What exactly is electric potential?

Question

My book says:

When a small positive test charge is placed in the electric field due to another charge, it experiences a force. So work has to be done on the positive test charge to move it against this force of repulsion. The electric potential at a point in an electric field is defined as the amount of work done in bringing a unit positive charge from infinity to that point without acceleration.

I have several questions about this:

1. What is a "test charge"?
2. "Work has to be done on the positive test charge to move it against this force of repulsion", but what if the charge in the electric field is negative?
3. Work done = Fs
   Since we are moving the charge from infinity to a point in the electric field, s = infinity. So, work done = infinity, which would imply that electric potential is always infinity. This is clearly not the case, so what exactly am I missing here?

I'd appreciate if the answer is more theoretical than mathematical, since I haven't studied advanced mathematics.

Answer 1

1. A test charge is a charge with a magnitude so small that placing it at a point has a negligible affect on the field around the point.
2. If the test charge would be negative then the work done would have the opposite sign, but the same magnitude as in the positive test charge case
3. Work done = F s only applies to constant forces, the correct equation is $\text{work} = \int \mathbf{F} \cdot d\mathbf{s}$

Answer 2

One intuitive interpretation of electric potential is that it's the work the electric field at a point in space does per unit charge. This can be seen by observing that:

$$V = \int{E\cdot ds}= \int{\frac{F}{q} \cdot ds} = \frac{W}{q}$$

Where $V$ is electric potential, $E$ is electric field, $F$ is the force due to Coulomb's law on a charge of arbitrary magnitude $q$, and $ds$ is some infinitesimal segment. In other words, the electric potential is defined to be almost the same as the work done on a charge by an electric field, but divided by all the charge. Hence, it is the work done by an electric field per unit charge.

This equation is different from the one you have in your question. Let's say you're bringing a charge in from $\infty \rightarrow a$. Our integral equation $V = \int{E\cdot ds}$ tells us to :

1. Imagine a path from $\infty \rightarrow a$. Now the explicit path turns out not to matter here, only that you start and end at the two boundary points ($\infty$ and $a$). [Btw, this is because the electric field is conservative, i.e., its curl is 0].
2. Now at each very very small segment (infinitesimally small) in the path, find the electric field.
3. Take the dot product between the infinitesimally small "path" ($ds$) and the electric field at every single point along the path. Sum all of these contributions up. You're done.

This computation is much more general that what you provided in your question. The formula you are using assumes that the $E$ is uniform everywhere along the path.

Aside:

Also, multivariable calculus is pretty integral (pun intended) for studying electricity and magnetism. It is a majority of the language used to formulate E&M in undergraduate and above textbooks.

I'm not sure what level you're at/textbook you are using. But, it is better to start familiarizing oneself with calculus earlier than later. It takes a while (I think) to get a good grasp of using it.

Answer 3

The best way I've found to visualize voltage i.e. electric potential is as height/elevation. You can picture charges and current as marbles rolling or as water flowing. Positive charges and current flow from high voltage to low voltage, exactly like marbles or water do under gravity. Negative charges and currents flow from low to high voltage, as if under anti-gravity.

Any voltage diagram is telling you a topological map of how charges will flow in that region.

This analogy is a mathematically accurate one, because elevation itself is simply gravitational potential, telling us how masses move.