# Voltage and potential

I just need a confirmation I've understood this concept and help in a particular problem.

So, what I understand under "voltage" is basically a difference between two electrical potentials (positions) which are actually just ratios of electrical energy per charge; if we knew how strong the probe charge is (how much coulombs does it have), and knowing the electrical potential - we would know the "exact" potential electrical energy at that point for that probe charge, measured from the middle of the bigger charge that emits the electrical field, to the probe charge. I understand the concept of potential energy and all. So voltage is proportional difference between two positions from a charge.

Ok, now if I were to calculate the voltage ( $U$)between 2 charges ($q1$, $q2$; $r$), where $φ$ is electrical potential, and $k$ a constant ($9 \times 10^9 \frac{N\times m^2}{C^2})$:

$$U=φ_1 - φ_2$$

$$φ=k\frac{q}{r}$$

I'm stuck. I can't think of it... I only understand voltage according to one charge... The best thing I could do is to make the starting electrical potential $0$ (if we don't care about the size of a charge... since it's very small, otherwise it would be big according to its radius), and finishing one with the distance $r$, then the voltage would be $U=φ_2$. All my textbook examples either give me electrical potentials, or the size of a charge which emits the electrical field, where the probe charge is moving, but none give me this kind of problem. Ty in advance.

Addition: When we look at voltage in a battery, which has some excess of + and - charge at "both sides", so how can you say a battery has a voltage at all, do you fixate the + charge and calculate voltage (electrical potential of - charge at the distance from +)?

• Distance (path length) doesn't matter. Voltage is a derived unit for talking about the energy required to move charges from one place to another. Charge and energy are the fundamental properties. Voltage has units joules/coulomb. Isn't your electric potential the energy per unit charge? Relative to some point defined as zero - like at q1, q2, or infinity? – C. Towne Springer May 13 '14 at 23:27
• V=k(q1q2)/rq1? I think you are done if you label the terms (but I'm tired. Mind wandered five minutes before amending the comment and had to add another. Thus comment not answer). – C. Towne Springer May 13 '14 at 23:38
• Yes I expressed myself badly, you substract that distance anyways. I am not sure what you mean by your question... my formula is correct and units can be derived from it. Unit of voltage and el. potential are volts. I didn't know I could take infinity as a reference point. Ok so the formula and solution is still correct, U=fi. Edit- this was written before seing your second comment. – user30948 May 13 '14 at 23:43

Both "Voltage" (or Potential) and Potential Energy are defined - in your case of point charges - with reference to $r=\infty$, where $V=0$:

$$V = k\frac{q}{r}$$

It doesn't make sense to ask "What is the potential difference between 2 charges?". In a given system, it makes more sense to ask "What is the potential difference between 2 points in space?" If so happens you have 2 point charges at those 2 points, then your answer is going to be undefined and messy.

You should see Potential as a scalar field in space created by a configuration of charges. Potential Energy, on the other hand, is in the interaction of an extra charge with this system.

• So, that's not what I should be worrying about, voltage between 2 point charges (thanks for the terms)... What if I said, according to only one, bigger charge? So the bigger charge is fixed in place, and only the other one is moving, then I could calculate fi2, the electrical potential (=voltage) and U=fi2-fi1, where fi1 is 0, voltage would be equal to that electrical potential at the distance r from the bigger charge? – user30948 May 14 '14 at 6:20
• Also, referring to the previous comment I've made -- when we look at voltage in a battery, which has some excess of + and - charge at "both sides", so how can you say a battery has a voltage at all, do you fixate the + charge and calculate voltage (electrical potential of - charge at the distance from +) according to it? – user30948 May 14 '14 at 6:34

The language can be confusing, both electric potential and potential difference are measured in volts. It can also be confusing because the term potential energy can refer to either the potential energy of a SYSTEM of charges or the potential energy of a single charge within that system. When dealing with voltage we use the latter. It can be even more confusing because electric potential and electric potential energy are not the same thing. The electric potential at a point in space is defined to be the potential energy a point charge of 1 coulomb would have if it was placed there. It is measured in Volts(Joules/Coulomb). So if the electric potential at some point in space was , say,

$5$V = $5$ J/C, it would mean that a point charge of $1$C would have $5$ J of potential energy if placed there. The relationship between electric potential and electric potential energy is given by$$U=qV$$

The electric potential energy of a point charge is defined to be the negative of the work done against the electric field in bringing that charge out from infinity. The formula for the electric potential due to a single point charge $q$ is given by:

$$V(\vec r) = \frac{kq}{r}$$

What this means is if you were to take some other point charge $q'$ and bring it out from infinity to a

distance $r$ from $q$. You will have done$-\frac{kqq'}{r}$J of work on $q'$ in bringing in there.

It turns out that to find the potential due to multiple charges one can just add the potential due to each charge. So, for two charges $q_1$ and $q_2$

$$V(x,y) = \frac{kq_1}{r_1} + \frac{kq_2}{r_2}$$

where $r_1$ is the distance from $q_1$ to the point $(x,y)$ and $r_2$ is the distance from $q_2$ to the point $(x,y)$

Say that $q_1$ is located at $(x_1,y_1)$ and $q_2$ is at $(x_2,y_2)$

The using the distance formula/Pythagorean theorem, $r_1 = \sqrt{(x-x_1)^2+(y-y_1)^2}$ and similarly for $r_2$, the electric potential at the point $(x,y)$ is then given by:

$$V(x,y) = \frac{kq_1}{\sqrt{(x-x_1)^2+(y-y_1)^2}} + \frac{kq_2}{\sqrt{(x-x_2)^2+(y-y_2)^2}}$$

Generally what we worry about is the potential difference between 2 points.This is also measured in Volts. if $$\vec a = (x_1,y_1)$$ $$\vec{b} = (x_2,y_2)$$

Then the potential difference between points $\vec{a}$ and $\vec{b}$ is defined to be

$$\Delta V = V_{ab} = V(\vec{b}) - V(\vec{a})$$

What this means is that if you move a charge $q$ from $\vec{a}$ to $\vec{b}$

its potential energy will have changed by $$\Delta U = q\Delta V$$

When you here about a battery being $12$V, what they mean is $\Delta V = V_+ - V_- = 12$V

In short, potential energy is a property of a charge and electric potential is a property of a point. Moving a charge $q$ between 2 points with a potential difference of $\Delta V$ changes the potential energy of $q$ by an amount equal to $\Delta U = q\Delta V$