<span>Let & be the angle between B und A. We have to show that B - |B|cos & A/|A| is normal to A. Using the dot product we must have (B - |B|cos & A/|A| )·A =0.
Since the dot product is distributive with respect to addition, we can write it as:
B·A - |B|cos & A/|A|·A = |B| |A|cos & - (|B|cos & A/|A|)|A|^2 = 0, since (|B|cos & A/|A|)|A|^2 = |B| |A|cos &.
hope this helps</span>
Units is the correct answer
Answer:
s = 6.25 10⁻²² m
Explanation:
Polarizability is the separation of electric charges in a structure, in the case of the atom it is the result of the separation of positive charges in the nucleus and the electrons in their orbits, macroscopically it is approximated by
p = q s
s = p / q
let's calculate
s = 1 10⁻⁴⁰ / 1.6 10⁻¹⁹
s = 0.625 10⁻²¹ m
s = 6.25 10⁻²² m
We see that the result is much smaller than the size of the atom, therefore this simplistic model cannot be taken to an atomic scale.
