In a plane electromagnetic wave the electric field osciallates sinusoidally at a frequency of `2.0 xx 10^(10)` and amplitude `48 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 ` vecE ` field equals to the average energy density of the ` vecB ` field .
` [ c= 3.0 xx 10^8 m s^(-1)`]

Explanation

To solve the given problem step by step, we will address each part of the question systematically.

### Given Data:
- Frequency of the electric field, \( f = 2.0 \times 10^{10} \, \text{Hz} \)
- Amplitude of the electric field, \( E_0 = 48 \, \text{V/m} \)
- Speed of light, \( c = 3.0 \times 10^8 \, \text{m/s} \)

### (a) Finding the Wavelength of the Wave
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