A thin disk with a circular hole at its center, called an a...

A thin disk with a circular hole at its center, called an $a n-$ nulus, has inner radius $R_{1}$ and outer radius $R_{2}$ (Fig. $\mathbf{P 2 1 . 8 7}$ ). The disk has a uniform positive surface charge density $\sigma$ on its surface. (a) Determine the total electric charge on the annulus. (b) The annulus lies in the $y z$ plane, with its center at the origin. For an arbitrary point on the $x$ -axis (the axis of the annulus), find the magnitude and direction of the electric field $\overrightarrow{\boldsymbol{E}}$. Consider points both above and below the annulus. (c) Show that at points on the $x$ -axis that are sufficiently close to the origin, the magnitude of the electric field is approximately proportional to the distance between the center of the annulus and the point. How close is "sufficiently close"? (d) A point particle with mass $m$ and negative charge $-q$ is free to move along the $x$ -axis (but cannot move off the axis). The particle is originally placed at rest at $x=0.01 R_{1}$ and released. Find the frequency of oscillation of the particle. (Hint: Review Section 14.2. The annulus is held stationary.)

A thin disk with a circular hole at its center, called an $a n-$ nulus, has inner radius $R_{1}$ and outer radius $R_{2}$ (Fig. $\mathbf{P 2 1 . 8 7}$ ). The disk has a uniform positive surface charge density $\sigma$ on its surface. (a) Determine the total electric charge on the annulus. (b) The annulus lies in the $y z$ plane, with its center at the origin. For an arbitrary point on the $x$ -axis (the axis of the annulus), find the magnitude and direction of the electric field $\overrightarrow{\boldsymbol{E}}$. Consider points both above and below the annulus. (c) Show that at points on the $x$ -axis that are sufficiently close to the origin, the magnitude of the electric field is approximately proportional to the distance between the center of the annulus and the point. How close is "sufficiently close"?
(d) A point particle with mass $m$ and negative charge $-q$ is free to move along the $x$ -axis (but cannot move off the axis). The particle is originally placed at rest at $x=0.01 R_{1}$ and released. Find the frequency of oscillation of the particle. (Hint: Review Section 14.2. The annulus is held stationary.)

Key Concepts

Surface Charge Density and Charge Distribution

This concept involves understanding how a uniformly charged surface distributes its total charge over an area and how to calculate the total charge by integrating the surface charge density over the region of interest. In problems involving annular regions, the integration is performed between the inner and outer radii, emphasizing the importance of correctly setting limits in polar coordinates and accounting for the geometry of the charged object.

Electric Field from Symmetric Charge Distributions

The determination of the electric field along the axis of symmetry relies on the principle of superposition and the exploitation of symmetry properties of the charge distribution. In the case of an annulus or ring, the electric field is derived by integrating the contributions of infinitesimal charge elements, which involves cancelations in directions perpendicular to the axis and yields an expression for the field component along the axis. This concept is foundational in electrostatics and is applied to various symmetric configurations.

Small Distance Approximations and Series Expansions

When examining the behaviour of the electric field in regions close to a point of symmetry, it becomes useful to employ series expansions (such as Taylor series) to approximate functions about that point. This technique simplifies complex expressions to linear terms, revealing that the field can be proportional to distance from the center, which is crucial for analyzing stability and small oscillation phenomena. The idea of 'sufficiently close' indicates a regime where higher order terms can be neglected.

Simple Harmonic Motion in Electrostatic Systems

When a negatively charged particle is influenced by an electric field that varies linearly with displacement near an equilibrium position, it undergoes simple harmonic motion (SHM). Understanding SHM entails determining the effective force constant from the linear approximation of the field and using it to calculate the oscillation frequency. This concept ties together the electrostatic environment with classical mechanics, illustrating how an electric force can create a restoring force analogous to that in mechanical oscillators.

Transcript

00:01 We're going to find the electric field of a uniform charge distribution using a variety of kulam's law for the electric field.

00:14 And in general, this is how you want to approach finding the electric field of a distribution of charge that is of a single monopole nature.

00:26 But we have a annulus, which is just kind of like a steel washer, a ring with a hole in it.

00:36 And it has a uniform charge distribution.

00:40 We could write the total charge on that ring simply by taking the surface charge and multiplying by the area of that ring.

00:54 And that area is like two different circles, an inner and an outer.

00:59 One and we have to subtract the inner area of the hole from the outer portion to wind up with the full area.

01:11 So we could find the total charge on that ring.

01:16 But more importantly, in order to find the electric field, what we're going to do is take the kulam law for the electric field, which is a one over, r squared type dependence.

01:33 I'll write this a little bit differently in its vector form.

01:39 So a reminder that r vector has magnitude of r and points in the direction of r.

01:48 So a radial field, if the point charges at the origin, of course, the electric field would point radially outward everywhere.

01:59 But what we're going to do with that is we're going to think about that ring of charge, broken up into little pieces of charge that we'll call dq, infinitesimally small.

02:13 And we're going to say that each of those is going to contribute to the electric field at some observation point.

02:23 And here we're going to put our observation point up along the x -axis in our coordinate system.

02:30 Now what you want to do with that electric field, that differential electric field is electric fields superimpose.

02:40 So we are going to have to sum up all the contributions of the infinitesimal little chunks of charge inside that annulus.

02:51 And that's, of course, what we call it integral.

02:54 But dq is simply, if we want to write down what dq is, we can write dq in terms of the surface charge times some area element for that ring.

03:16 You can see that by looking at our first equation up there.

03:21 And we're going to have to use polar coordinates because the area of the ring is a circle.

03:28 And so let me kind of say what's going on there.

03:31 The area of the ring is going to be equal to some very variable we'll call it r prime, and that's distinct from the r in the equation.

03:49 And in fact, i'm going to use kind of a funny little r, a cursive r, and i'll say a little bit more about what that cursive r is.

04:01 It is not the radius of a polar coordinate in the yz plane.

04:07 The area looks like drrr prime d -fi, where phi, of course, is the azimuthal direction.

04:24 And i'll just kind of show those coordinates schematically.

04:28 R -prime goes out from the origin in the yz plane to some point inside of that disk.

04:36 Inside the material, and phi, of course, wraps around in the azimuthal direction.

04:45 So the cursive r is slightly different.

04:48 What cursive r is, is it's the distance or the vector that joins the source point to the observer point.

05:00 So let me show what that looks like on the diagram.

05:05 And if i wanted to write it as a vector, it is the observation coordinate minus the source coordinate.

05:14 It.

05:18 And we can be a little bit more explicit here.

05:23 The observer is up along the x -axis, fixed.

05:30 So we're not going to let the observer move around.

05:34 But the source is going to be anywhere in that washer, and that's what we're going to integrate over.

05:47 And the source point is somewhere in the circle in the yz plane.

05:57 And so we're going to write that as r -prime cosine phi for the y direction.

06:05 So polar coordinates and r -prime sine of phi for the z direction.

06:15 And so we can write down explicitly what that r is.

06:30 And if we now wanted to find out what the magnitude of the that r is because our electric field depends on that.

06:41 We can simply find that magnitude by taking the sum of the squares, and i'll not bore you with that.

06:49 It is basically x squared plus r prime squared to the one -half power.

06:59 So now we're ready to write down what our electric field is when we integrate.

07:08 And i will point out that we know that both the ey and the ez are both going to be zero.

07:20 And we know that basically by symmetry, so there are a couple of ways we can check that out.

07:25 But one is by symmetry because that washer goes all the way around...

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