# Potential on the axis of a ring of charge - no need for directional component?

When calculating the net electric field of a ring charge, one will look at only the components that are contributed to the electric field (in this case it's only the x components).

However, when calculating potential, we don't consider the components of the electric field that a given charge on the ring $dq$ contributes. All we do is calculate $\frac{kdq}{r}$ for each piece of the ring and sum them up. But I don't see why this works. Moving a charge along the axis going through the center of the ring from infinity to a point $P$ away from the center of the ring, each charge $dq$ will only contribute a horizontal component of electric field, not a vertical component. So why do we take two separate approaches when calculating field and potential?

Adding different answer since I'm trying different approach. Perhaps, this one is a more detailed and illustrated version of "physical answer" from the previous answer.

Lets choose small $dq$ on the ring and remove the rest of the ring completely. We want to find electric potential in some specific dot on x axis (red dot on pict) for this situation. To do so, we put unitary charge in this dot and measure work of the field on moving charge into infinity.

Well we can move charge via different paths. Electric field is conservative, so regardless of chosen path we will get the same amount of work. The difference is that on some paths work can be calculated easily, while on others resulted integral will be quite tricky to solve. Most convenient way to move is directly from $dq$ as on pict 2:

In this trivial case work is $\frac{kdq}{r}$. What will we see on choosing path along x axis? Work will be the same, but which forces will lead to it?

$E_y$ doesn't contribute into the work, since it's normal to our shift, so we need to integrate only $E_x$. So integral of $E_x$ along x in this situation will give us the same $\frac{kdq}{r}$. But this is exactly how you were going to calculate potential in the original task.

Ok, back to the original task. Now we know, that integrating only $E_x$ along x axis gives us $\frac{kdq}{r}$, so we can use this formula in calculation. Generally speaking, math proves, that we can ALWAYS use this formula, because potential is additive since electric field is additive. But this time we explicitly proved it to adjust our intuitive view of the task, which was originally against this formula.

UPD: So the unclear question is why $\int E_xdx = \frac {kq}{r}$ for moving along x to infinity. This result arises from fact, that electric field is conservative (so amount of work depends only on the endpoints of that path, not the particular route taken) while we know this work for different (trivial) path.

However, this result looks weird to you, isn't it? You expect integration of $E_x$ instead of full $E$ to give lesser final value. General reason for these integrals to be the same is that replacement of $E$ with $E_x$ is compensated by differences in path. Well, it can be checked directly by solving integral, but I have another pict to illustrate the differences more friendly.

We are comparing work on green path and red one. Lets draw two circles with centre at $dq$, one of arbitrary radius $r$ and the second with slightly larger radius $r+dr$. They crosses green path with $AB$ and red path with $CD$. You may have noticed, that $CD = dr$, while $AB$ is longer, since it isn't parallel to radius going through $A$ to $O$. Considering $dr << r$, the work done on $AB$ is $E_x \cdot AB = E\cos\theta \cdot AB = E \cdot AO = E \cdot dr$ which is the work on $CD$ ($\theta$ denotes $\measuredangle BAO$). So both paths can be divided into intervals of the same work, thus full work on them is the same as well.

• Thank you so much! You are so incredibly helpful. The only thing I don't get is how the integral of $E_x$ along x is the same $kdq/r$. I realize that E field is conservative but I think moving from infinity on the x axis to the red point on the diagram is different than moving from infinity to the red point along the red line in your first figure. – rb612 Feb 10 '17 at 15:21
• And in addition, integrating the x component only of electric field should give you a smaller potential than integrating both components – rb612 Feb 10 '17 at 15:25

Electric potential is amount of work needed to bring unitary positive charge from infinity to some point. For one charge it's $$V_1 = -\int{\vec{E_1}(\vec{r})\vec{dr}}$$for two charges its $$V = -\int{(\vec{E_1}(\vec{r})+\vec{E_2}(\vec{r}))\vec{dr}} = -\int{\vec{E_1}(\vec{r})\vec{dr}} -\int{\vec{E_2}(\vec{r})\vec{dr}}= V_1 + V_2$$

You see, even if fields are compensating each other, potential can be simply summed from components disregarding field interaction (or actually summing means "calculate fields interaction")

So for the ring, you have bunch of $dq$ and you simply sum their potentials disregarding field interaction.

1. Ring of charge with chosen single $dq$
2. The same $dq$ as in point one, but no ring exists
Wait, we'll get exactly the same work as in situation 1, where orthogonal component of $dq$ field is removed by other charges and only tangential is counted. And what amount of work will we get? In situation 2 it's obviously potential of destination point, which is well known: $\frac{kdq}{r}$. So we proved, that in situation 1, component of potential brought by $dq$ is the same as if no other charges were presented.
• Hi! Thank you so much for your answer. I'm still unclear though as to how we use $kdq/r$ because our work is the integral of force composed of the tangential component of the field only. – rb612 Feb 10 '17 at 6:11
• Ok, lets review the first part. If that doesn't help, we can review the second one :) Our work is the integral of sum of forces, this sum doesn't have orthogonal component, since it's compensated, while each member of sum (full force from $dq$) has orthogonal component. We know, that integral of sum identical to sum of integrals. Thus our work can be represented as sum of integrals of each force. Integral of full force from any chosen $dq$ is $\frac {kdq}{r}$, so this formula is valid. – nnovich-OK Feb 10 '17 at 7:02