Question

In 1990, a pizza with a radius of 18.0 m was made in South Africa. Suppose you make an air-filled capacitor with parallel plates whose area is equal to that of the pizza. If a potential difference of 575 V is applied across this capacitor, it will store just 3.31 J of electrical potential energy. What are the capacitance and plate separation of this capacitor?

Answer 1

Step-by-step explanation:

We have,

Radius of pizza, r = 18 m

Potential difference, V = 575 V

Stored electrical potential energy, E = 3.31 J

The area if an air-filled capacitor with parallel plates is equal to that of the pizza. It means,
`A=\pi r^2=\pi (18)^2=1017.87\ m^2`

The stored potential energy in a capacitor is given by :

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

C is capacitance

`C=(2E)/(V^2)\\\\C=(2* 3.31)/((575)^2)\\\\C=2* 10^(-5)\ F`

The formula of the capacitance of a parallel plate capacitor is given by :

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

d is plate separation of this capacitor

`d=(\epsilon_o A)/(C)\\\\d=(8.85* 10^(-12)* 1017.87)/(2* 10^(-5))\\\\d=4.5* 10^(-4)\ m`