Find the magnitude of the electric field due to a charged...

Key Concepts

Coulomb's Law for Continuous Charge Distributions

This concept involves calculating the electric field produced by a distribution of charge by integrating the contributions of infinitesimal charge elements. It extends Coulomb's law, which is normally applied to point charges, to systems where the charge is spread out over a region of space.

Integration Over Charge Distributions

Instead of summing the effects of discrete point charges, this technique requires setting up an integral to add the contributions from each small element of the continuous charge distribution. This method is essential for correctly computing fields and potentials in non-uniform or extended charge configurations.

Symmetry in Charge Distributions

Symmetry is a powerful tool in solving electrostatic problems because it can simplify the integration process. In a symmetric charge distribution, certain vector components of the electric field may cancel out, reducing the problem to finding only the net contribution along the direction where symmetry is not complete.

Electric Field on the Axis of a Charged Ring

This specific application illustrates how the integration and symmetry concepts work together. For a ring of charge, symmetry dictates that the perpendicular components cancel, leaving only the axial component of the field. The technique then involves using the geometry of the ring to integrate and determine the resultant axial electric field.

Transcript

00:01 This question asks us to find the magnitude of the electric field due to a charged ring of radius a and total charge q at a distance a from the ring's center.

00:11 So i've got a picture drawn here.

00:13 We've got a ring with charge q, radius a, and the point q0, our test point for finding the magnitude of the electric field.

00:20 I'm just going to be placed on the z -axis at a distance a away.

00:24 Now i'm choosing a particular test point on the ring.

00:29 I'm going to break it up into chunks of the ring, ds, but this particular chunk, if we draw the line to the test charge and draw the magnetic field line in between, we know that the distance between them is a over a root 2 because this is an isosceles triangle and that's the hypotenuse, right? so if we're looking at the charge element, the element de from the formula for the electric field is going to be k times dq over a root 2 all squared.

01:02 We also know that for every point on this side of the circle, i'll draw that in yellow here, this side of the circle, there's a point on the opposite side and so for this de here, there's going to be one over here, which means that the x and y components are going to cancel by symmetry.

01:21 So let me write that down.

01:24 Dex and dey, those are equal to zero by symmetry.

01:35 It's important to note that and it means that the problem is going to be a little bit simpler for us.

01:43 So basically, the only component that we need to look at is the z component and that's just going to be the cosine times this, the entire vector.

01:57 So the cosine is this angle here and we know because we have all the side lines that the cosine is equal to the adjacent over hypotenuse, so that's going to be a over a root 2.

02:18 Great.

02:19 And so when we're finding dec, we just need to plug everything in.

02:24 We've got k dq over a root 2 squared times a over a root 2.

02:39 Great.

02:41 The last thing that we need to figure out is what dq is.

02:45 In order to do that, i'm going to use the linear charge density of the ring.

02:50 This is something that is usually given if it's a number, but in the type of problem that this is, it's just going to be theoretical.

02:58 So we know that it is each charge element dq per the arc length ds.

03:05 We also know that ds, just from geometry, is the radius times d phi.

03:16 In our case, the radius is a.

03:20 So if we take this and just rearrange it a little bit to get what dq is equal to, we have that it's equal to mu times a d phi.

03:34 Great.

03:35 And just before i forget, i'll write down that mu is the total charge on the ring divided by the circumference of the ring 2 pi a.

03:46 I'll plug that in at the very end...

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