# Calculating the kinetic energy of an electron in a given potential [closed]

Let $$\Phi(x,y,z)$$ with $$U_0$$ be a potential.

$$\Phi(x,y,z)=\begin{cases}-U_0, & y<-d \\ \frac{U_0}{d}\cdot y, & -d\leq y \leq d \\ U_0, & y>d\end{cases} \tag{1}$$

Further assume that we are in an isolated system, so there is just this potential and nothing else.

So if we e.g. place an electro at $$(0,0,0)$$ with $$v(t=0)=0$$ we have $$E_{\text{pot}}=E_{\text{kin}}=0 \Rightarrow E_{\text{tot}}=0\tag{2}$$

If now that electron, at some time $$t>0$$, somehow got to the position $$y_0>d$$, its kinetic energy would be

$$E_{\text{kin}}(y_0>d)=|e|\cdot U_0 \tag{3}$$

I can't see why we multiply it with $$|e|$$ that. Here's what I'd do:

We know that $$E_{\text{tot}}=E_{\text{kin}}+E_{\text{pot}}=0 \Rightarrow E_{\text{kin}}=-E_{\text{pot}}\tag{4}$$

so for $$y_0>d$$ we get

$$E_{\text{kin}}=-U_0\tag{5}$$

Now there are two problems:

1. This isn't correct

2. I have a negative kinetic energy, which also indicates that my approach is flawed.

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## 1 Answer

So you seem to be getting confused about notation and moving between potential and potential energy. The potential is just the potential energy per unit charge, so you don't want to use the absolute value of the charge, you want to use the actual value of the charge.

Another issue is that you are treating $$U_0$$ as a potential energy in your energy equations, but based on how you define $$\Phi$$, $$U_0$$ is actually just an electric potential. To make things less confusing, you should probably use $$\Phi_0$$ instead of $$U_0$$. Then you can define $$U$$ to be the potential energy and so (taking $$e>0$$) $$\Delta U=-e\Delta\Phi$$

Then you have no issues. As the electron moves to the right it loses potential energy and therefore gains kinetic energy.

• What's $V$? Also we have $F=qE$ and $E=-\nabla \Phi$ so for $-d\leq y \leq d$ we have $E=-U_0/d$ so we get $F=e\cdot(-U_0/d)>0$ No? – xotix Feb 10 at 11:43
• @xotix All of those equations use $E$ to refer to the magnitude of the electric field, not the energy. – probably_someone Feb 10 at 11:46
• @xotix Ah yes I have misunderstood your notation and I have a sign error. Let me fix my answer accordingly – Aaron Stevens Feb 10 at 12:12
• @xotix I have fixed my answer – Aaron Stevens Feb 10 at 12:20
• @xotix Yes that is right – Aaron Stevens Feb 10 at 12:47