The energy stored in a capacitor is

PHYSICS · Multi-part · Fermi University Physics · Mathematical Methods for Physics · Calculus of change and accumulation · Calculus of change and accumulation

Mathematical Methods for Physics, Ch Partial differentiation, chain rules and Jacobians

Problem

The energy stored in a capacitor is U(C,ε)=12Cε2U(C,\varepsilon) = \tfrac12 C\varepsilon^2, where CC is its capacitance and ε\varepsilon the voltage across it. At some instant C=2.0×10−6 FC = 2.0\times10^{-6}\,\mathrm{F} and ε=12.0 V\varepsilon = 12.0\,\mathrm{V}. ((a)) Find (∂U/∂C)ε(\partial U/\partial C)_\varepsilon and (∂U/∂ε)C(\partial U/\partial \varepsilon)_C at this instant, naming the held variable in each. ((b)) The capacitance is increased by ΔC=+0.1×10−6 F\Delta C = +0.1\times10^{-6}\,\mathrm{F} while the voltage drops by Δε=−0.5 V\Delta \varepsilon = -0.5\,\mathrm{V}. Use the total differential to estimate the change in stored energy. ((c)) Suppose instead C(t)=2.0×10−6+0.05×10−6t FC(t) = 2.0\times10^{-6} + 0.05\times10^{-6}t\,\mathrm{F} and ε(t)=12.0−0.3t V\varepsilon(t) = 12.0 - 0.3t\,\mathrm{V} (with tt in seconds). Find dU/dt\mathrm{d}U/\mathrm{d}t at t=0 st=0\,\mathrm{s} using the chain rule along this path. ((d)) New labels u=2Cu = 2C and v=3εv = 3\varepsilon are introduced. Find the Jacobian ∂(C,ε)/∂(u,v)\partial(C,\varepsilon)/\partial(u,v).

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