Final answer:
The electric field amplitude is 5.8 x 10^5 V/m and the magnetic field amplitude is 1.9 x 10^-3 T for an average energy density of 1.5 J/m^3. To double the energy density to 3.0 J/m^3, the field amplitudes must be increased by a factor of √2, approximately 1.41.
Step-by-step explanation:
To find the electric field amplitude (E) from the average energy density (u), one can use the following relationship in the context of an electromagnetic wave in free space:
u = ½ε0E2, where ε0 is the permittivity of free space (ε0 = 8.854 x 10-12 C2/N·m2). By rearranging this formula and solving for E, we find that:
E = √(2u/ε0)
Substituting the given energy density of 1.5 J/m3 into the equation, we get:
E = √(2 * 1.5 J/m3 / 8.854 x 10-12 C2/N·m2)
After calculating, the electric field amplitude is 5.8 x 105 V/m (rounded to two significant figures).
To find the magnetic field amplitude (B), we can use the relationship between the electric and magnetic fields in an electromagnetic wave: B = E/c, where c is the speed of light in vacuum (c = 3 x 108 m/s).
B = (5.8 x 105 V/m) / (3 x 108 m/s)
Therefore, the magnetic field amplitude is 1.9 x 10-3 T.
To double the average energy density to 3.0 J/m3, the factor by which the field amplitudes must be increased is √2, or approximately 1.41. This is because the energy density is proportional to the square of the field amplitudes, so to double the energy density, the field amplitudes must be increased by the square root of 2.