A capacitor is not the most efficient device for storing energy. Batteries can s

Question

A capacitor is not the most efficient device for storing energy. Batteries can store more energy in much less space. For example, a typical 12 V automobile battery stores on the order of 1.00 106 J.
(a) Find the capacitance necessary to store 1.00 106 J with a potential difference of 1.00 104 V across the capacitor's terminals.
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(b) Suppose that such a capacitor was made in the form of a parallel-plate capacitor with a vacuum between the plates and an electric field no greater than 7.00 106 V/m. What is the minimum area of the plates?
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And.........

The plates of a parallel-plate capacitor each have an area of 0.1 m2 and are separated by a 1 mm thick layer of paper. The capacitor is connected to a 11 V battery. (The dielectric constant for paper is 4.)
(a) Find the capacitance.
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(b) Find the charge stored.
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(c) Find the electric field between the plates.
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Explanation / Answer

given v = 12v energy strod in the battery = u = 1.00106J (a) potential difference = v = 1.00104v    energy stored = 1.00106J enegy stored = U = 1/2 * cv2                          c = 2u/v2 = 2 * 1.00106J/((1.00104v)2 = 1.99F b) electric field between the plates = E = 7.00106v/m         (c) area = A = 0.1m2    d = 1mm = 1* 10-3m v = 11v k = 4 capacitence = c = ke0A/d                            = 4 * 8.9 * 10-12 c2/N.m2 * 0.1m2/1 * 10-3m                             = 0.003µF charge stored = Q = cv = 0.003µF * 11v = 0.033µc c) electric field between the plates                   E = v/d = 11v/1 * 10-3 = 11 * 103v/m

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