# 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.

## closed as off-topic by John Rennie, Chair, Jon Custer, ZeroTheHero, Kyle KanosFeb 12 at 11:05

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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 $$e>0$$) $$\Delta U=-e\Delta\Phi$$
• 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