The plates of a parallel-plate capacitor are apart, and each carries a charge of magnitude . The plates are in vacuum. The electric field between the plates has a magnitude ofn . What is (a) the potential difference between the plates; (b) the area of each plate; (c) the capacitance?

A parallel plate capacitor is an arrangement of two metal plates connected in parallel separated from each other by some distance. A dielectric medium occupies the gap between the plates

u:ven

We are given the distance between the two parallel plates

and the charge on each plate is . Also, We are given the electric field between them

Required

to find (a) The potential difference between the plates

(b) The area A of each plate.

(c) The capacitance C

Solve For A

(a) Due to the charge on the two parallel plates, there is electrical energy stored between thenu where the potential difference due to the electric field between two conducting plates (plates a and b) as We studied in chapter 23 is given by

(1) Where E is the electric field due to the charge on the conductive plates and d is the distance between them. Now We can plug our given values for E and d into equation (l) to get

Therefore the the potential difference between two plates is

(b) Example 21_12 is related to ourproblem, where between the two plates there is a uniform electric field between them depends on the area of each plate, where this relation is given by

(Solve for A)

(2) Where Eo is the electric constant and equals (See Appendix F) and Q is the charge on the plates_ Now We can plug our values for E, Q and Eo into equation (2) to get the area of the plate A

Therefore the area of each plates is

(c) The electrical energy stored between the two plates called the capacitance, where the capacitance depends on the area of the plates and it is given by equation 242 in the form

s for d, eo and A into equation (3) to get the capacitance in

Therefore the capacitance is