Question

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 displacement current densities, and the frequency at they have equal magnitudes.

Answer 1

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