91Ó°ÊÓ

We are given that resistance $12KΩ$, diameter of parallel plate capacitor as$15cm$.The voltage$12V$and frequency$36kHz$of source is given . We are given that peak current equals to $0.65mA$.We need to find spacing between capacitor plates.

We know that capacitance between parallel plate capacitor is given by the formula ,

$$\begin{gathered} C=\frac{{\mathrm{Î\mu }}_{0}A}{d}=\frac{{\mathrm{Î\mu }}_0{\mathrm{Ï€°ù}}^2}{d} \\ ⇒d=\frac{{\mathrm{Î\mu }}_0{\mathrm{Ï€°ù}}^2}{C} \end{gathered}$$

Here $d$is separation between plates. We need to find the value of $d$, for that matter , we need to calculate capacitance , $C=\frac{1}{\mathrm{Ó\neg }{X}_C}$

$Z^2=R^2+{X_C}^2⇒X_C=Z^2-R^2$

Now we need to find impedance,

$$\begin{gathered} Z=\frac{V}{I}=\frac{\sqrt{2}V}{I_p} \\ {I}_p=0.00065A \end{gathered}$$

Putting the value of impedance , we get

$X_C=\sqrt{\frac{2V^2}{{I_p}^2}-R^2}=2617222.2$

We get capacitance as,

$C=\frac{1}{\left(2\mathrm{Ï€´Ú}\right)C}=\frac{1}{226194.67×2617222.2}=1.68×{10}^{-12}$$F$

So separation comes out to be $d=\frac{8.85×{10}^{-12}×\mathrm{π}×{0.075}^2}{1.68×{10}^{-12}}=0.093m=93mm$