Answer:
c = 1 / √(ε₀*μ₀)
Explanation:
The speed of the electromagnetic wave in free space is given in terms of the permeability and the permittivity of free space by
c = 1 / √(ε₀*μ₀)
where the permeability of free space (μ₀) is a physical constant used often in electromagnetism and ε₀ is the permittivity of free space (a physical constant).
Answer:
-54.12 V
Explanation:
The work done by this force is equal to the difference between the final value and the initial value of the energy. Since the charge starts from the rest its initial kinetic energy is zero.

The change in electrostatic potential energy
, of one point charge q is defined as the product of the charge and the potential difference.

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To develop this problem it is necessary to apply the concepts related to the Cross Product of two vectors as well as to obtain the angle through the magnitude of the angles.
The vector product between the Force and the radius allows us to obtain the torque, in this way,





Therefore the torque on the particle about the origen is 50k
PART B) To find the angle between two vectors we apply the definition of the dot product based on the vector quantities, that is,





Therefore the angle between the ratio and the force is 103.88°
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