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

CP A thin disk with a circular hole at its center, called an annulus, has inner radius $R_{1}$ and outer radius $R_{2}$ (Fig. P21. 104 ). 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 $\vec{E} .$ Consider points both above and below the annulus in Fig. P21. $104 .$ (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.)

CP A thin disk with a circular hole at its center, called an annulus, has inner radius $R_{1}$
and outer radius $R_{2}$ (Fig. P21. 104 ). 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 $\vec{E} .$ Consider points both above and below the annulus in Fig. P21. $104 .$ (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

Oscillatory Motion in an Electric Field

The concept of oscillatory motion in an electric field involves analyzing the motion of a charged particle subjected to a restoring force that is linearly proportional to its displacement. By equating the effective electric force to the mass times the acceleration, one can derive the equations of simple harmonic motion and thereby determine the frequency of oscillation. This technique is a staple in both classical electromagnetism and mechanics, linking electric field effects to oscillatory dynamics.

Symmetry in Electric Field Calculations

Exploiting symmetry greatly simplifies the analysis of electric fields produced by extended charge distributions. For instance, when calculating the electric field along the axis of a circularly symmetric charge distribution, many components of the field cancel out due to symmetry, reducing the problem to integrating only the non-cancelling component. This principle is widely used to streamline computations in electrodynamics.

Small Displacement Approximation and Harmonic Oscillator

Near points of equilibrium, many force fields can be approximated by a linear relationship between force and displacement. In the context of electric fields, when a charged particle is displaced slightly from an equilibrium point, the restoring force can often be linearized, leading to simple harmonic motion. This approximation is valid as long as the displacement is small enough that higher order terms in the expansion are negligible.

Surface Charge Density and Integration

This concept involves understanding that a uniformly distributed charge over a surface gives rise to a surface charge density, and the total charge on the surface can be determined by integrating this density over the area. In many problems, such as those involving disks or annuli, setting up the integration in an appropriate coordinate system (often polar coordinates) simplifies the calculation of the total charge present in the configuration.

Electric Field from Continuous Charge Distributions

When dealing with continuous distributions of charge, the electric field at a given point is calculated by integrating the contributions of infinitesimal charge elements using Coulomb's law. This approach is essential in solving complex problems where the charge is spread out over a region of space, and the integration takes into account the vector nature of the field as well as the geometry of the charge distribution.

Transcript

00:01 In this question we have an analyst with inner radius r1, auto radius r2 on the xy plane, there is a surface charge density sigma and there are four parts in this question we want to find the photo charge electric fuel along the x -axis and for electric fuel near the origin and the frequency of oscillation if there were if there is a point of charge released near the origin, okay, along the x -axis.

00:36 Okay, so in part a, we want to find the total charge.

00:43 So since the charge density is uniform, right? so the total charge will just be sigma times the area of the annulus, so sigma times i r2 square minus r1 square okay now that will be the area of the annulus okay so this is our q okay then in private b we want to find the electric field along the x -axis okay so i will be what we are going to do first is consider the electric fuel due to a ring of radius r at distance x from the center okay so you refer to the book the electric field due to a ring of charge okay this is the formula okay so e is equal to one one over four pi e qx over x squared plus r square to the three half in the i head direction okay right so we are going to consider the annulus okay consider the annulus as a superposition as a not say super position but it's made out of meaning rings, with radius ranging from r1 to r2.

03:36 So we are going to make the, make some replacement or substitution.

03:44 We'll write q into dq and this will be our sigma da, which is going to be sigma, times 2 pi r, we are.

03:59 And then in the above equation, the r will be replaced as r.

04:06 So we are going to integrate.

04:08 So e, again, this is in the x direction.

04:13 It would be equal to 1 over 4 pi epsilon 0.

04:17 And then integrate from r1 to r2.

04:23 And then the q, you write it as sigma to pi r, d r, okay, and then with an x, then divide by x squared plus r squared to the three half.

04:42 So you can pull out the constants, such as sigma x over to epsilon not, okay? then out 1 to r2, and then you have rdr, divide by x squared plus, r square to the three half.

05:05 And then if you can refer to the integration, the table of integrals at the back of the book, okay, and this integration will give you minus one over square root x squared plus r square.

05:27 Okay, and then r2, one...

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