I am studying the electric potential energy and for the first time i red about the electric potential. The book Halliday et al. states that the electric potential can be defined as V = U/q where q is the test charge in the electric field, so the first question is: can the electric potential V be negative? In which cases this happens?

The second question is related to the electronvolt. The book states the electronvolt is the energy gained by an elementary charge, such as a proton or electron, to pass through a potential difference of 1 V. It says also that the eV can be obtained by the following formula: ΔV*q = -L where q is the test charge in the electric field and L is the work done on the test charge by the electric field. Now I tried to calculate the electronvolt for the electron and the proton but i am not sure if it is correct.

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    $\begingroup$ Please do not post formulae as screenshots, but use MathJax instead. $\endgroup$ – Emilio Pisanty Feb 17 at 11:19
  • $\begingroup$ @Emilio Pisanty Sorry buy I am new here and I don't know how to use it. $\endgroup$ – Angelo Giannuzzi Feb 17 at 11:27
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    $\begingroup$ ... which is why I included a link to a tutorial where you can learn how to use it. $\endgroup$ – Emilio Pisanty Feb 17 at 12:06

Take care not to confuse the quantity or variable $V$ (voltage) with the unit $\mathrm{V}$ (volt).

Voltage is measured in volts ($\mathrm{V}$). Energy is measured in joules ($\mathrm{J}$) or electronvolts ($\mathrm{eV}$). Charge is measured in coloumbs ($\mathrm{C}$). One volt is equal to one joule per coulomb.

Here's how you do it: We know $q = -e = -1.6 \times 10^{-19} \mathrm{C}$. One electronvolt is defined as the energy gain of an electron that passes through a voltage of $V = 1\,\mathrm{V}$: $$ 1\,\mathrm{eV} = -qV = eV = (1.6 \times 10^{-19} \mathrm{C}) \cdot (1\,\mathrm{V}) = 1.6 \times 10^{-19} \mathrm{J}. $$

  • $\begingroup$ So is it correct how I solved the exercises? What about the electric potential? Can it be negative according to the formula V = U/q (where U is the electric potential energy in a point of the space)? $\endgroup$ – Angelo Giannuzzi Feb 17 at 11:55
  • $\begingroup$ What if you have to calculate the electron volt for a proton instead of an electron? $\endgroup$ – Angelo Giannuzzi Feb 17 at 12:08
  • $\begingroup$ @AngeloGiannuzzi It would be the same process as for the electron since a proton and electron have the same charge but the electron is just negative. $\endgroup$ – MrMineHeads Feb 17 at 13:18
  • $\begingroup$ @MrMineHeads So is it correct how I have done in the picture above? I set the voltage ΔV = -1 V because the proton starts from a point with a higher electric potential and ends to a point with a lower electric potential. $\endgroup$ – Angelo Giannuzzi Feb 17 at 13:26
  • $\begingroup$ @AngeloGiannuzzi In this case, the proton would be losing energy since it starts at a higher potential and ends at a lower potential. So $\Delta V = -1 \text{ V}$ and $e = 1.602 \times 10^{-19} \text{ C}$ yields a negative answer (exactly $-1 \text{ eV}$ or $-1.602 \times 10^{-19} \text{ J}$). $\endgroup$ – MrMineHeads Feb 17 at 13:49

To answer your first question, yes voltage is allowed to be negative. This happens when the reference voltage is at a higher potential than the measured voltage.

Voltage is just another name for electric potential difference, the important part being difference. This means voltage is calculated as a difference between two points meaning you need a reference, or a point where we define the potential to be $0$.

The definition of voltage actually is:

$$V_2 - V_1 = -\int_{P_1}^{P_2}{\mathbf{E}\,\cdot d\mathbf{l}}$$

where $\mathbf{E}$ is the electric field.

If we set the voltage at $P_1$ to be $0$, and it happens that the potential at $P_1$ is higher than $P_2$, $P_2$ will be negative.

Another way to find a voltage at a point it so set the voltage at $P_1 = \infty$ to $0$ and measure relatively that way. This has the consequence such that if you are in a negative electric field, any point $P_2$ will be at a lower voltage than at $P_1$, in other words, negative voltages.

  • $\begingroup$ My question was related to the electric potential V = U/q, not to the voltage. Here I use V to indicate the electric potential and ΔV to indicate the voltage. $\endgroup$ – Angelo Giannuzzi Feb 17 at 13:20
  • $\begingroup$ They are the same. $\endgroup$ – MrMineHeads Feb 17 at 13:45

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