Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Integral and physics

  1. Oct 21, 2003 #1
    I'm doing electric flux, potental, fields and charges right now in my physics class and I can't make the connection between integral and physics. For example, my teacher used dq with a uniformly charge distrubution problem. My calculus class uses integral to find area under a graph (I get that) and now my physics teacher is use it to find total charges which i don't understand why. Does integral have a different meaning in term of physics?
     
  2. jcsd
  3. Oct 21, 2003 #2

    chroot

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    The integral is just a sum. For example, let's say you want to figure out how much charge there is in a charged circle of metal. The circle of metal is one meter in radius and has a linear charge density [rho] of, say, 1 coulumb per meter (yes, it's HUGELY charged!:wink: ).

    What is the total charge? Obviously, it's just the circumference of the circle times its charge density:

    Q = 2 pi R [rho]
    = 2 pi (1) (1)
    = 2 pi coulombs

    However, you could also obtain the same result by integrating. What you're going to do conceptually is break down the circle into a lot of little tiny pieces, each with it's own little tiny charge, and then sum up the charge over all the little tiny pieces. In the limit as the pieces get very small, you're doing an integration.

    So, let's investigate a small piece of the circle. If an arc length is called s, a little tiny piece of an arc length is called ds. If charge is called q, then a little tiny piece of charge is called dq. How much charge does a tiny piece of the circle have on it?

    dq = [rho] * ds

    It has a charge equal to the linear charge density [rho] times its length, ds.

    Now we can integrate all these little tiny pieces of charge over the whole arc length of the circle (2 pi R), like this:

    Integral from 0 to (2 pi R) of ([rho] ds)

    which is just 2 pi R [rho], exactly as we found before.

    Does this make sense? Let me know if anything is confusing.

    - Warren
     
  4. Oct 22, 2003 #3

    turin

    User Avatar
    Homework Helper

    If you're just in first semester calculus, then it can lead you to the wrong answer to consider the integrals the same way. The fundamental theorem of calculus can be a tricky thing to apply to physics. Really, for purposes of calculation, the integral that explicitly contains "dq" is useless to you, but it has the most direct physical meaning. You need to get dq in terms of dx (and probably eventually you will be dealing with dy and dz, and sometimes in polar coordinates you will have some dr or dθ). The main thing that you want to make sure is that you're integrating "in the right direction." For instance, if you have a line charge, λ, on the x-axis, then dq becomes λdx. Sticking this into the integral leaves you with the decision to chose your limits. If you chose x = [a,b] you will get the negative of what you would get if you integrated on x = [b,a].
     
  5. Oct 22, 2003 #4

    chroot

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    That's true, most integrals in physics are over some spatial dimension(s).

    - Warren
     
  6. Oct 23, 2003 #5
    Integral from 0 to (2 pi R) of (P ds)

    which is just 2 pi R P, exactly as we found before.


    I don't see how u go from 0 to 2pi R |p ds to 2pi R P.
    Did u integra with respect to ds? Dont u have to change it to
    something else?
     
  7. Oct 23, 2003 #6
    Integral from 0 to (2 pi R) of (P ds)

    which is just 2 pi R P, exactly as we found before.


    I don't see how u go from 0 to 2pi R |p ds to 2pi R P.
    Did u integra with respect to ds? Dont u have to change it to
    something else?

    Btw, U did turn on a light bulb inside my head. it is just a little
    dim.

    :smile:
     
  8. Oct 23, 2003 #7

    chroot

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Yes, I integrated with respect to ds. What's the value of this integral:

    Integral from 0 to x of (ds)?

    It's just x, of course. The same thing applied here. [rho] is a constant and can be pulled out of the integral.

    - Warren
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Integral and physics
  1. Integral in physics (Replies: 4)

  2. Integration in physics (Replies: 4)

Loading...