Question

Paper has a dielectric constant K = 3.7 and a dielectric strength of 15 ×106V/m. Suppose that a typical sheet of paper has a thickness of 0.11 mm. You make a "homemade" capacitor by placing a sheet of 25 × 17 cm paper between two aluminum foil sheets of the same size. What is the capacitance C of your device? About how much charge could you store on your capacitor before it would break down?

Answer 1

Given

k= 3.7 and  

dielectric strength is = 15*10^6 V/m

thickness is d = 0.11 mm

the area of the plates is A = 25*17 cm = 25*17*10^-4 m^2

we know that the capacitance of A Parallel plate capacitor is  

C = epsilon not*A/d

C = (8.854*10^-12*25*17*10^-4)/(0.11*10^-3) F

C = 3.4208636*10^-9 F

with dielectric material the capacitance increases by k times so

the capacitance is C' = k*C = 3.7*3.4208636*10^-9 F = 1.265719532*10^-8 F = 12.65719532*10^-9 F

the breakdown voltage given is V = E*d = 15*10^6*0.11*10^-3 V

so the relation between Q,C,V is Q = C*V

Q = 12.65719532*10^-9 *15*10^6*0.11*10^-3 C

Q = 2.0884372278*10^-5 C