Electromagnetic Waves

Physics

NCERT

1   In the below figure shows a capacitor made of two circular plates each of radius $12 cm$, and separated by $5.0 cm.$ The capacitor is being charged by an external source (not shown in the figure). The charging current is constant and equal to $0.15 A$.$\\$ (a) Calculate the capacitance and the rate of charge of potential difference between the plates.$\\$ (b) Obtain the displacement current across the plates.$\\$ (c) Is Kirchhoff’s first rule (junction rule) valid at each plate of the capacitor? Explain.

Solution :

Radius of each circular plate,$ r = 12 cm = 0.12 m$$\\$ Distance between the plates, $d = 5 cm = 0.05 m$$\\$ Charging current, $I = 0.15 A$$\\$ Permittivity of free space, $\varepsilon_0 = 8.85*10^{-12} C^2 N^{-1} m^{-2} $$\\$ (a) Capacitance between the two plates is given by the relation,$\\$ $C=\dfrac{ \varepsilon_0A}{d}$$\\$ Where,$\\$ $A=$Area of each plate $=\pi r^2.$$\\$ $C=\dfrac{\varepsilon_0\pi r^2}{d}\\ \dfrac{8.85*10^{-12}F}=80.032pF$$\\$ Charge on each plate, $q = CV$$\\$ Where, $V =$ Potential difference across the plates$\\$ Differentiation on both sides with respect to time (t) gives:$\\$ $\dfrac{dq}{dt}=C\dfrac{dV}{dt}$$\\$ But,$\dfrac{dq}{dt}=$current(I)$\\$ $\therefore \dfrac{d V}{dt}=\dfrac{I}{C}\\ \implies \dfrac{0.15}{80.032*10^{-12}}=1.87*10^9V/s$$\\$ Therefore, the change in potential difference between the plates is $1.87 *10^ 9 V / s .$$\\$ (b) The displacement current across the plates is the same as the conduction current. Hence, the displacement current, $i_d$ is $0.15 A.$$\\$ (c) Yes$\\$ Kirchhoff’s first rule is valid at each plate of the capacitor provided that we take the sum of conduction and displacement for current.

2   A parallel plate capacitor (figure) made of circular plates each of radius $R = 6.0 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^{-1 }.$$\\$ (a) What is the rms value of the conduction current?$\\$ (b) Is the conduction current equal to the displacement current?$\\$ (c) Determine the amplitude of $B$ at a point $3.0 cm$ from the axis between the plates.

Solution :

Radius of each circular plate, $R = 6.0 cm = 0.06 m$$\\$ Capacitance of a parallel plate capacitor, $C = 100 pF = 100 × 10 -12 F$$\\$ Supply voltage, $V = 230 V$$\\$ Angular frequency, $\omega= 300 \text{rad} s^{-1}$ (a) Rms value of conduction current,$I=\dfrac{V}{X_c}$$\\$ Where,$\\$ $X_c=$Capacitive reactance$\\$ $=\dfrac{1}{\omega C}\\ \therefore I=V*\omega C\\ =230*300*100*10^{-12}\\ =6.9*10^{-6}A\\ =6.9 \mu A$$\\$ Hence, the rms value of conduction current is $= 6.9 \mu A .$$\\$ (b) Yes, conduction current is equal to displacement current.$\\$ (c) Magnetic field is given as:$\\$ $B=\dfrac{\mu_0 r}{2 \pi R^2}I_0$ where,$\\$ $\mu_ 0 =$ Free space permeability $=4\pi *10{-7} N A^{- 2}$$\\$ $I_0 =$ Maximum value of current$ = \sqrt{2I}$$\\$ $r =$ Distance between the plates from the axis $= 3.0 cm = 0.03 m$$\\$ $\therefore B=\dfrac{4 \pi*10^{-7}*0.03*\sqrt{2}*6.9*10^{-6}}{2\pi *(0.06)^2}\\ =1.63*10^{-11}T$$\\$ Hence, the magnetic field at that point is $1.63*10^{-11 }T .$