The dielectric to be used in a parallel-plate capacitor has a dielectric constant of 3-60 and a dielectric strength of 1-60- 107V-m- The capacilnr is to have a capacitance of 1-25- 10-9 F and must be able to u withstand a maximum potential difference of 5500 V- The minimum area the plates of the capacitor may have is equal to 135 - 10-x m2- The value of x is-

The capacitance with dielectric with dielectric constant K is C=AKϵ_0d and maximum potential difference , V_max=E_maxd Thus, area of plate A=CdK ...

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Standard XII

Physics

Introducing Epsilon

The dielectri...

Question

The dielectric to be used in a parallel-plate capacitor has a dielectric constant of 3.60 and a dielectric strength of $1.60 \times 10^{7} V / m$ . The capacilnr is to have a capacitance of $1.25 \times 10^{- 9} F$ and must be able to u withstand a maximum potential difference of 5500 V. The minimum area the plates of the capacitor may have is equal to $135 \times 10^{- x} m^{2}$ . The value of $x$ is:

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Solution

The capacitance with dielectric with dielectric constant K is $C = \frac{A K ϵ_{0}}{d}$ and maximum potential difference , $V_{m a x} = E_{m a x} d$
Thus, area of plate $A = \frac{C d}{K ϵ_{0}} = \frac{C V_{m a x}}{K ϵ_{0} E_{m a x}} = \frac{(1.25 \times 10^{- 9}) (5500)}{3.60 \times (8.85 \times 10^{- 12}) (1.60 \times 10^{7})} = 0.0135 m^{2}$

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