1. Frequency:
The frequency of a light wave is given by:
where
is the speed of light
is the wavelength of the wave
In this problem, we have light with wavelength
Substituting into the equation, we find the frequency:
2. Period:
The period of a wave is equal to the reciprocal of the frequency:
The frequency of this light wave is (found in the previous exercise), so the period is:
Answer:
Approximately (given that the magnitude of this charge is
.)
Explanation:
If a charge of magnitude is placed in an electric field of magnitude
, the magnitude of the electrostatic force on that charge would be
.
The magnitude of this charge is . Apply the unit conversion
:
.
An electric field of magnitude would exert on this charge a force with a magnitude of:
.
Note that the electric charge in this question is negative. Hence, electrostatic force on this charge would be opposite in direction to the the electric field. Since the electric field points due south, the electrostatic force on this charge would point due north.