Question

2
9) In a material for which sigma is 5 S/m and Er is 1 the electric field intensity is E =250 Sin10^10t V/m . find the conductor and displacement current densities, and the frequency at which they have equal magnitude.

Answer 1

Given that,

$\sigma= 5\ S/m$

Electric field intensity is

$E=250\sin10^{10}t$

We need to calculate the conductor current

Using formula of conductor current

$I_{c}=\sigma E$

Put the value into the formula

$I_{c}=5\times250\sin(10^{10}t)$

$I_{c}=1250\sin(10^{10}t)\ C/m^2$

We need to calculate the displacement current densities

Using formula of displacement current density

$I_{d}=\epsilon_{0}\dfrac{dE}{dt}$

Put the value into the formula

$I_{d}=8.854\times10^{-12}\times250\times10^{10}\cos(10^{10}t)$

$I_{d}=22.14\cos(10^{10}t)\ C/m^2$

The frequency at which they have equal magnitude.

$I_{c}=I_{d}$

Then, $\sigma=\omega\epsilon_{0}$

We need to calculate the frequency at which they have equal magnitude

Using formula of frequency

$\omega=\dfrac{\sigma}{\epsilon_{0}}$

Put the value into the formula

$\omega=\dfrac{5}{8.854\times10^{-12}}$

$\omega=5.64\times10^{11}\rad/sec$

$\omega=89.8\ GHz$

Hence, The conductor current is $1250\sin(10^{10}t)\ C/m^2$

The displacement current density is $22.14\cos(10^{10}t)\ C/m^2$

The frequency is 89.8 GHz.