Q. The electric field associated with an $e$.m. wave in vacuum is given by $\overrightarrow{E} = \widehat{i}\ 40\cos{(kz - 6 \times 10^{8}t)}$, where $E,z$ and $t$ are in $volt/m$, meter and seconds respectively. The value of wave vector $k$ is
A) $2m^{- 1}$
B) $0.5m^{- 1}$
C) $6m^{- 1}$
D) $3m^{- 1}$
Explanation
(a) $\omega = 6 \times 10^{8}$
$k = \frac{\omega}{v} = \frac{6 \times 10^{8}}{3 \times 10^{8}} = 2m^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.