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

Answer:

$A=2.49\times 10^6\ m^2$

Explanation:

Given that

Stored energy E

$U=10^6\ J$

a)

We know that stored energy in capacitor given as

$U=\dfrac{1}{2}CV^2$

Given that

$V=10^4\ V$

$U=\dfrac{1}{2}CV^2$

$10^6=\dfrac{1}{2}\times C\times (10^4)^2$

C= 0.02 F

b)

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

We know that

V = E .d

$10^4=9\times 10^6\times d$

d=1.11 mm

We know that

$C=\dfrac{\varepsilon _oA}{d}$

$0.02=\dfrac{8.89\times 10^{-12}A}{1.11\times 10^{-3}}$

$A=2.49\times 10^6\ m^2$

This is area  area of the plates.