Recent content by mattkuk

So using dr/dt=v and dv/dt=a and a=kq^2/mr^2 you could integrate a with respect to t twice to find v and then r in terms of a time t?

Yes assuming a classically described proton. What is known about position in relation to acceleration is at time t = zero seconds, acceleration is equal to zero. At the instant the protons are released acceleration would be F/m which would be (k*q^2)/(m*r^2) where k = 8.89*10^9, q =...

Homework Statement If two protons start out at rest relative to each other, separated by 0.1 nanometers in vacuum, and then they are released, calculate their relative positions after 10 seconds. Homework Equations F = (kq^2)/r^2 = m*a The Attempt at a Solution I am fairly...