The dielectric to be used in a parallel-plate capacitor has a dielectric constant of 3.60 and a dielectric strength of 1.60×107Vm. The capacitor is to have a capacitance of 1.25×10−9F and must be able to withstand a maximum potential difference of 5500 V. What is the minimum area the plates of the capacitor may have?

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reinosodairyshm · Accepted Answer

Given We are given the dielectric constant K=3.60. The dielectric strength is the electric field that the dielectric could tolerate before breakdown is occur and equals E=1.60×107Vm. Also, we are given the capacitance of the capacitor C=1.25×10−9F and the potential difference V=5500V Required We are asked to find the minimum area A of the plates. Solution The capacitor withstands till the electric field equals 1.60×107Vm, after this a breakdown occurs. This electric field is related to the distance d between the two plates E=Vd As shown, the electric field is inversely proportional to the distance d and E represents the maximum electric field between the two plates before the breakdown, therefore the distance d represents the minimum distance between the two electrodes before a breakdown. As we are given Emax and V, we can calculate the minimum distance dmin​​​​​​​ as next dmin=VEmax =5500V1.6×107Vm =3.44×104m The minmum area is related to the minimum distance as given by equation 24.19 in the next form A=CdKϵ0 Where ϵ0, is the electric constant and equals 8.854×10−12C2N⋅m2 (See Appendix F) and C is the capacitance of the capcitor in the presence of the dielectric. Now we can plug our values for C, K, d and ϵ0, into equation (1) to get the minimum area A A=CdKϵ0 =(1.25×10−9F)(3.44×10−4m)(3.60)(8.854×10−12C2N⋅m2) =0.0135m2 =135cm2

Neunassauk8 · Accepted Answer

The capacitance with dielectric with dielectric constant K is C=AKϵ0d and maximum    potential difference , Vmax=Emaxd    Thus, area of plate A=CdKϵ0=CVmaxKϵ0Emax=(1.25×10−9)(5500)3.60×(8.85×10−12)(1.60×107)=0.0135m2

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