Physic Labs

Problem 2

An ideal capacitor of capacitance C=10 μFC=10\,\mu\mathrm F is connected to an ideal source holding voltage V=12 VV=12\,\mathrm V. After electrostatic equilibrium is reached, energy is stored in the electric field between the plates; neglect fringing. (a) Find the first requested quantity. (b) Find the second quantity. (c) Determine the final result and check its units (a) Find the charge on the capacitor. (b) Find its stored electric-field energy. (c) Evaluate both results in SI units.
U=12CV2=7.20×10−4 JU=\frac12CV^2=7.20\times10^{-4}\,\mathrm J
problems.proof.analysis

Substituting the result into the relevant definition or law gives the next quantity. This connects the parts of the problem instead of treating their answers separately.