# Why potential due to system of charges is scalar sum?

Suppose there are three point charges $$q_1,q_2,q_3$$. We have to find potential at point $$P$$. So we do scalar sum of all potentials at that point taking one charge at a time. But I am saying that, when work is done suppose to move a $$q_0$$ charge from infinity to point $$P$$, then it will not only experience the force from suppose $$q_1$$ but also from $$q_2$$ and $$q_3$$ at the same time. So more work has to be done in bringing the charge from infinity to $$P$$. We also have to take components of forces. I.e., suppose, if we bring charge $$q_0$$ from infinity to $$P$$ in a straight line path towards $$q_1$$ then I think we will have to take components of forces in this direction from charges $$q_2$$ and $$q_3$$. And then we find the net force, total work and finally the total potential at $$P$$.

We are talking here about a rather general principle of superposition, which reflects the fact that in linear physics theories the forces/fields/etc. add up.

Newtonian mechanics (and by extensionc classical mechanics) is one such theory: Newton's second law explicitly states that the acceleration of the object is proportional to the net force acting on it. The first and the third law equally make reference to net forces: $$\mathbf{F}_{net} = \sum_i\mathbf{F}_i$$ If we are talking about potential forces, then they can be represented as gradients of the corresponding potentials (e.g., the potentials of each charge), $$\sum_i\mathbf{F}_i = -\nabla U_i(\mathbf{r}),$$ and therefore $$\mathbf{F}_{net} = -\sum_i \nabla U_i(\mathbf{r})=-\nabla\sum_i U_i(\mathbf{r}) =-\nabla U_{net}(\mathbf{r}),$$ where we use the fact that the summation and differentiation are linear operations that are interchangeable. The rest is wrapping one's mind around the different ways to do the math.

Another way to look at this is to note that we are dealing with two-body interactions: the effect of a charge on the test charge depends only on the charge in question and the test charge itself, not on any other charges. If instead we had to deal with three-body or four-body interactions, the order in which the charges are brought from infinity would indeed matter.

• the effect of a charge on the test charge depends only on the charge in question and the test charge itself, not on any other charges. But we know there are other charges too whose electric field will affect the test charge too. Isn’t it? A component of force should be considered in the direction opposite to test charge movement. Commented Jun 14, 2021 at 12:35
• @cOnnectOrTR12 The force produces by $q_1$ does not depend on the force produced by $q_2$ and vice versa. So the net force is $\mathbf{F}_1+\mathbf{F}_2$ Commented Jun 14, 2021 at 12:48
• I know both forces are independent. I am just saying both the forces will effect the test charge simultaneously. So two forces act at same time which means more work means more potential. There are two fields in which test charge is placed. Both will effect the charge. Both should be considered when finding potential at a point due to system of charges. Commented Jun 14, 2021 at 15:29
• They are all considered - have you tried doing calculations with some simple charge distributions? Commented Jun 14, 2021 at 15:58

The potential $$V$$ obeys superposition principle because the Coulomb's force obeys superposition principle:

1. Coulomb force: $$F = F_1 + F_2 + F_3 +...$$
2. Since ,$$F = qE$$, therefore $$E = E_1 + E_2 + E_3 +....$$
3. And $$V = - \int E.dl$$, hence $$V = V_1 + V_2 + V_3 +.....$$
4. However, work done to built a configuration of charge is $$W = \frac{1}{2} \sum_{n}^{i=1}q_i V(r_i),$$ here the potential is due to all the other charges except the $$i^{th}$$ charge.
5. Therefore,the work done in building a charge configuration does not obey superposition principle. Please do not confuse the potential $$V$$ and work done $$W$$. They are two different quantities.
6. Another formula for work done to built a configuration of charges is $$W = \frac{\epsilon_0}{2} \int E^2 d \tau$$, which is quadratic in nature implying it will not obey superposition principle due to the cross terms: $$W = \frac{\epsilon_0}{2} \int (E_1+E_2)^2 d \tau$$ $$W = \frac{\epsilon_0}{2} \int((E_1)^2+(E_2)^2+2E_1E_2) d \tau$$ $$W= W_1 + W_2 + \int 2E_1E_2 d\tau$$
• We know force is vector. So why dont we take components of forces when finding net force on test charge q0. Force component should be taken also so net force due to three charges in one direction should be more. Then w/q should be calculated. That will give the potential at p due to q1,q2,q3. In this way we find two more potentials taking q2 and q3 as centre charge. Then we will sum all the V. Commented Jun 14, 2021 at 12:17

Consider a system if $$n$$ point charges (which I'll label from $$1$$ to $$n$$).

We want to find the electric potential, $$V$$, at a certain position $$\vec{r}$$.

This is defined as the electric potential energy of $$+1 \mathrm{C}$$ charge. As a reference, electrical potential energy is conventionally taken as zero at a position infinitely far away from our system of point charges (i.e. somewhere where the electric field they produce is basically zero).

And the electric potential energy, $$U$$, of a test charge, $$q$$, at some position $$\vec{r}$$ is defined as the negative* of the work done by the electric field, on the test charge, in bringing it, through an electric field, from infinity to the position, $$\vec{r}$$.

(* Because when the electric field is doing negative work on the test charge, it is transferring energy from its kinetic energy to some other store - we call it electrical potential energy - and the energy stored as electrical potential energy increases. And, conversely, when the electric field is doing positive work on the test charge, the electrical potential energy decreases as the energy that the electric force transfers into the test charge's kinetic energy came from the electrical potential energy store.)

So by our definition, $$U = - \int_P \vec{F_e} \cdot d\vec{r}$$ Where the subscript $$P$$ indicates we're taking a path from 'infinity' to the position $$\vec{r}$$. (Luckily - because electric field is 'conservative' - it doesn't matter which exact path we take between these two 'points' - we'll have to do the same amount of work in any case. So we don't need to be too specific about the path.)

$$\vec{F_e}$$ is the electric force due to the electric field. It is just the vector sum of the individual electric forces from the $$n$$ other charges on our test charge:

$$\vec{F_e} = \sum_{i=1}^{n}\vec{F_i}$$

So, because integration is a linear operation (and the number of point charges in our system, $$n$$, isn't affected by our position, $$\vec{r}$$) I'm pretty sure I'm justified in the second equality: $$U = - \int_P \sum_{i=1}^{n}\vec{F_i} \cdot d\vec{r} = \sum_{i=1}^{n} \left\{- \int_P \vec{F_i} \cdot d\vec{r} \right\}$$

And looking back at our definition of electrical potential energy, we can recognize $$- \int_P \vec{F_i} \cdot d\vec{r} = U_i$$ as the electric potential energy of the test charge, $$q$$, at position $$\vec{r}$$ if only the $$i^{\mathrm{th}}$$ test charge existed.

$$\therefore U = \sum_{i=1}^{n} U_i \tag{1}\label{1}$$

And we get what you were after if we let $$q=+1\mathrm{C}$$ (remembering the definition of electrical potential):

$$V = \sum_{i=1}^{n} V_i$$

Remember though, what we have defined as U in $$\eqref{1}$$ is not the total work needed to be done to construct the system of $$n$$ (or $$n+1$$) point charges by bringing them all from infinity to their respective positions. But it is the minimum work needed to be done to bring the extra text charge, $$q$$, to its new position at $$\vec{r}$$ from infinity. (This is an equivalent definition to my previous definition of electrical potential energy.)