be:

$\Psi '= \frac{I}{U_0}\Phi$ $\space$ $\space$ $\space$ and $\space$ $\space$ $\space$ $\Phi '= \frac{U_0}{I}\Psi$

and denote the separations of the neighbouring pairs of curves by

$\Delta \ell' = \Delta s$ $\space$ $\space$ $\space$ and $\space$ $\space$ $\space$ $\Delta s' = \Delta \ell$ .

The scaling factors are chosen so that the maximal value for $\Phi '$ (the new voltage-potential) is $U_0$ , and that for $\Psi '$ (the new current-potential) is $I$ . If these new (primed) variables are used to label the original (unrotated) plate, then we get the arrangement shown in the figure below.

Point $B$ , shown in the figure above, is the place to which point $A$ has been moved by the rotation. As noted earlier, $B$ is bound to be on a different coloured square than $A$ was, and the product of the corresponding conductivities is:

$\sigma_A \sigma_B = \sigma_1 \sigma_2$ $\space$ $\space$ $(2)$

This always has the same value, wherever $A$ , and hence $B$ , are situated on the cube.

Ohm’s law will be obeyed (as it needs to be) by the new voltage- and current-potentials, if an equation corresponding to $(1)$ is satisfied, that is, if

$d\sigma_B\frac{\Delta \Phi'}{\Delta \ell'}$ $= \frac{\Delta \Psi'}{\Delta s'}$

Using the connections between the primed and unprimed quantities, this requirement can be written in the form:

$d\sigma_B\frac{U_0}{I}\frac{\Delta \Psi}{\Delta s}$ $= \frac{I}{U_0}\frac{\Delta \Phi}{\Delta\ell}$ .

Substituting the equation $(1)$ into the previous:

$d\sigma_B\frac{U_0}{I}d\sigma_A\frac{\Delta \Phi}{\Delta \ell} = \frac{I}{U_0}\frac{\Delta \Phi}{\Delta\ell}$ ,

from which we obtain:

$d\sigma_B\frac{U_0}{I}d\sigma_A\frac{\Delta \Phi}{\Delta \ell} = \frac{I}{U_0}\frac{\Delta \Phi}{\Delta\ell} d\sigma_B\frac{U_0}{I}d\sigma_A\frac{\Delta \Phi}{\Delta \ell}$ $= \frac{I}{U_0}\frac{\Delta \Phi}{\Delta\ell} \Rightarrow d^2\sigma_A\sigma_B\frac{U_0}{I}$ $= \frac{I}{U_0}$ $\Rightarrow$

$I^2 = U_0^2d^2\sigma_A\sigma_B$ $\Rightarrow$ $I = U_0d\sqrt{\sigma_A\sigma_B}$

Using $(2)$ , this can be put in the form:

$I = U_0d\sqrt{\sigma_A\sigma_B}$ $\Rightarrow$ $I = U_0d\sqrt{\sigma_1\sigma_2}$

Substituting the specific values of $d = L = 2a$ , $\sigma_1 = \sigma$ and $\sigma_2 = 2\sigma$ initially found, we have:

$I = U_0 \cdot 2a \sqrt{\sigma \cdot 2 \sigma}$ $\Rightarrow$ $I = 2\sqrt{2}\cdot U_0a \sqrt{\sigma^2}$ $\Rightarrow$ $\color{Red} \boxed{\color{Black} I = 2 \sqrt{2}\cdot U_0 \cdot a \cdot \sigma}$

We have just proved that the current flowing through the inhomogeneous cube is the geometric mean of $I_1$ and $I_2$ . This statement (about the geometric mean giving the required current) is valid for any set of square plates of constant thickness that satisfy the requirement according to which for a pair of points, $A$ and $B$ , which can be transformed into each other by rotating the square through 90°, the product of the conductivities at those two points must have a value that is independent of how the point pair is chosen.
It is clear that, in our case, this condition will be satisfied, because a rotation by 90° will always transform a point on a light square into one on a dark square, and viceversa; i.e. points $A$ and $B$ will always be on different coloured squares and $\sigma_A\sigma_B$ will always have the value $\sigma_1\sigma_2$ .

Notes. $1)$ All of the arguments used in the solution are valid not only for the $2 \times 2$ board: more generally, it can be stated that the final result is true for all boards with $n \times n$ squares if $n$ is even, but not true if $n$ is an odd number. In the latter case, a rotation by 90°does not change the "colour" of the square on which any particular point is located, and further, since the numbers of light and dark squares are different, the result cannot be symmetric in $\sigma_1$ and $\sigma_2$ .
$2)$ Using numerical methods, the current-streamlines and the equipotential lines for any particular cube can be found and plotted. Such a computer-generated map, for a $2 \times 2$ "cube", is shown in figure below; in it, the conductivity $\sigma_2$ of the dark squares is twice that, $\sigma_1$ , of the light ones. Because of the boundary conditions, both the current-streamlines and the equipotential lines change direction at the interfaces – in much the same way that light is refracted at the boundary between two different media.

Thank you so much! You are amazing!

Tarapìa Tapioco · Messaggio da **Tarapìa Tapioco** » 31 ago 2023, 1:12

Pubblico la soluzione di questo interessantissimo problema anche in italiano, nell'improbabile eventualità che qualcuno non abbia particolare dimestichezza con la lingua inglese.

In primo luogo, si analizzi il modello fisico. Poiché, degli otto cubi totali, quattro hanno conduttività $2\sigma$ , mentre gli altri quattro possiedono conduttività $\sigma$ , si assuma - dalla figura in esame - che $\sigma_1 = \sigma$ sia la conduttività dei quadrati chiari e $\sigma_2 = 2 \sigma$ quella dei quadrati scuri. Assumendo che ai terminali venga applicata una tensione costante $U_0$ e trascurando le resistenze tra i quadrati all'interfaccia, si può notare come la distribuzione di corrente e il campo elettrico possano essere caratterizzati, rispettivamente, da linee di corrente e linee equipotenziali che a tutti gli effetti giacciono in un unico piano, formando serie di curve reciprocamente ortogonali; ruotando il cubo di $90^\circ$ si invertono le due serie di curve, ma la sua resistenza equivalente rimane inalterata. Si può notare come lo spessore comune $d$ dei cubi metallici non sia da considerare - come invece accade in una scacchiera composta da lastre - molto inferiore (dunque, quasi trascurabile) rispetto alla lunghezza $L$ del pannello: dal momento che i cubi formano un cubo di Rubik $2 \times 2 \times 2$ , lo spessore $d$ dei cubetti è pari alla lunghezza $L$ della faccia del cubo più grande che, essendo la somma delle lunghezze $a$ delle singole facce dei due cubi che lo compongono, è pari a $2a$ . Così: $d = L = 2a$ .

Se tutti i quadrati presenti sul cubo fossero costituiti da una lastra metallica omogenea con conduttività $\sigma_1$ , allora i moduli del vettore campo elettrico e della densità di corrente sarebbero rispettivamente $E = \frac{U_0}{L}$ e $j_1 = \sigma_1 E$ , e la corrente totale che scorre $I_1$ attraverso la piastra quadrata sarebbe:

$I_1 = j_1S$ , con $S = Ld$ . Perciò : $I_1 = j_1 Ld$

Sostituendo $j_1$ and $E$ in $I_1$ , si ottiene:

$I_1 = j_1 Ld$ $\Rightarrow$ $I_1 =\sigma_1 ELd$ $= \frac{U_0}{L}\sigma_1 Ld$ $\Rightarrow$ $I_1 = U_0\sigma_1d$ .

Analogamente, in un'altra piastra metallica omogenea con conduttività $\sigma_2$ , sotto l'azione della stessa tensione $U_0$ , la corrente totale $I_2$ sarebbe:

$I_2 = j_2S$ , con $S = Ld$ . Quindi: $I_2 = j_2 Ld$

Sostituendo $j_2 = \sigma_2 E$ e $E = \frac{U_0}{L}$ in $I_2$ , si ottiene:

$I_2 = j_2 Ld$ $\Rightarrow$ $I_2 =\sigma_2 ELd$ $= \frac{U_0}{L}\sigma_2 Ld$ $\Rightarrow$ $I_2 = U_0\sigma_2d$ .
Si noti che la corrente non dipende dalla lunghezza $L$ della lastra.

Si considerino ora il campo elettrico instaurato nel pannello e le correnti che scorrono attraverso i quadrati della piastra. Un campo elettrico conservativo può essere completamente caratterizzato da un potenziale elettrico, indicato con $\Phi$ , e rappresentato graficamente mediante linee equipotenziali, disegnate con linee tratteggiate nella figura sottostante. Il bordo "inferiore" della piastra quadrata è scelto a potenziale zero, per cui il potenziale "superiore" è uguale alla tensione $U_0$ della batteria.

Immagine

Il campo elettrico, e conseguentemente il vettore densità di corrente ad esso proporzionale, sono entrambi perpendicolari alle curve equipotenziali. Ne consegue che anche le linee di corrente, disegnate in figura con linee continue, sono perpendicolari alle medesime linee. Si associ ad ogni linea di corrente un numero $\Psi$ pari alla corrente totale che scorre tra essa e il bordo sinistro della piastra.
Risulta evidente che il bordo sinistro della lastra si configuri come una linea di corrente corrispondente a $\Psi = 0$ , e che il bordo destro sia $\Psi = I$ . Da qui in poi si chiameranno $\Phi$ il potenziale di tensione e $\Psi$ il potenziale di corrente. Se, in prossimità di un punto arbitrario $A$ della piastra, si fa corrispondere una distanza $\Delta \ell$ tra due linee equipotenziali a una piccola differenza di potenziale $\Delta \Phi$ tra di esse, allora il valore del modulo del campo elettrico in $A$ è

$E_A = \frac{\Delta \Phi}{\Delta \ell}$ ,

e la densità di corrente elettrica misurata localmente è

$j_A = \sigma_A E_A$ $\Rightarrow$ $j_A = \sigma_A\frac{\Delta \Phi}{\Delta \ell}$ .

Qui, $\sigma_A$ indica la conduttività intorno al punto $A$ , che potrebbe essere $\sigma_1$ o $\sigma_2$ a seconda del "colore" del particolare quadrato in cui si trovi $A$ . Dunque, dalla definizione di potenziale di corrente, l'elemento di corrente $j_A d \Delta s$ che scorre attraverso una sezione trasversale $d \Delta s$ della piastra deve essere uguale alla variazione $\Delta \Psi$ del potenziale di corrente sulla distanza infinitesima $\Delta s$ :

$j_A d\Delta s = \sigma_A\frac{\Delta \Phi}{\Delta \ell} d \Delta s$ $= \Delta \Psi \Rightarrow \sigma_A\frac{\Delta \Phi}{\Delta \ell} d \Delta s = \Delta \Psi$ ,

da cui si ottiene

$d\sigma_A\frac{\Delta \Phi}{\Delta \ell}$ $= \frac{\Delta \Psi}{\Delta s}$ $\space$ $\space$ $(1)$

Si consideri poi di ruotare la prima figura di $90^\circ$ in senso antiorario nel piano della piastra, ottenendo la disposizione mostrata nella figura in basso.

Immagine

Ci si chiede ora se questa disposizione, dopo aver scambiato i ruoli dei potenziali di tensione e di corrente e averli moltiplicati per fattori opportunamente scelti, rappresenti una descrizione adeguata delle correnti che attraversano la piastra di partenza.
Siano

$\Psi '= \frac{I}{U_0}\Phi$ $\space$ $\space$ $\space$ e $\space$ $\space$ $\Phi '= \frac{U_0}{I}\Psi$

i nuovi potenziali introdotti, opportunamente rinominati e ridefiniti, e si denotino le separazioni di coppie di curve adiacenti con

$\Delta \ell' = \Delta s$ $\space$ $\space$ $\space$ e $\space$ $\space$ $\Delta s' = \Delta \ell$ .

I fattori di ridimensionamento sono scelti in modo che il valore massimo per $\Phi '$ (il nuovo potenziale di tensione) sia $U_0$ , e quello per $\Psi '$ (il nuovo potenziale di corrente) sia $I$ . Utilizzando queste nuove variabili introdotte per identificare la piastra originaria (non ruotata), si ottiene la disposizione mostrata nella figura seguente.

Immagine

Il punto $B$ , mostrato nella figura di cui sopra, è il luogo in cui il punto $A$ è stato spostato dalla rotazione. Come già notato, $B$ è destinato a trovarsi su un quadrato di colore diverso da quello di $A$ ; il prodotto delle conduttività corrispondenti è:

$\sigma_A \sigma_B = \sigma_1 \sigma_2$ . $\space$ $\space$ $(2)$

Esso ha sempre lo stesso valore, ovunque $A$ , e quindi $B$ , si trovino sul cubo.

La legge di Ohm sarà rispettata (come ci si aspetta) dai nuovi potenziali di tensione e di corrente se è soddisfatta un'equazione analoga alla $(1)$ , cioè se

$d\sigma_B\frac{\Delta \Phi'}{\Delta \ell'}$ $= \frac{\Delta \Psi'}{\Delta s'}$ .

Utilizzando le relazioni tra le quantità introdotte e quelle principali, questa condizione può essere scritta nella forma:

$d\sigma_B\frac{U_0}{I}\frac{\Delta \Psi}{\Delta s}$ $= \frac{I}{U_0}\frac{\Delta \Phi}{\Delta\ell}$ .

Sostituendo l'equazione $(1)$ nella precedente:

$d\sigma_B\frac{U_0}{I}d\sigma_A\frac{\Delta \Phi}{\Delta \ell} = \frac{I}{U_0}\frac{\Delta \Phi}{\Delta\ell}$ ,

da cui si ottiene:

$d\sigma_B\frac{U_0}{I}d\sigma_A\frac{\Delta \Phi}{\Delta \ell} = \frac{I}{U_0}\frac{\Delta \Phi}{\Delta\ell} \Rightarrow d^2\sigma_A\sigma_B\frac{U_0}{I}$ $= \frac{I}{U_0}$ $\Rightarrow$

$I^2 = U_0^2d^2\sigma_A\sigma_B$ $\Rightarrow$ $I = U_0d\sqrt{\sigma_A\sigma_B}$ .

Utilizzando la $(2)$ , l'equazione può essere avanzata nella forma:

$I = U_0d\sqrt{\sigma_A\sigma_B}$ $\Rightarrow$ $\boxed{I = U_0d\sqrt{\sigma_1\sigma_2}}$ .

Sostituendo gli specifici valori di $d = L = 2a$ , $\sigma_1 = \sigma$ e $\sigma_2 = 2\sigma$ trovati inizialmente, si ha:

$I = U_0 \cdot 2a \sqrt{\sigma \cdot 2 \sigma}$ $\Rightarrow$ $I = 2\sqrt{2}\cdot U_0a \sqrt{\sigma^2}$ $\Rightarrow$ $\color{Red} \boxed{\color{Black} I = 2 \sqrt{2}\cdot U_0 \cdot a \cdot \sigma}$ .

Si è appena dimostrato che la corrente che attraversa il cubo non omogeneo corrisponde alla media geometrica di $I_1$ e $I_2$ . Questa affermazione (relativa alla media geometrica che fornisce la corrente richiesta) è valida per qualsiasi insieme di piastre quadrate di spessore costante che soddisfino il requisito secondo il quale, per una coppia di punti, $A$ e $B$ , che possono essere trasformati l'uno nell'altro ruotando il quadrato di $90^\circ$ , il prodotto delle conduttività in quei due punti deve avere un valore indipendente da ogni scelta della coppia di punti.
Appare chiaro che, nel caso in esame, questa condizione sarà soddisfatta, perché una rotazione di $90^\circ$ trasformerà sempre un punto su un quadrato chiaro in uno su un quadrato scuro, e viceversa; cioè, i punti $A$ e $B$ si troveranno sempre su quadrati di colore diverso e $\sigma_A\sigma_B$ assumerà sempre il valore $\sigma_1\sigma_2$ .

Osservazioni. $1)$ Tutte le argomentazioni utilizzate nella soluzione sono valide non solo per il pannello $2 \times 2$ : più in generale, si può affermare che il risultato finale sia vero per tutti i pannelli con $n \times n$ quadrati se $n$ è pari, falso se $n$ è un numero dispari. In quest'ultimo caso, una rotazione di $90^\circ$ non cambia il "colore" del quadrato su cui si trova un particolare punto e, inoltre, poiché il numero di quadrati chiari e scuri è differente, il risultato non può essere simmetrico in $\sigma_1$ e $\sigma_2$ .
$2)$ Utilizzando metodi numerici, è possibile trovare e tracciare le linee di corrente e le linee equipotenziali per ogni determinato cubo. Una mappa di questo tipo, generata al computer per un "cubo" di $2 \times 2$ , è mostrata nella figura seguente; in essa, la conduttività $\sigma_2$ dei quadrati scuri è doppia rispetto a quella, $\sigma_1$ , dei quadrati chiari. A causa delle condizioni di contorno, sia le linee di corrente che quelle equipotenziali cambiano direzione alle interfacce, proprio come la luce viene rifratta al confine tra due mezzi diversi.

Immagine

Rudolf Ortvay 2015 Problem 13

Rudolf Ortvay 2015 Problem 13

Re: Rudolf Ortvay 2015 Problem 13

Re: Rudolf Ortvay 2015 Problem 13

Re: Rudolf Ortvay 2015 Problem 13