Question

electrostatics question

Answer 1

taking the specialised case of an equilateral triangle with one corner at (0,0) and set symmetrically on x axis with other corners at \((\dfrac{\sqrt{3}}{2}, \pm \dfrac{1}{2})\) , the potential of charge q at point (x,0) is

\(U(x) = k q Q \left( \dfrac{1}{x} + \dfrac{2}{\sqrt{(\frac{\sqrt{3}}{2} - x)^2 + \frac{1}{4}}}\right)\)

if you plot this you get the following

https://gyazo.com/abd58f564d5df4e996a115be0f64a374

so there is a mimimum energy level which coincides with \((\dfrac{\sqrt{3}}{3}, 0)\)..... ie the centroid of the triangle

So for small displacement, it is a kind of well, ........, ***and you should get oscillations***


for larger displacements that allow it to climb the peak at x = 0.7414, the particle will eventually shoot off to the right......

the calculus backs up these numbers -- this is the solution for the extrema for \(\dfrac{dU}{dx} = 0\)

https://www.wolframalpha.com/input/?i=x%5E2+2+(sqrt(3)+%2F+2+-+x)%3D+((sqrt(3)%2F2+-+x)%5E2+%2B+1%2F4)%5E(3%2F2)

.....though i think this was all pretty guessable anyway :p