A parallel plate capacitor has a separation b | vedclass

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Question

(B) The capacitance of a parallel plate capacitor is given by $C = \frac{K \epsilon_0 A}{d}$.
The resistance of the dielectric material is given by $

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) The capacitance of a parallel plate capacitor is given by $C = \frac{K \epsilon_0 A}{d}$.
The resistance of the dielectric material is given by $R = \frac{ho d}{A}$.
Multiplying these two expressions, we get $RC = \left( \frac{K \epsilon_0 A}{d} ight) \left( \frac{ho d}{A} ight) = K \epsilon_0 ho$.
Given values are $R = 17.7 	imes 10^{14}$ $\Omega$, $C = 1 	imes 10^{-6}$ $F$, $ho = 10^{13}$ $\Omega$m, and $\epsilon_0 = 8.85 	imes 10^{-12}$ $F$/m.
Substituting these values into the equation: $(17.7 	imes 10^{14}) 	imes (1 	imes 10^{-6}) = K 	imes (8.85 	imes 10^{-12}) 	imes (10^{13})$.
$17.7 	imes 10^8 = K 	imes 8.85 	imes 10^1$.
$17.7 	imes 10^8 = 88.5 K$.
$K = \frac{17.7 	imes 10^8}{88.5} = 0.2 	imes 10^8 = 2 	imes 10^7$.
Given that the relative permittivity $K = \alpha 	imes 10^7$, we find $\alpha = 2$.

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