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 x 10^6 J. (a) Find the capacitance necessary to store 1.00 x 10^6 J with a potential difference of 1.00 x 10^4 V across the capacitor's terminals.
(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 9.00 x 10^6 V/m. What is the minimum area of the plates?

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Netta00 Netta00 · Answer 1 · 2019-10-07T09:37:28+02:00

Answer:

[tex]A=2.49\times 10^6\ m^2[/tex]

Explanation:

Given that

Stored energy E

[tex]U=10^6\ J[/tex]

a)

We know that stored energy in capacitor given as

[tex]U=\dfrac{1}{2}CV^2[/tex]

Given that

[tex]V=10^4\ V[/tex]

[tex]U=\dfrac{1}{2}CV^2[/tex]

[tex]10^6=\dfrac{1}{2}\times C\times (10^4)^2[/tex]

C= 0.02 F

b)

Electric filed E = 9 x 10^6 V/m

We know that

V = E .d

[tex]10^4=9\times 10^6\times d[/tex]

d=1.11 mm

We know that

[tex]C=\dfrac{\varepsilon _oA}{d}[/tex]

[tex]0.02=\dfrac{8.89\times 10^{-12}A}{1.11\times 10^{-3}}[/tex]

[tex]A=2.49\times 10^6\ m^2[/tex]

This is area area of the plates.