Question

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

Answer 1

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

Answer 2

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