Question

A series $RC$circuit is built with a $12k\Omega$resistor and a parallel-plate capacitor with$15-cm-$diameter electrodes. Alocalid="1649923196754" $12V,36kHz$source drives a peak current of$0.65mA$through the circuit. What is the spacing between the capacitor plates?

Short answer

The spacing between capacitor plate is$93mm$

Step-by-step solution

Given Information

We are given that resistance $12K\Omega$, 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.

Simplify

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

$$\begin{gathered} C=\frac{{\epsilon }_{0}A}{d}=\frac{{\epsilon }_0{\mathrm{\pi r}}^2}{d} \\ \Rightarrow d=\frac{{\epsilon }_0{\mathrm{\pi r}}^2}{C} \end{gathered}$$

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

$Z^2=R^2+{X_C}^2\Rightarrow X_C=Z^2-R^2$

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{\pi f}\right)C}=\frac{1}{226194.67\times 2617222.2}=1.68\times {10}^{-12}$$F$

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