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EastWind [94]
3 years ago
10

In a material for which o=5.0 S/m and e, =1 the electric field intensity is E- 250 sin 10" (V/m). Find the conduction and displa

cement current densities, and the frequency at they have equal magnitudes.
Physics
1 answer:
Elodia [21]3 years ago
5 0

Answer:

Explanation:

Given that:

The conductivity of the material \sigma = 5.0 s/m

The  \ relative  \ permittivity \  of \  the \  material {\varepsilon_r} = 1 ; &

The electric field intensity of the material E = 250 sin(10¹⁰ t) V/m

(a) The conduction current density ( J_c) = \sigma E

= 5.0 \times 250 \ sin ( 10^{10} \ t )

\mathbf{ = 1250 \ sin (10 ^{10} t ) \ A/m^2 }

(b)  Displacement  current density (J_d) = \varepsilon _d * \dfrac{\delta E}{\delta t}

Recall that:

\varepsilon _o  = 8.854 \times 10^{-12}

∴

(J_d) =  8.854 \times 10^{-12} \times \dfrac{d}{dt} \times  (250  \ sin \ (10^{10} \ t))

(J_d) = 8.854 \times 10^{-12} \times 250 \times 10^{10} \times \ cos \ (10^{10} \ t)

(J_d) = 22.135 \ cos \  10^{10} \ t \ A/m^2

(c) The frequency at which J_c  \  and  \  J_d will have the same magnitude is:

f = \dfrac{\sigma}{2 \pi \varepsilon_o \varepsilon_r}

By substitution

f = 18 \times 10^9 \times \dfrac{\sigma }{\varepsilon_r}

f = 18\times 10^9 \times \dfrac{5}{1 }

f = 90 GHz

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