Final Speed of a Proton and Electron

[Bashyboy]

Homework Statement

(a) Calculate the speed of a proton that is accelerated from rest through an electric potential difference of 121 V.

(b) Calculate the speed of an electron that is accelerated through the same potential difference.

Homework Equations

The Attempt at a Solution

I actually understand how to solve this particular problem, for the most part; when it comes to calculating the speed of the electron, there is one detail that confuses me: " The electron, since it has a negative charge, gains speed moving from Vi = 0 to Vf = 121 V. Why does the electron exclusively gain speed while moving in an electric field where Vi = 0 to a point in the field where Vf = 121? Why can't the proton also do this?

EDIT: For the proton, the conditions are: $0 + qV_i = 1/2m_pv_p^2 + 0$

For the electron, the conditions are:[STRIKE] [STRIKE] $0 + 0 = 1/2m_ev_e^2 + eV_f[/STRIKE]$ [/STRIKE]

Why do the equations differ so much? Why do they reverse the potential energy in each case?

EDIT: Conditions for the electron: $0 + 0 = 1/2m_ev_e^2 - eV_f$

 

[rude man]

The problem does not specify the sign of the potential difference. Assume for the electron the potential difference is + and for the protn it's -.

You might have to consider relativity, particularly for the electron, since it's going to be zinging along at a good clip. My offhand guess is not, especially if you haven't covered relativity in your work thus far.

 

[vela]

To put it another way, electrons like to roll up potential hills while protons like to roll down potential hills. So the question is asking you to compare the speed of the electron at the top of the hill with the speed of the proton at the bottom of the hill.

In terms of the math, what you get is
\begin{align*}
K_i + U_i &= K_f + U_f \\
0 + eV_\text{top} &= \frac{1}{2}m_p v_p^2 + eV_\text{bottom} \\
0 + (-e)V_\text{bottom} &= \frac{1}{2}m_e v_e^2 + (-e)V_\text{top}
\end{align*} where $V_\text{top}-V_\text{bottom}=121\text{ V}$ and $e = +1.6\times 10^{-19}\text{ C}$.

 

[Bashyboy]

To put it another way, electrons like to roll up potential hills while protons like to roll down potential hills. So the question is asking you to compare the speed of the electron at the top of the hill with the speed of the proton at the bottom of the hill.

In terms of the math, what you get is
\begin{align*}
K_i + U_i &= K_f + U_f \\
0 + eV_\text{top} &= \frac{1}{2}m_p v_p^2 + eV_\text{bottom} \\
0 + (-e)V_\text{bottom} &= \frac{1}{2}m_e v_e^2 + (-e)V_\text{top}
\end{align*} where $V_\text{top}-V_\text{bottom}=121\text{ V}$ and $e = +1.6\times 10^{-19}\text{ C}$.

Yes, but to roll up a hill, wouldn't you need initial kinetic energy?

 

[Bashyboy]

Looking back, I see that I wrote the conservation of energy equation wrong for the electron. I'll fix that.

 

[voko]

Yes, but to roll up a hill, wouldn't you need initial kinetic energy?

You are taking this too literally. Electricity, unlike gravity, is of two kinds, one kind always eager to go uphill.

 

[vela]

You need kinetic energy to roll up a potential energy hill, but that's not what's going on here. The electron is rolling up an electric potential hill. Because of its negative charge, an electron's potential energy decreases as it moves to higher electric potential.