Sea water at frequency v = 4*108 Hz has permittivity ε≈80 , permeability µ≈ 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 ). What fraction of the conduction current density is the displacement current density?
Sea water at frequency v = 4*108 Hz has permittivity ε≈80 , permeability µ≈ 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 ). What fraction of the conduction current density is the displacement current density?
This is a long answer type question as classified in NCERT Exemplar
Suppose the distance between the plates is d and applied voltage is Vt= V02
Then electric field is E= sin(2 )
Jc=
=
=
Jd= =
= cos(2 )
= J0d cos 2
J0d=
= 2 = 2 v 0.25 = 4 v =4/9
Similar Questions for you
According to definition of displacement current, we can write
B0 = 42.9 × 10-9
42.9 Ans
s? = -? + k? ∴ s? = (-? +k? )/√2
I = P/ (4πr²) . (i)
and I = (1/2)ε? E? ²c . (ii)
From (i) and (ii)
P/ (4πr²) = (1/2)ε? E? ²c
E? = √ (2P/ (4πε? r²c) ; E? = √ (9×10? ×2×15)/ (15×15×10? ×3×10? ) = 2×10? ³ V/m
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Physics Ncert Solutions Class 12th 2023
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