Question

The world energy consumption was about 6*10^22 J. How much area must a parallel plate capacitor need to store this energy Assume we maintain the capacitor at delta V= 5 volts for safety reasons, and have a plate separation distance of 1 meter. 5 *10^32 m^2 5 *10^61 m^2 9.10^8 m^2 4*10^14 m^2 2 *10^23 m^2

Answer 1

`A = 5 * 10^(32) m^2`

Step-by-step explanation:

As we know that the energy stored in the capacitor is given as

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

here we know that

`Q = 6 * 10^(22) J`

also we know that

`V = 5 Volts`

now we have

`6 * 10^(22) = (1)/(2)C(5^2)`

`C = 4.8 * 10^(21) F`

now we know the formula of capacitance

`C = (\epsilon_0 A)/(d)`

`4.8 * 10^(21) = ((8.85 * 10^(-12))(A))/(1)`

`A = 5 * 10^(32) m^2`