Physic Labs

Problem 2

A parallel-plate capacitor of capacitance C0C_0 (air-filled) is charged to voltage U0U_0 then disconnected from the source. A dielectric slab of constant κ\kappa is then inserted, filling the gap. Find the capacitor's energy before and after insertion, and explain the change. (a) Find plate charge, capacitance after insertion, and final voltage. (b) Calculate initial and final field energies and their ratio in terms of κ\kappa. (c) Explain the mechanical work associated with the energy decrease as the slab is drawn in; distinguish field work from external work and state what is conserved after disconnection.
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Q=C0U0, Cf=κC0, Uf=U0/κ, Ei=12C0U02, Ef=Ei/κ, Wfield=Ei−Ef, Wext=−(Ei−Ef)\boxed{Q=C_0U_0,\ C_f=\kappa C_0,\ U_f=U_0/\kappa,\ E_i=\tfrac12C_0U_0^2,\ E_f=E_i/\kappa,\ W_{\rm field}=E_i-E_f,\ W_{\rm ext}=-(E_i-E_f)}
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This is the collected answer to all parts. It obeys the stated assumptions and conventions.