# Velocity of an Electron as it Passes through a Uniformly Charged Ring

I've been presented with a problem in which an electron is placed a certain distance x from the center of a positively charged ring and allowed to move freely. The ring has a known charge density λ. I am tasked with finding the velocity of the electron as it passes through the center of the ring.

First of all, I know that when the electron is initially positioned it will have a certain electric potential energy and that as it moves in towards the ring that potential energy will be converted into kinetic energy until it reaches the center, at which point all of the potential energy will be converted into kinetic energy.

The best way that I can think of to find the initial potential energy is to first find the electric potential at that point and then translate that into potential energy via the equation $V = \frac{U_e}{q}$. However, I am not certain what charges should be used in each equation. I have tried using the total charge of the ring (extrapolated from λ) in the electric potential calculation, and the dividing the charge of the electron out of that expression to produce the electric potential energy. Which produces the following equations:

$$V = k\int\frac{dq}{r} = k\int\frac{dq}{\sqrt{R^2 + x^2}} = \frac{kQ}{\sqrt{R^2 + x^2}}$$

$$U_e = \frac{V}{q_e} = \frac{kQ}{q_e\sqrt{R^2 + x^2}}$$

$$U_e = \frac{1}{2}mv^2$$

$$velocity = \sqrt{\frac{2U_e}{m_e}} = \sqrt{\frac{2kQ}{m_eq_e\sqrt{R^2+x^2}}}$$

I'm told that this is an incorrect solution. Could somebody please explain to me where I went wrong?

• This is a case of SHM. As the electron passed through the centre of the ring, it's velocity is maximum. You can try and apply that. – Tamoghna Chowdhury Sep 18 '15 at 16:44
• I have to disagree, @TamoghnaChowdhury. This would be SHM only for very small amplitudes. ($x$) Energy conservation is the way to go. Aaron, the potential energy at the initial point should equal the kinetic and potential energies at the final point. – Hritik Narayan Sep 18 '15 at 16:55

i.e. $$V(x)=V(0)+KE$$ ($x,0$ are along the axis of the ring.)
The potential at a point on the axis of the ring is given by: $$V(x)=\frac{kQ}{\sqrt{R^2+x^2}}$$
From this, we get: $$-\frac{keQ}{\sqrt{R^2+x^2}}=-\frac{keQ}{R}+\frac{1}{2}mv^2$$ From this you can obtain the velocity. The P.E. has a negative sign because the electron and the ring have opposite charges and are attracted to each other. The point at the center of the ring has the least potential energy locally, and that is why the electron moves towards it.
The first thing is that $U_e = q_e * V$. Then, electrostatic potential $V= \int {E.dr}$. For a ring, $E= \frac {kqx}{(x^2 + R^2)^{3/2}}$. The energy conservation otherwise is spot on.