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

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 :

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

C is capacitance

$C=\dfrac{2E}{V^2}\\\\C=\dfrac{2\times 3.31}{(575)^2}\\\\C=2\times 10^{-5}\ F$

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

$C=\dfrac{\epsilon_o A}{d}$

d is plate separation of this capacitor

$d=\dfrac{\epsilon_o A}{C}\\\\d=\dfrac{8.85\times 10^{-12}\times 1017.87}{2\times 10^{-5}}\\\\d=4.5\times 10^{-4}\ m$