You need to design a capacitor capable of storing 3.0 10-7 c of charge. at your disposal, you have a 100 v power supply and two metal plates, each of area 0.490 m2 each. what is the limit of the separation of the plates? mm

Capacitance = Q/V

Where Q is the charge

V is the voltage

C = 3 E - 7 C / 100V = 3 E - 9 F = 3nF

Capacitance is also equal to

C = (Eo*A) / d

Where A is the area of the plate

d is the separation between them

Eo is the permittivity (8.85E-12 for Air)

We find d

d = (8.85E-12 * 0.49) / 3 E-9 = 1.4455 E - 3 m

d = 1.4455 mm

The capacitor formula is given by

Capacitance = Q/V

(3 * 10-7) / 100 = 3*10-9

the relation between capacitance, Area and distance between the plates is given by

C = (EoA) / d - - > (EoA) / C = d

(8.85 * 10-12) * 0.490 / 3*10-9

d=.1.44mm