In the shown ideal-op-amp circuit, the parallel-plate capacitance is $C=\varepsilon A/x$, where ε is dielectric permittivity, A is plate area and $x=x_0+kt$ is plate separation. Here t is time and x₀,k are positive nonzero constants. Input vᵢ is constant. What is output v₀?
Virtual ground places the full constant input across the capacitor. Its current is $i=d(Cv_i)/dt=v_i\,dC/dt$. Since $dC/dt=-\varepsilon Ak/x^2=-Ck/x$, the inverting current-to-voltage stage gives $v_0=-Ri=Rv_iCk/x$. Thus A; a variable capacitor can carry current even at constant applied voltage.
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