# How does energy and speed affect sub-atomic particles (protons, neutrons, electrons) in an electric field [closed] I am having trouble understanding the problem above. Why does it matter if the particles are held at constant energy or speed? How does energy and speed even affect sub-atomic particles. Why do the different constants produce different diagrams?

• Such a question belongs rather to the physics SE site. – Poutnik May 6 at 5:23
• It literally came from a chemistry website – user78822 May 6 at 5:32
• Surrey, but it is a topic of physics. I am writing an answer. – Poutnik May 6 at 5:33
• ..but note that: We have a policy which states that ‎you should show your thoughts and/or efforts into solving the problem. It'll make us certain that ‎we aren't doing your homework for you. Basically any question with the wording your question has is considered homework; it needn't be literally one. Self-study questions, puzzles etc. also count as homework. Don't worry, they're not banned. But, we require a minimal effort. Otherwise, this question may get closed.‎ Please edit in your full reasoning or thoughts on this. See homework – Poutnik May 6 at 6:48
• I'm voting to close this question as off-topic because belongs to physics SE – Buck Thorn May 6 at 9:13

The energy and speed do not affect particles themselves, but their interactions with other diatomic objects.

And, of course, they affect their kinematics in electromagnetic field.

Kinetic energy of a particle is

$$E_\mathrm{k}=\frac 12 \cdot m \cdot v^2$$

The force on a charged particle in an electrostatic field of the strength $$E$$ is

$$\vec F=q\cdot \vec E$$

The perpendicular acceleration is

$$\vec a=\vec F/m=\frac{q\cdot \vec E}{m}$$

For the length of the path between the electrodes $$L$$, the time of flight is

$$t=\frac Lv=L \cdot \sqrt{ \left( \frac {m}{2\cdot E_\mathrm{k}}\right)}$$

The perpendicular deviation due the acceleration in the field is

\begin{align} \vec L_\mathrm{p}&=\frac 12 \cdot a \cdot t^2 \\ &= \frac 12 \cdot \frac{q\cdot \vec E}{m} \cdot \left( \frac {L^2\cdot m}{2\cdot E_\mathrm{k}}\right) \\ &= \frac 14 \cdot q \cdot L^2 \cdot \frac{\vec E}{E_\mathrm{k}}\\ \end{align}

One can see the bending with the equal kinetic energy is mass independent.

The scenario with the equal speed is a different story, as the perpendicular acceleration has the reciprocal proportionality to the mass, so the path of the much lighter electron is bent much more.