Question 178: EM waves

Question

Seawater at a frequency f = 9 $$	imes$$ 10
2
Hz, has permittivity $$\varepsilon $$ = 80$$\varepsilon $$
0
and resistivity $$ho$$ = 0.25 $$\Omega$$m. Imagine a parallel plate capacitor is immersed in seawater and is driven by an alternating voltage source V(t) = V
0
sin(2$$\pi$$ft). Then the conduction current density becomes 10
x
times the displacement current density after time t = $${1 \over {800}}$$s. The value of x is _____________. (Given : $${1 \over {4\pi {\varepsilon _0}}} = 9 	imes {10^9}$$ Nm
2
C
$$-$$2
)

Accepted Answer

Given f = 9 $$	imes$$ 10
2
Hz
$$ \in $$ = $$ \in $$
0
$$ \in $$
r
$$ \in $$ = 80 $$ \in $$
0
So $$ \in $$
r
= 80
$$ho $$ = 0.25 $$\Omega$$m
V(t) = V
0
sin (2$$\pi$$ft)
$${I_d} = {{dq} \over {dt}} = {{cdv} \over {dt}}$$
$${I_d} = {{{ \in _0}{ \in _r}A} \over d}{d \over {dt}}({v_0}\sin (2\pi ft))$$
$${I_d} = {{{ \in _0}{ \in _r}A} \over d}{V_0}(2\pi f)\cos (2\pi ft)$$ .......... (1)
& $${I_c} = {V \over R}$$
$${I_c} = {{{V_0}\sin (2\pi ft)} \over {ho {d \over A}}} = {{A{v_0}\sin (2\pi ft)} \over {ho d}}$$ ....... (2)
divide equation (1) and (2)
$${{{I_d}} \over {{I_c}}} = { \in _0}{ \in _r}2\pi f(ho )\cot (2\pi ft)$$
$${{{I_d}} \over {{I_c}}} = {1 \over {4\pi  	imes 9 	imes {{10}^9}}} 	imes 80 	imes 2\pi  	imes 9 	imes {10^2} 	imes (0.25) 	imes \cot (2\pi  	imes 9 	imes {10^2} 	imes {1 \over {800}})$$
$$ = {{{{10}^3}} \over {{{10}^9}}}\left( {\cot \left( {{{9\pi } \over 4}} ight)} ight)$$
$$ = {{{{10}^3}} \over {{{10}^9}}}$$
$${{{I_d}} \over {{I_c}}} = {1 \over {{{10}^6}}}$$
$${I_c} = {10^6}{I_d}$$
So x = 6

EM waves

Integer / Numeric Answer Type