In a coaxial transmission line (εr = 1), the electric field intensity is given by:
$$E = \frac{{100}}{\rho }\cos \left( {{{10}^9}t - 6z} \right){u_p}{\text{V/m}}$$
The displacement current density is
A. $$ - \frac{{100}}{\rho }\sin \left( {{{10}^9}t - 6z} \right){u_p}{\text{A/}}{{\text{m}}^2}$$
B. $$\frac{{116}}{\rho }\sin \left( {{{10}^9}t - 6z} \right){u_p}{\text{A/}}{{\text{m}}^2}$$
C. $$ - \frac{{0.9}}{\rho }\sin \left( {{{10}^9}t - 6z} \right){u_p}{\text{A/}}{{\text{m}}^2}$$
D. $$ - \frac{{216}}{\rho }\cos \left( {{{10}^9}t - 6z} \right){u_p}{\text{A/}}{{\text{m}}^2}$$
Answer: Option C
This Question Belongs to Electronics And Communications Engineering >> Electromagnetic Field Theory
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