Parallel Plate Capacitor

Question

A parallel-plate air capacitor has a capacitance of 500 pf and a charge of magnitude 0.2 uc on each plate. The plates are 2mm apart.

a) What is the potential difference between the plates?

b) What is the area of each plate?

c) What is the electric field magnitude between plates?

Answer 1

part A) You just need to know how Charge, capatitance, and voltage are related. It's the simple equation Q = CV, and in this case Q calls for the charge stored on one plate. Since you know C = 500*10^-12 farads and Q = 0.2*10^-6 coulombs, then you can find voltage easily.

Q = CV -------> 0.2*10^-6 = Voltage*500*10^-12

V = (0.2*10^-6)/(500*10^-12) = 400 Volts

Part B) For this, you'll need the formula to calculate Capacitance based on the physical setup of a parallel capacitor. This formula is C = ε₀*(A/d), and related Capacitance (C) to the Area of the plates (A, in m²) and the distance of separation (d, in meters), and also this physical constant ε₀ which is always 8.854*10^-12. In this case you know everything except Area, so you can solve for it.

C = ε₀*(A/d)

500*10^-12 = 8.854*10^-12 * (A/.002)

by rearranging, you can find A = .112943302 m²

Part C) For this you need to know how Electrical Field is related to the quantities you already know about this situation. You can combine the formulas for Q=CV (relating total charge to capacitance and voltage) and the formula for V=Ed (relating voltage to Electrical field and distance) to get Q = C(Ed)

Q = C*E*d

0.2*10^-6 = 500*10^-12 * E * .002

E = 200,000 N/c