A parallel-plate capacitor with area 0.300 m2 and plate separation of 2.00 mm is

Question

A parallel-plate capacitor with area 0.300 m2 and plate separation of 2.00 mm is connected to a 12.00-V battery. (a) What is the capacitance? How much charge is stored on the plates? What is the electric field between the plates? Find the magnitude of the charge density on each plate. Without disconnecting the battery, the plates are moved farther apart. Qualitatively, what happens to each of the previous answers?

Explanation / Answer

(a) C = ?o x A/d =>C = [8.854×10-12 x 40 x 10^-4]/[1 x 10^-3] =>C = 35.42 x 10^12 F =>C = 35.42 pF (b) By Q = CV =>Q = 35.42 x 10^-12 x 600 = 2.12 x 10^-8 Coulomb (c) By E = 1/2CV^2 =>E = 1/2 x 35.42 x 10^-12 x (600)^2 = 6.38 x 10^-6 J (d) BY E = V/d =>E = 600/(1 x 10^-3) = 6 x 10^5 V/m (e) energy density = Energy/Volume = 6.38 x 10^-6/(A x d) =>energy density = 6.38 x 10^-6/(40 x 10^-4 x 1 x 10^-3) =>energy density = 1.6 J/m^3

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