Sea water at frequency v = 4×108 Hz has permittivity ε≈80 <math> <msub> <mrow> <mrow> <mi mathvariant="bold-italic">ε</mi> </mrow> </mrow> <mrow> <mrow> <mn>0</mn> </mrow> </mrow> </msub> </math> , permeability µ≈ <math> <msub> <mrow> <mrow> <mi mathvariant="bold-italic">μ</mi> </mrow> </mrow> <mrow> <mrow> <mn>0</mn> </mrow> </mrow> </msub> </math> and resistivity ρ = 0.25 m. Imagine a parallel plate capacitor immersed in sea water and driven by an alternating voltage source V (t)= V0sin(2 <math> <mi mathvariant="bold-italic">π</mi> <mi mathvariant="bold-italic">v</mi> <mi mathvariant="bold-italic">t</mi> </math> ). What fraction of the conduction current density is the displacement current density?

This is a long answer type question as classified in NCERT ExemplarSuppose the distance between the plates is d and applied voltage is Vt= V02Then electric field is E= V d sin(2 π v t )Jc= 1 ρ V 0 d s i n ( 2 π v t ) = J 0 c s i n 2 π v t J 0 c = V 0 ρ d Jd= ε d E d t = V 0 ρ d = ε 2 π v V 0 d cos(2 π v t ) = J0d cos 2 π v t J0d= 2 π V ε V 0 d J 0 d J 0 c = 2 πVερ = 2 π80ε0 v × 0.25 = 4 πε0 v ×10 =4/9

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Electromagnetic Waves Physics Ncert Solutions Class 12th Physics Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Eight

Sea water at frequency v = 4×10⁸ Hz has permittivity ε≈80 $ε_{0}$ , permeability µ≈ $μ_{0}$ and resistivity ρ = 0.25 m. Imagine a parallel plate capacitor immersed in sea water and driven by an alternating voltage source V (t)= V₀sin(2 $π v t$ ). What fraction of the conduction current density is the displacement current density?

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Payal Gupta | Contributor-Level 10

4 months ago

This is a long answer type question as classified in NCERT Exemplar

Suppose the distance between the plates is d and applied voltage is V_t= V₀2

Then electric field is E= $\frac{V}{d}$ sin(2 $π v t$ )

J_c= $\frac{1}{ρ} \frac{V_{0}}{d} s i n (2 π v t)$

= $J_{0}^{c} s i n 2 π v t$

$J_{0}^{c}$ = $\frac{V_{0}}{ρ d}$

J_d= $ε \frac{d E}{d t}$ = $\frac{V_{0}}{ρ d}$

= $ε \frac{2 π v V_{0}}{d}$ cos(2 $π v t$ )

= J₀^d cos 2 $π v t$

J₀^d= $\frac{2 π V_{ε} V_{0}}{d}$

$\frac{J_{0}^{d}}{J_{0}^{c}}$ = 2 $π V_{ε} ρ$ = 2 $π 80 ε_{0}$ v $\times$ 0.25 = 4 $π ε_{0}$ v $\times 10$ =4/9

This is a long answer type question as classified in NCERT ExemplarSuppose the distance between the plates is d and applied voltage is Vt= V02Then electric field is E= <math> <mfrac> <mrow> <mrow> <mi>V</mi> </mrow> </mrow> <mrow> <mrow> <mi>d</mi> </mrow> </mrow> </mfrac> </math> sin(2 <math> <mi>π</mi> <mi>v</mi> <mi>t</mi> </math> )Jc= <math> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mi>ρ</mi> </mrow> </mrow> </mfrac> <mfrac> <mrow> <mrow> <msub> <mrow> <mrow> <mi>V</mi> </mrow> </mrow> <mrow> <mrow> <mn>0</mn> </mrow> </mrow> </msub> </mrow> </mrow> <mrow> <mrow> <mi>d</mi> </mrow> </mrow> </mfrac> <mi>s</mi> <mi>i</mi> <mi>n</mi> <mo>(</mo> <mn>2</mn> <mi>π</mi> <mi>v</mi> <mi>t</mi> <mo>)</mo> </math>   =  <math> <msubsup> <mrow> <mrow> <mi>J</mi> </mrow> </mrow> <mrow> <mrow> <mn>0</mn> </mrow> </mrow> <mrow> <mrow> <mi>c</mi> </mrow> </mrow> </msubsup> <mi>s</mi> <mi>i</mi> <mi>n</mi> <mn>2</mn> <mi>π</mi> <mi>v</mi> <mi>t</mi> </math> <math> <msubsup> <mrow> <mrow> <mi>J</mi> </mrow> </mrow> <mrow> <mrow> <mn>0</mn> </mrow> </mrow> <mrow> <mrow> <mi>c</mi> </mrow> </mrow> </msubsup> </math>   = <math> <mfrac> <mrow> <mrow> <msub> <mrow> <mrow> <mi>V</mi> </mrow> </mrow> <mrow> <mrow> <mn>0</mn> </mrow> </mrow> </msub> </mrow> </mrow> <mrow> <mrow> <mi>ρ</mi> <mi>d</mi> </mrow> </mrow> </mfrac> </math> Jd= <math> <mi>ε</mi> <mfrac> <mrow> <mrow> <mi>d</mi> <mi>E</mi> </mrow> </mrow> <mrow> <mrow> <mi>d</mi> <mi>t</mi> </mrow> </mrow> </mfrac> </math> = <math> <mfrac> <mrow> <mrow> <msub> <mrow> <mrow> <mi>V</mi> </mrow> </mrow> <mrow> <mrow> <mn>0</mn> </mrow> </mrow> </msub> </mrow> </mrow> <mrow> <mrow> <mi>ρ</mi> <mi>d</mi> </mrow> </mrow> </mfrac> </math>    = <math> <mi>ε</mi> <mfrac> <mrow> <mrow> <mn>2</mn> <mi>π</mi> <mi>v</mi> <msub> <mrow> <mrow> <mi>V</mi> </mrow> </mrow> <mrow> <mrow> <mn>0</mn> </mrow> </mrow> </msub> </mrow> </mrow> <mrow> <mrow> <mi>d</mi> </mrow> </mrow> </mfrac> </math> cos(2 <math> <mi>π</mi> <mi>v</mi> <mi>t</mi> </math> )   = J0d cos 2 <math> <mi>π</mi> <mi>v</mi> <mi>t</mi> </math> J0d= <math> <mfrac> <mrow> <mrow> <mn>2</mn> <mi>π</mi> <msub> <mrow> <mrow> <mi>V</mi> </mrow> </mrow> <mrow> <mrow> <mi>ε</mi> </mrow> </mrow> </msub> <msub> <mrow> <mrow> <mi>V</mi> </mrow> </mrow> <mrow> <mrow> <mn>0</mn> </mrow> </mrow> </msub> </mrow> </mrow> <mrow> <mrow> <mi>d</mi> </mrow> </mrow> </mfrac> </math> <math> <mfrac> <mrow> <mrow> <msubsup> <mrow> <mrow> <mi>J</mi> </mrow> </mrow> <mrow> <mrow> <mn>0</mn> </mrow> </mrow> <mrow> <mrow> <mi>d</mi> </mrow> </mrow> </msubsup> </mrow> </mrow> <mrow> <mrow> <msubsup> <mrow> <mrow> <mi>J</mi> </mrow> </mrow> <mrow> <mrow> <mn>0</mn> </mrow> </mrow> <mrow> <mrow> <mi>c</mi> </mrow> </mrow> </msubsup> </mrow> </mrow> </mfrac> </math> = 2 <math><mi>π</mi><msub><mrow><mrow><mi>V</mi></mrow></mrow><mrow><mrow><mi>ε</mi></mrow></mrow></msub><mi>ρ</mi></math> = 2 <math><mi>π</mi><mn>80</mn><msub><mrow><mrow><mi>ε</mi></mrow></mrow><mrow><mrow><mn>0</mn></mrow></mrow></msub></math> v <math><mo>×</mo></math> 0.25 = 4 <math><mi>π</mi><msub><mrow><mrow><mi>ε</mi></mrow></mrow><mrow><mrow><mn>0</mn></mrow></mrow></msub></math> v <math><mo>×</mo><mn>10</mn></math> =4/9

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