The dielectric to be used in a parallel-plate capacitor has a dielectr



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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 1.25×109F 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?

Answer & Explanation



Beginner2022-01-11Added 36 answers

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=1.25×109F and the potential difference V=5500V
We are asked to find the minimum area A of the plates.
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
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
The minmum area is related to the minimum distance as given by equation 24.19 in the next form
Where ϵ0, is the electric constant and equals 8.854×1012C2Nm2 (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



Beginner2022-01-12Added 30 answers

The capacitance with dielectric with dielectric constant K is C=AKϵ0d and maximum
potential difference , Vmax=Emaxd
Thus, area of plate A=CdKϵ0=CVmaxKϵ0Emax=(1.25×109)(5500)3.60×(8.85×1012)(1.60×107)=0.0135m2

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