**Problem 1 : charged bead sliding on a straight rod**

Your calculation is correct. The net work done in moving from one end A of the rod to the other end B is zero. But this does not necessarily mean that the minimum initial KE you should give to the bead is zero. If there is some point P along the path AB which is at a higher potential than A then the bead will need sufficient initial KE to get over this potential "hill", equal to the potential difference $\Delta V=V_P-V_A$ times the charge $Q$.

This is just like throwing an object over a wall. In order to reach a point B at the same height as A but on the other side of a wall, the particle must pass through a point M on the top of the wall. So you must give the particle an initial KE equal to the gravitational PE between A and M. The particle reaches M with zero velocity, and falls back to B with the same velocity with which it started at A.

In this problem we have to examine the shape of the potential function, to see if there are any points between A and B with higher potential than at A. Taking the origin as zero, the potential function is

$$V(x,y) = -\int E.dl = -\int (xdx+ydy) = -\frac12(x^2+y^2)$$

This is the expression for a circle. Equi-potential lines are circles, and the highest potential is at the origin.

You can easily see that the ends of the rod at A(L,0) and B(0,L) have the same potential, so the work done in moving between them is 0. You can also see that the potential increases toward the middle of the rod M. At the end A it is $V_A=-\frac12L^2$ while at the middle M $(\frac12 L, \frac12 L)$ it is $V_M=-\frac14L^2$.

In order to get the bead to move over the potential hill $\Delta V=V_M-V_A=\frac14L^2$ between end A and the middle of the rod M we have to give it some KE equal to $Q\Delta V$. So

$$\frac12 mv^2=\frac14 QL^2$$ $$v=\sqrt{\frac{QL^2}{2m}}$$

**Problem 2 : charged bead sliding around a fixed hoop**

The same issue arises here. The start and end points are at the same potential (because they are the same point). So the change in KE from start to finish will be zero. But this does not mean that the particle will move in a circle if given a very small initial speed.

As in problem 1, the particle has to climb over a potential hill and fall back down on the other side. It has to first reach the point R closest to charge $Q$, which has the highest potential. Then it "falls" back to P on the other side of R.

The potential at P is $V=\frac{kQ}{4a}$. The potential at R is $\frac{kQ}{a}=4\frac{kQ}{4a}=4V$. The difference in potential between P and R is $V_R-V_P=4V-V=3V$. The difference in potential energy is $3qV$. So the initial KE required for the particle to make a complete circle is $$\frac12 mv^2=3qV$$ $$v=\sqrt{\frac{6qV}{m}}$$

1. If the rod is straight and one end is at (0,L) then what are the co-ordinates of the other end?

2. In your comment you mention a circle. But there is no mention of a circle in your question.

Can you provide a link to the question and solution which you are asking about?

edited May 4 by sammy gerbil