An isolated parallel-plate capacitor in vacuum has plates of area separated by

PHYSICS · Subjective · Fermi University Physics · Electricity, Magnetism & Electromagnetism · Conductors, capacitance and boundary-value electrostatics · Conductors, capacitance and boundary-value electrostatics

Electricity, Magnetism and Electromagnetism, Ch Conductors, screening and capacitance

Problem

An isolated parallel-plate capacitor in vacuum has plates of area A=0.020 m2A = 0.020\,\mathrm{m^2} separated by d=0.80 mmd = 0.80\,\mathrm{mm}, carrying +Q+Q and −Q-Q with Q=24 nCQ = 24\,\mathrm{nC}. No battery is connected, and fringing is ignored. A student writes: "The surface charge density is σ=Q/A=1.2×10−6 C m−2\sigma = Q/A = 1.2\times10^{-6}\,\mathrm{C\,m^{-2}}, so the field in the gap is E=σ/ε0=1.36×105 V m−1E = \sigma/\varepsilon_0 = 1.36\times10^{5}\,\mathrm{V\,m^{-1}}. The energy density of the field is u=ε0E2=0.163 J m−3u = \varepsilon_0 E^2 = 0.163\,\mathrm{J\,m^{-3}}, so the stored energy is U=uAd=2.60 μJU = uAd = 2.60\,\mu\mathrm{J}. The positive plate carries charge QQ and sits in the field EE, so the plates attract with F=QE=3.25 mNF = QE = 3.25\,\mathrm{mN}." Two steps are wrong. Identify both, say why each is wrong, and give the correct stored energy and the correct force.

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