A capacitor is not the most efficient device for storingenergy.Batteries can sto

Question

A capacitor is not the most efficient device for storingenergy.Batteries can store more energy in much less space. Forexample, atypical 12 V automobile battery stores on the order of1.00 106J.
(a) Find the capacitance necessary to store 1.00 106 J withapotential difference of 1.00 104 V across thecapacitor'sterminals.
(b) Suppose that such a capacitor was made in the form ofaparallel-plate capacitor with a vacuum between the plates andanelectric field no greater than 2.00 106 V/m. What is theminimumarea of the plates?

Explanation / Answer

The energy stored in a capacitor is      U = (1/2)CV^2 Given V = 1x10^4 , U = 1x10^6 J   The capacitance of the capacitor is       C = 2U / V^2          = 2 (1x10^6) / (1x10^4)^2          = 0.02 F   The relation between electric field and potential is       E = V /d the distance between the plates of a capacitor is      d = V/ E         = 1x10^4 / 2x10^6          = 0.005m The capacitance of a capacitor is      C= o A /d The area of the plates is      A = C d / o         = (0.02)(0.005) / (8.85x10^-12)         = 1.129x10^7 m^2 The energy stored in a capacitor is      U = (1/2)CV^2 Given V = 1x10^4 , U = 1x10^6 J   The capacitance of the capacitor is       C = 2U / V^2          = 2 (1x10^6) / (1x10^4)^2          = 0.02 F   The relation between electric field and potential is       E = V /d the distance between the plates of a capacitor is      d = V/ E         = 1x10^4 / 2x10^6          = 0.005m The capacitance of a capacitor is      C= o A /d The area of the plates is      A = C d / o         = (0.02)(0.005) / (8.85x10^-12)         = 1.129x10^7 m^2

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