Electric potential and field

Question

I don't know how much sense this question makes, electric field is a vector, so there exists field due to negative and positive charges, they cancel each other when they are equal and opposite. Can electric potential get cancelled like that? Do positive and negative potentials exist?

Answer 1

Let $\mathbf{E}$ be an electric field, and its potential $V$, then

$$\mathbf{E} = -\mathbf{\nabla} V$$

Now, let $\mathbf{E}' = -\mathbf{E}$, then using the same reference for its potential $V'$,

$$\mathbf{E}' = -\mathbf{\nabla}V'$$

so $$ \nabla V = -\nabla V'$$

and finally $$V = -V'$$

Thus, electric fields and electric potentials behave similarly, considering the situtation you were mentionning.

Answer 2

Yes the electric potential can have both positive and negative sign. Potential at any point is defined as the amount of work done in bringing a unit positive charge from infinity to that point. Consider a situation in which we have two charged particles, say $q_{1} (+ve)$ $q_{2} (-ve)$. How do we arrange such a system?

First we need to bring the +ve charge $q_{1}$ from infinity to a location say $\vec{r}_{1}$, the work done in this process is 0 since there was no other external field against which we needed to do work. Now we bring the other $q_{2}$ -ve charge from infinity to a point say $\vec{r}_{2}$. In this case we need to do work against the field of already kept +ve charge $q_{1}$at $\vec{r}_{1}$, which will be -ve as the +ve charge $q_{1}$ will attract the -ve charge $q_{2}$. This causes a decrease in potential energy of the system of two charges as we are doing -ve work on the system.

Now suppose you want to calculate the potential at $\vec{r}$ (i.e. the amount of work done to bring a unit positive charge from infinity to a point $\vec{r}$ in presence of these charges $q_{1}$ and $q_{2}$), it will be somewhat less than what it would have been if only $q_{1}$ was present. This decrease in potential due to the -ve charge $q_{2}$ is what we call as the -ve potential due to $q_{2}$. Or stated otherwise in simpler terms, we say that the potential at any point due to a -ve charge is also -ve. We can calculate potential at any point due to a point charge $q$ as: $$V(\vec{r}) = \frac{q}{4\pi\varepsilon_{0}r}$$ So yes, just like Electric fields cancel out due to their direction, we can say that potential at any point due to a charge configuration can also cancel out due to their magnitude. Remember potential due to electric field is a scalar quantity.