In a plane electromagnetic wave,the electric field oscillates sinusoidally at a frequency of $2.0 \times 10^{10} \; Hz$ and amplitude $48 \; V m^{-1}$.
$(a)$ What is the wavelength of the wave?
$(b)$ What is the amplitude of the oscillating magnetic field?
$(c)$ Show that the average energy density of the $E$ field equals the average energy density of the $B$ field. $[c = 3 \times 10^{8} \; m s^{-1}]$.

(A) Given:
Frequency $v = 2.0 \times 10^{10} \; Hz$
Electric field amplitude $E_{0} = 48 \; V m^{-1}$
Speed of light $c = 3 \times 10^{8} \; m s^{-1}$
$(a)$ Wavelength $\lambda = \frac{c}{v} = \frac{3 \times 10^{8}}{2.0 \times 10^{10}} = 0.015 \; m$.
$(b)$ Magnetic field amplitude $B_{0} = \frac{E_{0}}{c} = \frac{48}{3 \times 10^{8}} = 1.6 \times 10^{-7} \; T$.
$(c)$ Energy density of the electric field is $U_{E} = \frac{1}{2} \epsilon_{0} E^{2}$ and for the magnetic field is $U_{B} = \frac{B^{2}}{2 \mu_{0}}$.
Using $E = cB$ and $c = \frac{1}{\sqrt{\epsilon_{0} \mu_{0}}}$,we have $E^{2} = c^{2} B^{2} = \frac{B^{2}}{\epsilon_{0} \mu_{0}}$.
Thus,$\epsilon_{0} E^{2} = \frac{B^{2}}{\mu_{0}}$.
Dividing by $2$,we get $\frac{1}{2} \epsilon_{0} E^{2} = \frac{B^{2}}{2 \mu_{0}}$,which implies $U_{E} = U_{B}$.