Question

series RC circuit is built with a 15 kΩ resistor and a parallel-plate capacitor with 18-cm-diameter electrodes. A 18 V, 36 kHz source drives a peak current of 0.65 mA through the circuit. Part A What is the spacing between the capacitor plates?

Answer 1

`d=1.84\ mm`

Step-by-step explanation:

Capacitance

A two parallel-plate capacitor has a capacitance of

`\displaystyle C=(\epsilon_o A)/(d)`

where

`\epsilon_o=8.85\cdot 10^(-12)\ F/m`

A = area of the plates =
`\pi r^2`

d = separation of the plates

`\displaystyle d=(\epsilon_o A)/(C)=(\epsilon_o \pi r^2)/(C)`

We need to compute C. We'll use the circuit parameters for that. The reactance of a capacitor is given by

`\displaystyle X_c=(1)/(wC)`

where w is the angular frequency

`w=2\pi f=2\pi \cdot 36000=226194.67\ rad/s`

Solving for C

`\displaystyle C=(1)/(wX_c)`

The reactance can be found knowing the total impedance of the circuit:

`Z^2=R^2+X_c^2`

Where R is the resistance,
`R=15 K\Omega=15000\Omega`. Solving for Xc

`X_c^2=Z^2-R^2`

The magnitude of the impedance is computed as the ratio of the rms voltage and rms current

`\displaystyle Z=(V)/(I)`

The rms current is the peak current Ip divided by
`√(2)`, thus

`\displaystyle Z=(√(2)V)/(I_p)`

`I_p=0.65\ mA/1000=0.00065\ A`

Now collect formulas

`\displaystyle X_c^2=Z^2-R^2=\left((√(2)V)/(I_p)\right)^2-R^2`

Or, equivalently

`\displaystyle X_c=\sqrt{(2V^2)/(I_p^2)-R^2}`

`\displaystyle X_c=\sqrt{(2\cdot 18^2)/(0.00065^2)-15000^2}`

`X_c=36176.34\ \Omega`

The capacitance is now

`\displaystyle C=(1)/(226194.67\cdot 36176.34)=1.22\cdot 10^(-10)\ F`

The radius of the plates is

`r=18\ cm/2=9 \ cm = 0.09 \ m`

The separation between the plates is

`\displaystyle d=(8.85\cdot 10^(-12) \cdot \pi\cdot 0.09^2)/(1.22\cdot 10^(-10))`

`d=0.00184\ m`

`\boxed{d=1.84\ mm}`