Dielectrics Capacitor Problem

It has only been fairly recently that 1.0-F capacitors have been readily available. A typical 1.0-F capacitor can withstand up to 5.00 V. To get an idea why it isn't easy to make a 1.0-F capacitor, imagine making a 1.0-F parallel plate capacitor using titanium dioxide (k = 90.0, breakdown strength 4.00 kV/mm) as the dielectric.

(a) Find the minimum thickness of the titanium dioxide such that the capacitor can withstand 5.00 V.
m

(b) Find the area of the plates so that the capacitance is 1.0 F.
m2

Find "d" by noting that in a pp capacitor E-field & voltage are related by;
V = Ed

Just put in the given values for V and maximum E as;
d = V/E = 5/(4000) =.00125 mm = 1.25 x 10^-6 m

The plate area necessary to produce 1 F capacitance with the calculated "d" is from the pp capacitor eq;
C = keoA/d

Solve for A and that should get it. (eo is the vacum dielectric constant)

(a) 5/dmin = 4 *10^6 or dmin = 1.25*10^-6 m = 1.25*10^-3 mm

(b) C = (8.85*(10^-12)*90)*(A)/(1.25*10^-6) = 1 F
or required A = 0.0001569*10^6 m^2 = 157 m^2