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What does it mean to have a negative electric potential? not talking about potential difference or voltage.

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    $\begingroup$ Electric potential is voltage. We can have negative voltage $\endgroup$ – Jim Mar 25 '15 at 15:02
  • $\begingroup$ I thought potential difference is voltage. $\endgroup$ – yashasvi2 Mar 29 '15 at 8:01
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Potentials are defined up to an arbitrary constant, so there is no particular meaning associated to a negative potential. It is the difference of potential that really matters, since the arbitrary constant is then washed away.

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Yes you can have negative electrostatic potentials. Consider a system of 4 charges $q$ and four $-q$ organized in a cube of lengths $d$ such that no charges of the same sign are adjacent to each other. This system has potential energy $$U=-\frac{q^2}{\pi\epsilon_0 d}\left(3+\frac{1}{\sqrt{3}}-\frac{3}{\sqrt{2}}\right)\le 0$$ This is a model of a crystal of salt and is negative due to nature always wanting to be in the state of lowest possible energy and the convention that at infinity potential energy due to electromagnetic interaction is null

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For a distribution of charges in space, the electrical potential, $\phi$ at a location in space can be calculated as the change in potential energy $\Delta U$ of a system of charges caused by adding a small ($\lim q\rightarrow 0$) charge $q$ to the system in that location, divided by the charge, i.e., $$\phi= {\tiny \lim q\rightarrow 0} \frac{\Delta U}{q}. $$ The potential energy of the system is usually defined to be zero when all charges are infinitely separated. Obviously, one could change the potential energy level to be zero with some specified configuration (e.g. with a cylindrical electrode to avoid having a logarithm of zero).

If the potential energy of the new system is reduced ($\Delta U< 0$) by adding a positive charge $q$, the potential at that location was negative.

Imagine a single negative charge (electron). Adding a positive charge,$q$ at a point 1 nm away will decrease the electrical potential energy of the system from zero (by definition) to $$\frac{\mathcal{k}(-e\,q)}{1\, [nm]}= -1.44(q)\, eV$$. The potential at the location where the positive charge was added was $\Delta U/q$ = -1.44 V.

Keep in mind that the term voltage usually refers to a difference of potentials within a system, so one could have a positive or negative voltage independent of whether the actual potentials or positive or negative. If point A has a potential of 25000 V and point B has a potential of 25005 V, the voltage of B relative to A (the voltage "across" BA) would be 5 volts. Or if point A has $\phi_A$= -10 V and point B has $\phi_B$=-5 V, the voltage $V_{BA}=\phi_B-\phi_A=+5\,V$. Or $V_{AB}=-5\,V$.

Note: Some texts use $V$ for potential and voltage, some use $V$ for potential and $\Delta V$ for voltage, and some use $\phi$ for potential and $V$ for voltage. Be sure you understand what convention a text is using.

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