# Why is it that protons and electrons undergo the same amount of deflection in an electric field if they have the same energy?

The question I have is with respect to this diagram. Which depicts a stream of protons and electrons entering a proton field with equal energy.

Why is it that in the case where the energy of protons and electrons are equal, their angle of deflection is exactly the same? However when speed is kept the same, the deflection of electrons is greater. Why is this so?

• a better quality image would be appreciated and appreciated! In an electrostatic field, the force undergone is worth $\vec{F} = q \vec{E}$ therefore the force depends only on the charge of the particle Mar 1 at 8:14
• I’m voting to close this question because it's not about chemistry. Mar 1 at 14:08
• @Mithoron it's about mass spectrometry, isn't it? Mar 1 at 22:21
• @LevelRiverSt: It's a question about basic electromagnetism, which'd commonly be accepted as a straightforward Physics topic.
– Nat
Mar 2 at 12:47
• @Nat it had not escaped my notice that was about basic electromagnetism. I was pointing out that OP had presumably posted it due to its application in mass spectrometry, an important technique in analytical chemistry. If so, I think it has a place here. Mar 2 at 14:23

Let's say a particle with mass $$m$$, charge $$q$$, and initial velocity $$v$$ enters an area of length $$L$$ where an electric field $$E$$ starts to deflect it sideways. This is a clear example of uniformly accelerated motion, and its laws are well known: $$x=vt,\;y={at^2\over2}$$, where the acceleration $$a={qE\over m}$$. By the time the particle reaches the right side of the picture, it gets deflected by $$y={at^2\over2}={1\over2}{qE\over m}\left({L\over v}\right)^2={1\over2}{qEL^2\over mv^2}$$.
That pretty much sums it up. $$E$$ and $$L$$ are the same for everyone, $$q$$ is the same in absolute value for electron and proton, so the only difference is $$mv^2$$. Now, considering your two situations:
1. Same speed: same $$v$$, but a proton is slightly (that is, a thousand-and-something times) heavier, hence lower deflection.
2. Same energy: same $$mv^2$$, same deflection (just to the opposite sides).