# Intro - Inverse-Square Recap Welcome to E&M! I'll be your host, and today we'll start off with a bang - just like an overloaded capacitor. Do you remember anything from [[Newton's Law of Gravitation - The Intro|gravitation?]] A prevailing theme in this topic is the concept of the *inverse-square law* - where the intensity of a force from a point source decreases proportionally to the distance from the source - as shown: $F \propto \frac{1}{r^2}$ We'll assume this force is between two point sources - no [[Canonical Pertubation Theory (Mechanics)|chaos theory]] just yet! Let's forge on - we can imagine each length we use as giving us an arbitrary sphere. This'll give us a discrepancy of around $4\pi$ every time $r$ increases by 1 - showing us that the value for our constant is $\frac{1}{4\pi}$. %%arbitrary diagram%% *Feel free to take a look at gravity if you haven't - we'll use it there too!* [[Newton's Law of Gravitation - The Intro]] # Coulomb's Law Back to E&M. Simply put, Coulomb's law is a revised version of [[Newton's Law of Gravitation - The Intro#Newton's Law of Gravitation|Newton's Law of Gravitation]], as you can see below: $F = k \frac{q_{1}q_{2}}{r^2}$ Where $k$ is Coulomb's constant, an arbitrary value also given as: $k = \frac{1}{4\pi\mu_{0} \epsilon_{0}}$ Where $\mu_0$ and $\epsilon_0$ are the permeabilities and permittivities of free space - more on those later. If they're point charges (for most cases), we'll take the constant as: $k = \frac{1}{4\pi\epsilon_{0}}$ Simple as that! There are other common forms of this we might choose to employ - being: $F = \frac{q^2}{4\pi\epsilon_{0}r^2}$ Ignoring the constants for now (we'll be back - swear!) that's practically it! You'll see other relationships with [[|the intensity of light]] among other things. # Example Problems >[!Example]- Module 21 Problem 42 >In Fig. 21-39, two tiny conducting balls of identical mass m and identical charge q hang from nonconducting threads of length L.Assume that u is so small that tan u can be replaced by its approximate equal, sin u. >(a) Show that >$x = (\frac{q^2L}{2\pi \epsilon_{0}mg})^\frac{1}{3}$ >gives the equilibrium separation x of the balls. >(b) If L = 120 cm, m = 10 g, and x = 5.0 cm, what is |q|? >**Solution:** >*a)* >Let's logic this first. Each of the particles have its own mass $mg$ and an electrostatic repulsive force acting perpendicular to it. Diagram for clarity: > >![[Pasted image 20231023230902.png]] > >Point made. We're given a bunch of nice assumptions by the question, so let's use them to create an identity: >$\tan \theta \approx \sin \theta \approx \theta \approx \frac{x/2}{L}$ >With the forces we have we can substitute $\tan \theta$ for: >$\tan \theta = \frac{F}{mg}$ >Use Coulomb's law! The electrostatic force $F$ is therefore: >$F = k \frac{q^2}{x^2} = \frac{1}{4\pi \epsilon_{0}} \frac{q^2}{x^2}$ >Plug this into the fraction above to get: >$\frac{{\frac{1}{4\pi \epsilon_{0}} \frac{q^2}{x^2}}}{mg} = \frac{x/2}{L}$ >$\frac{2{\frac{1}{4\pi \epsilon_{0}} \frac{q^2}{x^2}}L}{mg} = x$ >$\frac{2q^2L}{4\pi\epsilon_{0}x^2mg} = x$ >Isolate the $x$ - remember, we're proving the expression for it! >$\frac{2q^2L}{4\pi\epsilon_{0}mg} = x^3$ >Cleaning up our expression gives us our answer: >$\sqrt[3]{ \frac{q^2L}{2\pi\epsilon_{0}mg} } = x$ > >*b)* >Substitute numbers! Let's # What's Next? If you don't feel satisfied with how I've ignored the two constants (permeability and permittivity of free space), you can skip straight to where I go over them in more detail, namely: 1. Permeability: [[Current (ElectroMagnetism)]] 2. Permittivity: [[Capacitance (ElectroMagnetism)]]