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 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?

Answer 1

`A=2.49* 10^6\ m^2`

Step-by-step explanation:

Given that

Stored energy E

`U=10^6\ J`

a)

We know that stored energy in capacitor given as

`U=(1)/(2)CV^2`

Given that

`V=10^4\ V`

`U=(1)/(2)CV^2`

`10^6=(1)/(2)* C* (10^4)^2`

C= 0.02 F

b)

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

We know that

V = E .d

`10^4=9* 10^6* d`

d=1.11 mm

We know that

`C=(\varepsilon _oA)/(d)`

`0.02=(8.89* 10^(-12)A)/(1.11* 10^(-3))`

`A=2.49* 10^6\ m^2`

This is area area of the plates.