A parallel-plate air capacitor is to store charge of magnitude on each plate when the potential difference between the plates is . (a) If the area of each plate is , what is the separation between the plates? (b) If the separation between the two plates is double the value calculated in part (a), what potential difference is required for the capacitor to store charge of magnitude on each plate?

A parallel plate capacitor can only store a finite amount of energy before dielectric breakdown occurs. It can be defined as: When two parallel plates are connected across a battery, the plates are charged and an electric field is established between them, and this setup is known as the parallel plate capacitor

Given We are given the stored charge area of the plates is

Required

(a) We are asked to find the separated distance d lof3 C where the potential difference is and the

(b) We want to determine the potential difference when the separated distance is doubled.

(a) This problem uses the relationship among the charge Q, plate separation d, plate area A and the potential difference So We can apply equation 242 in the next form to get the separated distance d

Therefore The separated distance is

(b) If the distance d is doubled and becomes d = 22 mm, then the voltage will be doubled when the area A and the charge Q are constant Where d and Vab are directly proportional to each other as shown by equation (1)

So, the voltage will be doubled and becomes This will affect the capacitance where , so increasing V will keep C constant

Therefore The potential difference of 84V required for the capacitor to store the charge