Answer
(a) $\lambda = 10.0~nm$
(b) $f = 3.00\times 10^{16}~Hz$
(c) $B_0 = 6.67\times 10^{-8}~T$
Work Step by Step
We can write the general equation for an electric field:
$E = E_0~cos(kx-\omega t)$
We can write the equation for this electric field:
$E_y = (20.0~V/m)~cos[(6.28\times 10^8)x-\omega t]$
(a) We can find the wavelength:
$\lambda = \frac{2\pi}{k}$
$\lambda = \frac{2\pi}{6.28\times 10^8}$
$\lambda = 1.00\times 10^{-8}~m$
$\lambda = 10.0~nm$
(b) We can find the frequency:
$f = \frac{c}{\lambda}$
$f = \frac{3.0\times 10^8~m/s}{1.00\times 10^{-8}~m}$
$f = 3.00\times 10^{16}~Hz$
(c) We can find the magnetic field amplitude:
$B_0 = \frac{E_0}{c}$
$B_0 = \frac{20.0~V/m}{3.0\times 10^8~m/s}$
$B_0 = 6.67\times 10^{-8}~T$
