A sinusoidal electromagnetic wave is propagating in a vacuum in the z-direction. Part A If at a particular instant and at a certain point in space the electric field is in the x-direction and has a magnitude of 4.70 V/mV/m, what is the magnitude of the magnetic field of the wave at this same point in space and instant in time

the magnitude of the magnetic field is 1.23 × 10⁻⁸ T  { in the +y direction }

Explanation:

Given the data in the question;

the relation between the electric field and the magnetic field in an electromagnetic wave can be written as follows;

c = E₀/B₀

where c is speed of light, E₀ is the amplitude of the electric field and B₀ is the amplitude of the magnetic field.

given that;  E₀ = 4.70 V/m, we know that; speed of light c = 3 × 10⁸ m/s

so from our relation; c = E₀/B₀

B₀ = E₀/c

we substitute

B₀ = 4.70 / 3 × 10⁸

B₀ = 1.23 × 10⁻⁸ T  { in the +y direction }

Therefore, the magnitude of the magnetic field is 1.23 × 10⁻⁸ T  { in the +y direction }

two charges are fixed in place on the x-axis as shown. the charge at x = 0 is negative and has magnitude q = 2 μcoulombs (1 μcoulomb = 10-6 coulombs). the charge at x = d, (d = +12 cm), is positive and has magnitude q = 4 μcoulombs. find a point x = xo along the x-axis at which the electric field is zero.