Past exam of the thermal and statistical physics course of the University of Cambridge / 2013 / Paper 1 / 1 / a / Solution

In the classical equilibrium circuit, the capacitor stores $C{V}^2/2$. Equipartition assigns this quadratic term mean energy $k_{B}T/2$, soThe mean voltage is zero and its variance is $k_{B}T/C$. Resistance determines the equilibration time and noise spectrum, not this full-band equilibrium variance. A finite-bandwidth or appreciably loading readout can measure less than the ideal result.