Two parallel plates are made of two metals 1 and 2 inside a vacuum chamber (Fig. 9.11). The chemical potentials of electrons in metals 1 and 2 are μ₁ and μ₂. When contact is established between the two metals, conduction electrons flow until equilibrium is reached. As a result of this electron flow, there is a charge Q and −Q at the surface of the electrodes. These electric charges produce an electric field E between the plates. According to relation (9.67) the corresponding electrostatic potential difference Δφ  is related to the charge Q by,where C is the capacitance of the capacitor. The plates have a surface area A and theyare separated by a distance d. It can be shown that the capacitance C is given by,A parallel plate capacitor is made of metals 1 and 2. The metallic plate 1 is fixed mechanically. The metallic plate 2 vibrates under the effect of a spring of elastic constant k.When it vibrates, conduction electrons flow through the resistance R and the spring. The electrostatic potential differenceΔφ is maintained constant by the power supply. where ε₀ is the electric permittivity of vacuum. For practical applications, the flow of electric charges is too difficult to detect. It is easier to make the separation distance d oscillate in time, which causes C to vary, hence Q oscillates in time, which implies that an electric current of intensity I flows through the resistance R. Determine the voltage V detected between the plates as a function of d˙, μ₁ and μ₂.

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Q=C Δρ