Question #329532

The plates of a parallel-plate capacitor are 2.50 mm apart, and each carries a charge of


magnitude 80.0 nC. The plates are in vacuum. The electric field between the plates has a magnitude


of 4.00 X 106 V/m(a) What is the potential difference between the plates? (b) What is the area of


each plate? (c) What is the capacitance?



1
Expert's answer
2022-04-19T12:05:33-0400

a) The potential difference is:


V=EdV = Ed

where E=4.00×106V/mE= 4.00 \times 10^6 V/m and d=2.50mm=2.50×103md=2.50mm = 2.50\times 10^{-3}m. Thus, obtain:


V=4.00×106V/m2.50×103m=1.00×104VV = 4.00 \times 10^6 V/m\cdot 2.50\times 10^{-3}m = 1.00\times 10^4V

c) The capacitance is given as follows:


C=qVC = \dfrac{q}{V}

where q=80.0×109Cq = 80.0\times 10^{-9}C. Obtain:


C=80.0×109C1.00×104V=8.00×1013FC = \dfrac{80.0\times 10^{-9}C}{1.00\times 10^4V}=8.00\times 10^{-13} F

b) On the other hand the capacitance of a parallel plate capacitor is:


C=ε0AdC = \dfrac{\varepsilon_0 A}{d}

where ε08.85×1012F/m\varepsilon_0\approx 8.85\times 10^{-12} F/m is the vacuum permittivity and AA is the are of the plate. Expressing AA and substituting CC, obtain:


A=dCε0=2.50×103m8.00×1013F8.85×1012F/m2.26×104m2A = \dfrac{dC}{\varepsilon_0} = \dfrac{2.50\times 10^{-3}m\cdot 8.00\times 10^{-13} F}{8.85\times 10^{-12} F/m} \approx 2.26\times 10^{-4}m^2

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