# Minimum velocity to be given to the ring

1 vote
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Electric field is given by the vector $\vec{E}= x \hat i + y \hat j$ is present in the XY plane. A small ring carrying charge +Q, which can freely slide on a smooth non conducting rod, is projected along the rod from the point $(0,L)$ such that it can reach the other end of the rod. What minimum velocity should be given to the ring? (Assume zero gravity)

Attempt:

$\dfrac{1}{2}mv^2 = \text{Work done}$ //Work energy theorem

$\implies \dfrac{1}{2}mv^2 = \int _0^l x dx + \int_l^ 0 y dy = 0$

Which means you can just leave the ring and it will slide all by itself!

However, my answer doesn't match the given answer given to be $\sqrt{\dfrac{QL^2}{2m}}$

I would like to know the mistake in my reasons.

Also, I searched for solutions online and saw that people integrated till half of the rod and got the right answer. Another problem asked for minimum velocity that should be given to a particle so that it can traverse the complete circle in the presence of an electric field. There also, I worked this way and got $0$. But when I searched for a solution, again people had found the change in potential energy till half of the track (i.e.) till$\theta = \pi$ instead of $2\pi$. I really don't get why we are supposed to integrate till half of the given traversal surface, I would like to know the concept(s) involved in this.

There seems to be information missing from your question :
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.

@sammygerbil the other end is on $(l,0)$. Here is the image of the second question: https://imgur.com/a/qGNGyVA
You are mixing up two different problems here. I understand the first but not the 2nd, given in the image. You have not included the diagram. Maybe this also gives the potential function $V$ or the electric field function $E$, and the position of the ring. Without this information the problem cannot be solved.

If the electric field is the same as in Problem 1, so that the equi-potential lines are circles centred on the origin, and the ring is also centred on the origin, then all points on the ring are at the same potential, so the KE required to launch the bead around the ring is zero.

1 vote

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 $$\frac12 mv^2=3qV$$ $$v=\sqrt{\frac{6qV}{m}}$$

answered May 4, 2018 by (26,660 points)
edited Jun 1, 2018