Recent content by jo2jo

Thanks Philip, The reason I went to First Order DE is I wanted to understand your way of solving DE rather than just get an answer, and that is why I asked for any online link to help in that. The main part that I don't get in your Second Order answer is the lhs in: \int{...

Thanks Philip, May I just go a step back and check the DE, really all the online sources I checked say follow certain format (blindly), but your method seems more generic with logical steps. I just need to clarify the step after multiplying both sides of the (=) by dr/dt, let's take first...

Many thanks Philip, I did actually the integration now, and that looks sooo messy, I mean because the result of the integration is very mixed up. The rhs is just , right? \sqrt{\frac{2kq^2}{m}}t + c I rearranged the lhs to (I just used b instease of r0 less confusion)...

@ Philip,, wow! If I progress from where you stopped: sqrt(m/2kq2)v= sqrt(1/r0–1/r) Lets just focus on v= sqrt(1/r0–1/r) dr/dt = sqrt(1/r0–1/r) (just to check, r0 is constant now, right?) Integrating both sides with respect to (t) gives...

Thanks everyone, To be honest I thought calculating the drift of two moving charged particles is something that has been done already. I could not find any online source to discuss this yet! @tiny-tim, how comes (quoting: that's the solution to d2r/dt2 = kq2r2/m)? I mean the...

Gosh, I went to do the second order differential equation, do you agree with this?: d2r/dt2 = kq2/mr2 Second order DE: r(t) = A e√(m/kq^2)t + B e-√(m/kq^2)t Notice (t) is outside the squared root, i just could not do it in the notation here. I feel I missed it up! especially the...

I am trying to workout the drift of a charged particle from another particle using coulomb law. but the problem is the further the particles move, the less the force between them, so how can I work out the drift in such case? We know that the force between two charged particles is: F = (k Q1...

Think of salty water does NOT resist the flow of current, while tap water resist the flow of current. So when you touch a live wire via salty water you get a full shock because this salty water does not resist the current flowing toward your body. However if you have tap water (more resistive)...