In a coaxial transmission line (εr = 1), the electric field intensity is given by:\(E = \frac{{100}}{\rho }\cos \left( {{{10}^9}t - 6z} \right){u_p}V/m\)The displacement current density is:

\( - \frac{{216}}{\rho }\cos \left( {{{10}^9}t - 6z} \right){u_p}A/{m^2}\)

\( - \frac{{100}}{\rho }\sin \left( {{{10}^9}t - 6z} \right){u_p}A/{m^2}\)

\(\frac{{116}}{\rho }\sin \left( {{{10}^9}t - 6z} \right){u_p}A/{m^2}\)

\( - \frac{{0.9}}{\rho }\sin \left( {{{10}^9}t - 6z} \right){u_p}A/m\)

In a coaxial transmission line (εr = 1), the electric field intensity

1.	In a coaxial transmission line (εr = 1), the electric field intensity is given by:\(E = \frac{{100}}{\rho }\cos \left( {{{10}^9}t - 6z} \right){u_p}V/m\)The displacement current density is:
A.	\( - \frac{{100}}{\rho }\sin \left( {{{10}^9}t - 6z} \right){u_p}A/{m^2}\)
B.	\(\frac{{116}}{\rho }\sin \left( {{{10}^9}t - 6z} \right){u_p}A/{m^2}\)
C.	\( - \frac{{0.9}}{\rho }\sin \left( {{{10}^9}t - 6z} \right){u_p}A/m\)
D.	\( - \frac{{216}}{\rho }\cos \left( {{{10}^9}t - 6z} \right){u_p}A/{m^2}\)
Answer» D. \( - \frac{{216}}{\rho }\cos \left( {{{10}^9}t - 6z} \right){u_p}A/{m^2}\)

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In a coaxial transmission line (εr = 1), the electric field intensity is given by:\(E = \frac{{100}}{\rho }\cos \left( {{{10}^9}t - 6z} \right){u_p}V/m\)The displacement current density is:

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