I'm trying to do this exercise, but I don't understand how the textbook does it:

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I don't understand, how they get a positive $27V$ and a got a few more doubts:

First thing, you can only measure differences in potential energy, and therefore you must choose a point where the potential energy is $0$. Say the potential is $0$ where the proton is. The electron will move and go near the proton, therefore the final potential energy $U_f=0$.

Now we need to find what the potential energy is when the electron is far away from the proton.

The electric potential energy for point charges is: $U=k\frac{q'q}{r}$

Therefore we have:
$U_i=k\frac{e^+ e^-}{r}$
This number is negative since the charges have opposite sign. But how is this possible? How can the potential energy far away from the proton be smaller than the potential energy near the proton?
At the same time $\Delta U= 0 - k\frac{e^+ e^-}{r}$ is positive.

Now Since $\Delta V= \frac{\Delta U}{q}$

We have $\Delta V= \frac{\Delta U}{e^-}=\frac{0-k\frac{e^+ e^-}{r}}{e^-}=-k\frac{e^+}{r}$

Substituting the value for $e^+$, $r$ and $k$ we get: $-k\frac{e^+}{r}=-27V$

Why is my voltage the opposite of their voltage?
How can they calculate only "$U$" without a reference point? Where is their $0$ potential energy?
And why is my $\Delta U$ positive, and their $U$ negative? Where is my mistake?

Could you help me out please? Thank you!


1 Answer 1


First, It's rather conventional (in most of the cases unless the charge is extended to infinite) to use infinite as a zero potential point. So $V(\infty)=0$ as convention.

The potential in this case given by $$V=\frac{1}{4\pi \epsilon_0}\frac{q}{r}$$ which on substituting all the value as done in the text gives $27$ V.

The problem with your calculation is that you have taken the wrong convention (it's true it's up to you what you use but even though choosing zero potential points at the point where there is already a point charge is not right as the potential is not defined at that point).

The rest of your mistakes follows from here.

  • $\begingroup$ I don't get it.. if $\Delta V = (V_{final} - V_{initial})q$, in this case none of the two positions are at infinity.. I don't know if my misunderstandings come from the fact that i view this as some charge falling towards another charge, just like it happens on earth with gravity, but what does it have to do with $V(\infty)=0$? $\endgroup$ Mar 12, 2021 at 21:10

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