Chapter 8 Class 12 Physics NCERT Solutions: Important Topics, Questions with Downloadable PDF

Q.8.1 Figure 8.6 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 0.15A.

(a) Calculate the capacitance and the rate of change 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.

Ans.8.1 Radius of the each circular plate, r = 12 cm = 0.12

Distance between the plates, d = 5 cm = 0.05 m

Charging current, I = 0.15 A

Permittivity of free space, ${ϵ}_0$ = 8.85 $\times {10}^{-12}$ $C^2{N}^{-1}{m}^{-2}$

Capacitance between two plates is given by the relation,

C = $\frac{{ϵ}_{0}A}{d}$  , where A = Area of each plate = $\pi r^2$ = $\pi \times {0.12}^2$

C = $\frac{8.85\mathrm{}\times {10}^{-12}\times \mathrm{}\pi \times {0.12}^2}{0.05}$  = 8.007 $\times {10}^{-12}$

F

Charge on each plate, q = CV, where V = potential difference across plates

Differentiating both sides w.r.t. t, we get

$\frac{dq}{dt}$ = C  $\frac{dV}{dt}$

But $\frac{dq}{dt}$  = I, therefore

$\frac{dV}{dt}=\frac{I}{C}$  = $\frac{0.15}{8.007\mathrm{}\times {10}^{-12}}$  = 1.87 $\times {10}^{10}$  V/s $$

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The displacement current across the plates is the same as the conduction current. Hence the displacement current is 0.15 A

Kirchhoff’s first rule is valid at each plate of the capacitor provided that we take the sum of conduction and displacement for current.

Q.8.2 A parallel plate capacitor (Fig. 8.7) 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 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.

Ans. 8.2 Radius of each circular plate, R = 6.0 cm = 0.06 m

Capacitance of parallel capacitor, C = 100 pF = 100  $\times {10}^{-12}$ F

Supply voltage, V = 230 V

Angular frequency,  $\omega$ = 300 rad/s

rms value of the conduction current, I =  $\frac{V}{X_c}$ , where $X_c=\mathrm{c}\mathrm{a}\mathrm{p}\mathrm{a}\mathrm{c}\mathrm{i}\mathrm{t}\mathrm{i}\mathrm{v}\mathrm{e}\mathrm{}\mathrm{r}\mathrm{e}\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{c}\mathrm{e}=\mathrm{}\frac{1}{\mathrm{\omega }\mathrm{C}}$

Hence, I = V  $\times \omega \times C$ = 230  $\times 300\times$ 100  $\times {10}^{-12}$ = 6.9  $\times {10}^{-6}$ A = 6.9 $\mu A$

Yes, the conduction current is equal to the displacement current.

Magnetic field is given as, B =  $\frac{{\mu }_{0}r}{2\pi R^2}{I}_0$ , where

${\mu }_0$ = Free space permeability = 4  $\pi \times {10}^{-7}$ N  $A^{-2}$

$I_0$ = Maximum value of current =  $\surd 2I$

r = distance between two plates on the axis = 3.0 cm = 0.03 m

then, B =  $\frac{4\pi \times {10}^{-7}\times 0.03}{2\pi \times {0.06}^2}$  $\times \surd 2\times$ 6.9  $\times {10}^{-6}$ = 1.626  $\times {10}^{-11}$ T

Ans.8.4 The electromagnetic wave travels in a vacuum along the z- direction. The electric field (E) and the magnetic field (H) are in the x-y plane. They are mutually perpendicular.

Frequency of the wave,  $\nu$ = 30 MHz = 30  $\times {10}^6$ /s

Speed of light in vacuum, c = 3  $\times {10}^8$ m/s

Wavelength of a wave is given as

$\lambda =\frac{c}{\nu }$ =  $\frac{3\mathrm{}\times {10}^8}{30\mathrm{}\times {10}^6}$ = 10 m

• $\oint \mathbf{E}\cdot d\mathbf{A}=Q/{ϵ}_0$
• $\oint \mathbf{B}\cdot d\mathbf{A}=0$
• $\oint \mathbf{E}\cdot d\mathbf{l}=-d{\Phi }_B/dt$
• $\oint \mathbf{B}\cdot d\mathbf{l}={\mu }_0{I}_c+{\mu }_0{ϵ}_0(d{\Phi }_E/dt)$

• $E_x=E_0\sin (kz-\omega t)$
• $B_y=B_0\sin (kz-\omega t)$