Coulomb's law with an $r^3$, not $r^2$, in the denominator

Question

I am reading an older physics book that my professor gave me. It is going over Coulomb's law and Gauss' theorem. However, the book gives both equations with an $r^3$, not $r^2$, in the denominator. Can somebody please explain why it is given as r^3? An image is attached for reference.

Also for equation 1-24, can somebody please explain how the middle side is equal to the right side with the del operator?

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possible duplicate of Why do we say that in Coulomb's law the force is proportional to $\frac{1}{r^{2}}$ and not $\frac{1}{r^{3}}$? — Kyle Kanos, Jul 19 '15 at 19:46

score 7 · Accepted Answer · answered Jul 19 '15 at 19:39

7

It does give "Coulomb's law" with $\frac{1}{r^3}$, it gives it in its proper vectorial form $$ \vec E \propto \frac{\vec r}{r^3}$$ which, when taking the absolute values, yields the form you are probably more familiar with $$ E \propto \frac{1}{r^2}$$ since $\lvert \vec r \rvert = r$.

answered Jul 19 '15 at 19:39

ACuriousMind

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Great, thank you for the help. Also, in the figure at the bottom of the attached image, do you know what the d(omega) stands for? – Dynamical Jul 19 '15 at 19:45
1

@DevinSchneider: Differential solid angle. – ACuriousMind Jul 19 '15 at 19:46

Coulomb's law with an $r^3$, not $r^2$, in the denominator

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