Question

A) A parallel plate capacitor made of circular plates each of radius $R=6.0\text{cm}$ has a capacitance C = 100 pF. The capacitor is connected to a 230 V ac supply with a (angular) frequency of $300\text{rad/s}$. What is the $\text{rms}$ value of the conduction current?


B) A parallel plate capacitor made of circular plates each of radius $R=6.0\text{cm}$ has a capacitance $C=100pF$. The capacitor is connected to a $230V$ ac supply with a (angular) frequency of $300\text{rad/s}$ Is the conduction current equal to the displacement current?


C) A parallel plate capacitor made of circular plates each of radius $R=6.0\text{cm}$ has a capacitance $C=100pF$. The capacitor is connected to a 230 V ac supply with a (angular) frequency of $300\text{rad/s}$. Determine the amplitude of B at a point $3.0\text{cm}$ from the axis between the plates? (rms value of conduction current $6.9\mu A)$?

Answer 1

A) Formula used: $I=\frac{v}{X_c}$
Given, radius of each circular plate,
$R=6\text{cm}=0.06m$
Capacitance of parallel plate capacitor,
$C=100pF=100\times {10}^{-12}F$
Supply voltage, $V=230V$
Angular frequency, $\omega =300\text{rad/s}$
Rms value of conduction current,
$I=\frac{V}{x_C}$ here, $X_c$ = Capacitive reactance
$X_c=\frac{1}{\omega C}$
$\Rightarrow I=V\times \omega C$
$I=230\times 300\times 100\times {10}^{-12}$
$I=6.9\times {10}^{-6}A$
$I=6.9\mu A$ Hence, the rms value of conduction current is $6.9\mu A$.
Final answer: $6.9\mu A$

B) The displacement current across the plates, $i_d={ϵ}_0\left(\frac{d{\varphi }_E}{dt}\right)$
Using Gauss's law, electric flux ${\varphi }_E=\frac{q}{{ϵ}_0}$
$i_d={ϵ}_0\left(\frac{1}{{ϵ}_0}\frac{dq}{dt}\right)=\frac{dq}{dt}$
$i_d=\frac{dq}{dt}$ = conduction current
Hence, conduction current is equal to displacement current.
Final answer: Yes

C) Formula use: $B_0=\frac{{\mu }_{0}d{I}_0}{2\pi R^2}$
The formula goes through even if $i_d$ ( and therefore B) oscillates in time. The formula shows they oscillate in phase.
Since $i_d=i$, we have $B_0=\frac{{\mu }_{0}d{I}_0}{2\pi R^2}$ where $B_0$ and $i_0$ are the amplitudes of the oscillating magnetic field and current, respectively.
Step 1: Find maximum value of current.
Rms value of conduction current,
$I=6.9\mu A$
Maximum value of current, $I_0=I\sqrt{2}$
$I_0=\sqrt{2}\times 6.9\times {10}^{-6}A=9.7566\times {10}^{-6}A$
Step 2: Find amplitude of magnetic field.
Given, Distance between the plates from the axis $d=3\text{cm}=0.03m$
Amplitude of magnetic field, $B_0=\frac{{\mu }_{0}d{I}_0}{2\pi R^2}$
Here, ${\mu }_0=4\pi \times {10}^{-7}N/A^2$
$B_0=\frac{4\pi \times {10}^{-7}\times 0.03\times 9.7566\times {10}^{-6}}{2\pi \times {0.06}^2}$
$=1.63\times {10}^{-11}T$
Hence, the magnetic field at that point is
$1.63\times {10}^{-11}T$.
Final answer : $1.63\times {10}^{-11}T$