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 2.00 cm2? (Assume air has a dielectric strength of 3.00 ✕ 106 V/m and dielectric constant of 1.00.) nC

(b) Find the maximum charge if polystyrene is used between the plates instead of air. (Assume polystyrene has a dielectric strength of 24.0 ✕ 106 V/m and dielectric constant of 2.56.) nC

Answer 1

Part (a) We know that max charge is given by:

Qmax = C*Vmax

Vmax = E*d

C = k*e0*A/d

Qmax = (k*e0*A/d)*(E*d)

Qmax = k*e0*A*E

Now using given values:

k = dielectric constant of air = 1.00

E = dielectric strength of Air = 3.00*10^6 V/m

A = Area of each plate = 2.00 cm^2 = 2.00*10^-4 m^2

e0 = 8.854*10^-12

Qmax = 1.0*8.85*10^-12*2.00*10^-4*3*10^6

Qmax = 5.31*10^-9 = 5.31 nC

Part B.

for polystyrene material:

k = dielectric constant of polystyrene = 2.56

E = dielectric strength of polystyrene = 24.0*10^6 V/m

A = Area of each plate = 2.00 cm^2 = 2.00*10^-4 m^2

e0 = 8.854*10^-12

Qmax = 2.56*8.85*10^-12*2.00*10^-4*24.0*10^6

Qmax = 108.75*10^-9 = 108.75 nC

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