The area of the plates of a parallel plate ca | vedclass

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(C) The capacitance $C$ of a parallel plate capacitor with multiple dielectric slabs is given by the formula:$C = \frac{{\varepsilon _0 A}}{{\sum \fra

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$\frac{5000}{7}{\varepsilon _0}A$

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(C) The capacitance $C$ of a parallel plate capacitor with multiple dielectric slabs is given by the formula:$C = \frac{{\varepsilon _0 A}}{{\sum \frac{t_i}{k_i}}}$Given:$t_1 = 6\,mm = 6 	imes 10^{-3}\,m$,$k_1 = 10$$t_2 = 4\,mm = 4 	imes 10^{-3}\,m$,$k_2 = 5$Substituting the values:$C = \frac{{\varepsilon _0 A}}{{\frac{6 	imes 10^{-3}}{10} + \frac{4 	imes 10^{-3}}{5}}}$$C = \frac{{\varepsilon _0 A}}{{0.6 	imes 10^{-3} + 0.8 	imes 10^{-3}}}$$C = \frac{{\varepsilon _0 A}}{{1.4 	imes 10^{-3}}}$$C = \frac{1000}{1.4} \varepsilon _0 A = \frac{10000}{14} \varepsilon _0 A = \frac{5000}{7} \varepsilon _0 A$

The area of the plates of a parallel plate capacitor is $A$ and the distance between the plates is $10\,mm$. There are two dielectric sheets in it,one of dielectric constant $10$ and thickness $6\,mm$ and the other of dielectric constant $5$ and thickness $4\,mm$. The capacity of the capacitor is

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