Recent content by Rob2024

The component.

hmm there cannot be any net torque either, this implies the point mass cannot exert horizontal force on the rod. This means the point mass travels at constant horizontal velocity until the rod becomes horizontal. I think I see what's happening.

Because the system has no friction, when the point mass falls, I think both angular momentum and energy are conserved. Further because the rod is massless, there cannot be a vertical force from mass on the rod. The difficulty is to determine the property of tension. It determines the subsequent...

I don't understand what the book meant as 'longitudinal current'. Is this in the axial direction (z direction) or the azimuthal direction (\phi direction)? It would only make sense if the current is in the axial direction. A confirmation of my guess would be appreciated. A picture from the book...

Thanks, this worked. I used the potential energy incorrectly.

I initially tried energy method, I realized this cannot be minimized. ##U = \frac{kq^2}{2r} - \frac{2 k qq'}{r} , q' = \frac{2e r^3}{R^3},q = e ## ##U = \frac{kq^2}{2r} - \frac{2 k q\frac{2e r^3}{R^3}}{r} ## ##= k q^2 ( 1/2r -4 r^2/R^3) ## ##= \frac{1}{2} k q^2 ( r - 8 r^2/R^3) ## ##U' \sim...

I am familiar with image charge. It's what I used in the reasoning since image charge for spherical conductor is well understood.

I know this question is not constrained too well. Since it's not constrained too well. I thought I could just use a ball conductor, it induces two image charges which would increase the force experienced by either one of the hanging charges. Therefore the answer is that they will move closer to...

#5 has the answer to your question. I understand you are alluding to use the solution from part a) but I am not too sure if that's correct since part a)'s solution uses the condition ##Q## is constant. Thanks for the help. I'll think about this some more.

The problem I have with the argument of trying to use the first solution is that the first problem uses a fixed point charge. I am not too sure if we can turn a linear charge distribution into the point charge...i.e. I am not comfortable to say if we have a curve embedded in a uniformly charged...

they need to be symmetric.

For part a) we have $$ E = \frac{Qy}{4 \pi \epsilon_0 (x^2 + y^2)^{3/2} } = \text{constant} $$ I am stuck on part b). What should the shape of the volume?...

The net force on each charge is 0 at equilibrium.

Thanks, yeah, I don't see how that'll work at all. In fact I already wrote it down in #8, how does that become 0 without using the work argument?

I understand the given solution. I am wondering specifically if it's possible to determine the total potential energy without using the work detour.