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Where do the electric charges sit, which generate the thermoelectric potential?

As we have seen that the thermoelectric potential measured in a thermoelectric circuit is of purely electrostatic origin, we ask by which electric charges this potential is generated.

As is shown below, inside a homogeneous conductor charge neutrality is maintained even in the presence of a stationary temperature gradient. Therefore the electric charges generating the thermoelectric potential can only sit at the surface of the conductor, including the interface at the junctions with another conducting material.

The distribution of these charges can be inferred from the variation of the electrostatic potential (Fig. 10).

Figure 10: Variation of temperature $ T$, temperature gradient $ dT/dz$, electric field $ E_z$ (in direction of wire) and electrostatic potential $ \varphi $ in the thermoelectric circuit shown in Fig. 1 (schematic; $ Q_{B}^{\prime}<Q_{A}^{\prime}<0$).

The electrostatic potential follows via

$\displaystyle \vec E = -\vec\nabla \varphi$ (28)

from the result (31) derived below as

$\displaystyle \varphi (\vec r) = -Q^\prime \cdot \delta T (\vec r) + \varphi_0
 \, .$ (29)

Here the coefficients $ Q^\prime$ and the constants of integration $ \varphi_0$ are different for the two conductors. The potential $ \varphi $ must fulfil the condition of continuity (7) at the interfaces. This potential pattern is explained qualitatively by a charge distribution as drawn in Fig. 11 (assuming $ Q_B^\prime < Q_A^\prime < 0$). At the interfaces a positive or negative charge cloud is superimposed on the electric double layer. Charges are also located at the surfaces of the conductors, especially near the points of contact.

Figure 11: Distribution of electric charge at the surfaces and interfaces of a thermoelectric circuit (schematic).

The magnitude of the total electric charge generating the thermoelectric potential, according to Fig. 12, can be estimated as

$\displaystyle q=\varepsilon_0 L (Q_A-Q_B) (T_1-T_2)\, .$ (30)

For $ Q_A-Q_B = 10^{-3} \,\,\rm {V/deg}, T_1-T_2 = 100 \,\,\rm {K}$ and $ L = 0.1 \,\,\rm {m}$ one obtains $ q=10^{-13} \,\,\rm {Cb} $ as an order of magnitude. This is a very small charge indeed, amounting to about $ 10^6$ electron charges. To my knowledge, so far the electric (and upon rotation: magnetic) field generated by this charge has not been measured directly outside the conductors.

Figure 12: Estimate of total electric charge in a thermoelectric circuit.

next up previous
Next: Proof of charge neutrality Up: The origin of the Previous: Compensation of the thermodiffusion
Klaus Froboese 2000-11-07