Concerning differentials in electric field strength calculations

  1. Hi guys, first time posting here, but I have a question that I have been thinking about for quite a while, and I hope someone can help out with it.

    Assume a line of charge (with overall charge of +Q and of length L) that is lying on the x-axis. You want to calculate the electric field strength E a distance away from the line of charge, yet along the same axis. To do this you make use of integration to find the electric field strength emitted by an infinite amount of infinitesimally small "point charges" alone this line of charge.

    This is denoted as dE since the field strength is also infinitesimally small at a given theoretical point that is a distance X away from the point of interest where we want to calculate E. When you calculate this you get the formula:
    dE=(k dQ)/(X^2).

    This is the small E emitted by a small point of charge dQ. The problem is that we cannot integrate that since it is our value for X that is changing, yet we have the differential of our charge. So, we need to convert dQ into dX. From books I have found that the way to do this is by equating the ratio to the charge density along the line of charge: Q/L. The result is: dQ/dX = Q/L. This is where I have my problem with the logic of this equality. While the charge is distributed over the length L, X is the length from the point of interest to the point charge, which is definitely not the same as L. I understand that the differential of something causes it to become infinitesimally small, but dividing dQ/dX is dividing 2 small numbers, hence still giving a potentially large fraction. Is anybody able to explain the logic behind this?
     
  2. jcsd
  3. Simon Bridge

    Simon Bridge 15,259
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    Dividing any two numbers is potentially a large, or small, or undefined, fraction. Nothing special about infinitesimals in that regard. What is important is the relationship between them.

    The way to understand infinitesimals is as a small change in one variable.
    So dq/dx is how q changes as x changes. This is called "the derivative of q(x)".
    By definition: $$\frac{dq}{dx}=\lim_{\delta x \rightarrow 0}\frac{q(x+\delta x)-q(\delta x)}{\delta x}$$

    But you can probably understand it better in terms of fractions:

    Imagine we know that dq = 2dx, then dq/dx = 2 ... that didn't blow up on us!

    Usually, for charge distributions that occur in Nature, the fraction will be well behaved.
     
  4. Thanks for the response, but I'm still a bit confused about why these ratios are equal. It makes sense that we need to find the change that occurs between charge Q and distance X, but I don't know why that would be equal to Q/L. The proportion of X that dX is does not seem to be the same as that taken up by dQ.

    To maybe help explain why I'm having trouble with this I have another example. The following link is a photo of the first example, the line of charge, and another example, a semicircle of charge with a point of interest along the x-axis.
    https://imagizer.imageshack.us/v2/660x495q90/905/7a12a8.jpg
    In the second example I encountered the same problem. It is claimed that Q/∏, which is the charge density over the length of the semicircle, is equal to dQ/dΘ. The problem I have with this is that, once again, dQ is a small portion of the entire semicircle, while dΘ is only of that smaller angle inside the circle. It seems that for the equality to be true, it should be: Q/∏ = dQ/d∏. But that wouldn't make sense because it is Θ that is changing. I also know that dΘ is an arbitrary angle measure, and that it could be of any size, but wouldn't that mean that dΘ is always of a different size as well? That is primarily my question, because it seems that that's where I'm having trouble. dQ appears to be a constant, since it is always the same small amount, but simply on a different location on the line of charge or the semicircle, while dX or dΘ is changing.

    I hope that I'm making sense with this. It's possible that I'm simply looking at this entirely wrong. Questions involved with this problem could be answered by following this convention, but I'm really trying to understand the logic behind what I'm learning. Thanks again for helping out.
     
  5. Simon Bridge

    Simon Bridge 15,259
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    It helps to use a lower case for the (possibly) variable and upper case for specific charges.

    Since there is a total charge Q evenly distributed along the line length L, then the linear charge density is a constant Q/L. That's just the definition of "density".

    This means that the charge between x=a and x=b is (b-a)Q/L right?

    It follows that the charge between x and x+dx is the charge in length dx which is dq=(Q/L)dx

    Naturally the proportion of dq is not the same as dx ... that is why dq/dx = Q/L. There is Q/L more charge than there is length. If they were the same, then dq/dx=1.

    If the region if interest was a volume with charge Q evenly distributed through it, then I have to use a volume charge density Q/V and the charge in a small volume would be (Q/V)dV. A rectangular volume would be dV=dxdydz.

    Over a semi-circle - the linear charge density is ##Q/ \pi r## ... just the charge divided by the length of the semi-circle so ##dq= (Q/ \pi r) ds## where ##ds = rd\theta## ... it's just different ways to represent the same concept. Thus ##dq = (Q/\pi)d\theta## if we use the angles instead.

    Thus ##Q/\pi## is not the charge density over the length of the semicircle, it is the charge density over the angles subtended by the semicircle. ##dQ/d\theta = Q/\pi##

    You cannot have a ##d\pi## because pi is 180degrees while the "d" represents an infinitesimal portion.
    Think of it as ##\Delta Q/\Delta\theta##.

    Perhaps it will help to go back to motion.
    If the position-time function is x(t) then v=dx/dt would be the speed and a=dv/dt would be the acceleration.
    dv/dt is the slope of the a vs t graph.

    You are used to kinematics?
    If the acceleration is a constant from rest over time T to a top speed of V, then dv/dt=V/T (notice the change in the label? T is a constant while t is a variable.)

    Relate back to ##dq/d\theta = Q/\pi##
     
  6. I understand what you're saying with this.

    We know that x (variable) is changing. It is reaching from the point of interest to an arbitrary point on the line charge. This means that x could be any length. Is dx then always a specified amount (infinitesimal in size) or is it a ratio of the total length? If it is always the same amount, then wouldn't that have to correspond to the dq?

    I am simply confused about how the sizes of these infinitesimally small components of X and Q are determined, and how they relate. I don't see any reason why they should have to follow the ratio of charge density.

    What I understood from what you said is that we first determine dx, and then multiply it by charge density, which makes sense. However, from that we get a value, which you called dq. However it seems like even without having dx ever, from books I have found that you can simply call a small part of the line charge dq, regardless of it necessarily corresponding to the dx.

    I hope you see that I am trying to find out how these two relate to each other, and how we can then call them a ratio of charge density. Unless charge density is established before, and one of the two is the product of multiplying dq/dx by the density to get the other one.

    Thanks
     
  7. Simon Bridge

    Simon Bridge 15,259
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    That's right - dq is the amount of charge between x and x+dx in the example. In general it is what you call the amount of charge in some infinitesimal region - by convention. The reason for calling it that is because the label plays nice with Leibnitz notation.
    It would make no sense to define dq in terms of x and dx if the Cartesian coordinates were not being used, for example.

    ... but the relationship between charge and length is the linear charge density. That is what "linear charge density" means.

    I suspect what you are having trouble with, though, is that the notation is not mathematically rigorous.

    I was thinking about this.
    Go back to the charge Q on a line length L.
    Instead of going to infinitesimals, try it this way: divide the length L into N segments.
    Then you would say that ##\Delta Q = Q/N## and ##\Delta L = L/N## and you can see that $$\frac{\Delta Q}{\Delta L} = \frac{Q}{L}$$
    And the deltas turn into "d"s in the limit that N approaches infinity.
    (BTW: I made a typo in the definition of the derivative before.)

    But what about the semicircle?
    In that case, divide the angle by N so is ##Q/N = \Delta Q## then ##\pi/N=\Delta\pi##

    ... is that how you are thinking?
     
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  8. Ahh yes of course, I'm starting to get the idea behind what you're saying. It actually makes a lot of sense that dx and dq will always follow the charge density Q/L.

    This really helped me to understand the idea better, I really appreciate the help Simon, thanks so much
     
  9. Simon Bridge

    Simon Bridge 15,259
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    One thing to bear in mind is that we have only been considering cases where the charge distribution is even over something (like length or angle). This gives rise to a constant charge density. In general this is not true - it is possible for dq to be a function of x. In those situations it is more important to keep your notation carefully.

    i.e. in the last line last post, the delta-pi is confusing because I had already used pi to represent a particular angle, while, in delta-pi it is a fraction of that angle so the "pi" in "delta-pi" means angle" not the ratio of the circumference to the diameter.

    It should be as confusing in the delta-Q too, since Q is a particular amount of charge and delta-Q is a smaller amount of charge. Best to use delta-q instead.

    If you've been thinking that dq is the smallest amount of q you can have without actually being zero - or something like that - then it is no wonder you are having trouble with dq varying and/or being different sized from dx. But only dx has been defined as having a constant amount (in your examples, you can do it the other way too), ... thus, the intuition is incomplete.

    It's OK - it takes a while to get used to it.
    It can help to remember that this way of treating infinitesimals is not mathematically rigorous... it's a kind-of sloppy notation that just happens to have many of the same rules as fractions.
     
    Last edited: Jul 27, 2014
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