# 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.