Question

(a) How much charge can be placed on a capacitor with air between the plates before it breaks down if the area of each plate is 8.00 cm²? __________ nC

(b) Find the maximum charge if paper is used between the plates instead of air. ___________ nC

Please show your work and steps for each part!

Answer 1

(a) Assume the dielectric strength of air, E = 3.0 x 10^6 V/m,

The field in the capacitor is σ/ε where σ = Q/A. For air ε = ε0 = 8.85 x 10^-12 F/m

Also, σ = k*ε0*E

The maximum charge is then

Q= A*σ = A* k*ε0*E --------------------------------------------------(i)

For air, k = 1

Area of each plate, A = 8.00 cm^2 = 8.00 x 10^-4 m^2

So -

Q = 8.00 x 10^-4 x 1 x 8.85 x 10^-12 x 3.0 x 10^6  = 21.24 x 10^-9 C = 21.24 nC

(b) The dielectric strength of paper is E = 16 x 10^6 V/m

The dielectric constant, k = 3,

Put these values in the above expression (i) -

Q = 8.00 x 10^-4 x 3 x 8.85 x 10^-12 x 16.0 x 10^6  = 339.84 x 10^-9 C = 339.84 nC (Answer)