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Why do we need to integrate to find electric field?

  1. Mar 24, 2013 #1
    Hi, I never really understood this fundamental concept of integration of rings to find the electric field for say a flat disk. In my book to find the electric field on a point charge on the axis of a disk a distance z away they just integrate all the rings of charge to the end of the disk. If it is a flat disk with no depth, then why can't we just use the area of the disk ∏R2dR. If we are given the charge per unit area σ then for our dQ why can't the dQ just be σ∏R2dR instead of the sum of all the rings 2∏RdRσ?
  2. jcsd
  3. Mar 24, 2013 #2


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    I'm sorry, I do not understand which problem you like to solve. Could you state this more clearly?

    To answer the general question: The fundamental laws of electromagnetics are in terms of a local relativistic field theory, which means that the most natural way to express these laws are in terms of partial differential equations, complemented by appropriate boundary conditions for a given problem. Thus, to get the fields from the charge-current distribution and the boundary conditions you have to integrate.
  4. Mar 24, 2013 #3
    Sorry I'll write out an example. "Charge is distributed uniformly over a thin circular disk of radius R. The charge per unit area is sigma. Calculate the electric field at a point P on the axis of the disk, a distance z above its center. The book solves this by finding the electric field due to a ring, and then summing up these rings to get the area of the disk. My question is why do I have the integrate these rings? Why can't I just use the circle area formula pi R^2?
  5. Mar 24, 2013 #4


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    its because each ring is the collection of points which are equal distance from the point P. And the electric field at P due to charges on the disk depends on the distance between the point P and the charges.

    In other words, because of symmetry around the z axis, it is fairly easy to calculate the electric field at P due to each infinitesimal ring. So then integrating over rings is the easiest way to find the electric field.

    You could integrate over the whole disk to get the total charge. This is fine. But this doesn't tell you anything about the electric field at P. It is not possible to use Coulomb's law with the total charge.
  6. Mar 25, 2013 #5
    By dividing the disk into infinitesimal rings you get a more exact result. Instead of the whole disk try 2 rings only. Then try 4. Then try 8. You will see you are getting closer and closer to the precise answer. Using the disk itself as you suggest would give a very crude (incorrect) approximation.

    What you are really doing is calculating the contribution to the field at that point using Coulombs law for each infinitesimal point on the disk. Because of the superposition principle, which says some physical phenomena combine additively, the total field will be the sum of the field from each point on the disk. Since they are points, you have an infinite sum. Calculus (integration) was invented to handle infinite sums like that.

    Rings are used rather than say, quadrants or pie slices, because of the way it simplifies the problem due to symmetry. Instead of worrying about 2 dimensions we just have to worry about 1.

    You might try some exercises in numerical summation, its what finally gave me intuition for this kind of stuff. Its tedious to do by hand but if you have/need some excel experience I would say go for it for the educational value.
  7. Mar 25, 2013 #6
    Maybe another way to say it is that you aren't summing dr's to get a total area as part of your expression, but rather you are looking at how distance to the point of interest varies with r. Remember from coulombs law that its that distance that goes in the denominator squared and plays a critical role in the strength of the field. If you draw a triangle for each point on the disk with one side the distance from the center of the disk to the point of interest and the other side r, you will see its the hypotenuse that actually matters to the physics of the situation, not the area. There is no way to incorporate the variability of this hypotenuse magnitude by just calling it a disk. And again the only reason we can get away with using rings is symmetry. Points on opposite sides of the ring cancel out (equal zero) in one dimension, leaving us just the other one to care about.
  8. Apr 9, 2013 #7
    It's just one possible technique of several. Symmetry says you do not have to compute any radial effect as they all cancel. So compute the axial-only effect a random point on the disk has, then integrate that over the disc whatever way is easiest. You could integrate to get a circle, then integrate again to get the disk. Or you can integrate a radial line, multiplying each radius by the length of the circumference at that radius before integration, which might get you the result faster. Or whatever suits you.
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