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

Answer:

$A = 5 \times 10^{32} m^2$

Explanation:

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

$Q = \frac{1}{2}CV^2$

here we know that

$Q = 6 \times 10^{22} J$

also we know that

$V = 5 Volts$

now we have

$6 \times 10^{22} = \frac{1}{2}C(5^2)$

$C = 4.8 \times 10^{21} F$

now we know the formula of capacitance

$C = \frac{\epsilon_0 A}{d}$

$4.8 \times 10^{21} = \frac{(8.85 \times 10^{-12})(A)}{1}$

$A = 5 \times 10^{32} m^2$