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Chapter 22: Problem 41

(II) A flat ring (inner radius \(R_{0},\) outer radius 4\(R_{0} )\) is uniformly charged. In terms of the total charge \(Q,\) determine the electric field on the axis at points \((a) 0.25 R_{0}\) and (b) 75\(R_{0}\) from the center of the ring. [Hint: The ring can be replaced with two oppositely charged superposed disks.]

Short Answer

Expert verified
At 0.25 \(R_0\), use superposed disc fields to find \(E\). At 75 \(R_0\), the field approaches zero.

Step by step solution

01

Understanding the System

The system consists of a uniformly charged flat ring. This can be considered by dividing it into two disks: one from 0 to \(R_0\) with charge \(Q_1\) and the other from 0 to 4\(R_0\) with charge \(Q_2\). Notably, \(Q = Q_2 - Q_1\) (net charge). The electric field is calculated using the concept of a charged disk.
02

Electric Field Contribution from Disks

The electric field along the axis of a uniformly charged disc with surface charge density \(\sigma\) at a distance \(z\) from the center is given by the formula:\[E_{disk} = \frac{2 \pi k \sigma}{R^2} \left(1 - \frac{z}{\sqrt{z^2 + R^2}} \right)\]where \(k = \frac{1}{4\pi\varepsilon_0}\). Here, \(R\) is the radius of the disk, and \(z\) is the distance along the axis.
03

Express Charge Densities for the Disks

The surface charge density \(\sigma_2\) for the larger disk (radius 4\(R_0\)) is \(\sigma_2 = \frac{Q_2}{\pi (4R_0)^2}\) and for the smaller disk (radius \(R_0\)) is \(\sigma_1 = \frac{Q_1}{\pi R_0^2}\). Solve for charges: \(Q_2 = Q + Q_1\), and \(Q_1 = \sigma \times \pi R_0^2\). Thus, \(\sigma_2 = \frac{Q + \sigma(\pi R_0^2)}{\pi (16R_0^2)}\).
04

Electric Field Calculation at 0.25 \(R_0\)

Calculate net electric field at 0.25 \(R_0\) by summing contributions:\[\begin{align*}E_{0.25R_0} &= E_{2, 0.25R_0} - E_{1, 0.25R_0} \E_{2, 0.25R_0} &= \frac{2 \pi k (Q/\pi (16R_0^2))}{(4R_0)^2}(1 - \frac{0.25R_0}{\sqrt{(0.25R_0)^2 + (4R_0)^2}}) \E_{1, 0.25R_0} &= \frac{2 \pi k (Q/\pi R_0^2)}{R_0^2}(1 - \frac{0.25R_0}{\sqrt{(0.25R_0)^2 + R_0^2}})\end{align*}\].
05

Electric Field Calculation at 75 \(R_0\)

For large \(z \gg R\), electric fields approach zero:\[E_{75R_0} \approx 0\]Conduct similar calculations but leverage approximations for small ratios \(\frac{R_0}{75R_0}\) simplifying to using inverse square relationship.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Charged Ring
A charged ring is a common configuration in electrostatics where a thin, circular object carries charge over its entire structure. Such a ring can be considered as composed of another basic shape: two superposed disks. One disk represents the inner section (from 0 to an inner radius) and another covers the whole ring (from 0 to an outer radius). By assigning charges to each disk, you can effectively determine the charge distribution on the ring.
  • The net charge of the ring is defined as the difference between the charges of these two disks.
  • This approach is useful because it helps simplify complex calculations, allowing us to analyze each disk independently.
  • The electric field produced by the ring can then be obtained by analyzing the contributions from each component disk on their own.
This technique transforms the intricate geometry of a ring into a more manageable calculation, allowing us to precisely pinpoint electric fields at various points along the axis of the ring.
Surface Charge Density
Surface charge density, denoted as \( \sigma \), refers to the amount of electric charge per unit area on a surface. When dealing with objects like charged disks, surface charge density is a crucial parameter to understand electric field behavior.
  • For a disk, surface charge density can be calculated as the total charge \( Q \) divided by the area (which is \( \pi R^2 \) for a disk with radius \( R \)).
  • In the scenario of a charged ring, surface charge densities of both component disks (inner and outer) need to be determined separately.
  • These densities help breakdown the problem into simpler components since each disk's electric field is proportional to its surface charge density.
Different surface charge densities account for variations in charge distribution and field strength, giving rise to the composite effect that can be calculated at any point along the axis of the structure.
Electric Field of a Disk
The electric field of a disk is a foundational concept central to understanding many electrostatic systems. When a disk is uniformly charged, the electric field along the axis passing through its center gets calculated using known formulas. Here is a breakdown of the approach:
  • The magnitude of the electric field from a uniformly charged disk can be described by the formula:\[E_{disk} = \frac{2 \pi k \sigma}{R^2} \left(1 - \frac{z}{\sqrt{z^2 + R^2}} \right)\]where \( k = \frac{1}{4\pi\varepsilon_0} \), \( \sigma \) is the surface charge density, \( R \) is the radius, and \( z \) is the perpendicular distance from the disk's center along the axis.
  • At small distances compared to the disk鈥檚 radius, the electric field is much stronger, showing a fast drop-off as you move further away.
  • As the observation point's distance \( z \) increases far beyond \( R \), the field decreases rapidly and can be approximated as nearly zero for large values such as \( 75R_0 \).
This analytical approach by dividing the charged ring into disks allows simplification in calculating the total electric field at designated points, improving understanding of electrostatic influence regions in circularly symmetric shapes.

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Most popular questions from this chapter

(1) The field just outside a 3.50 -cm-radius metal ball is \(6.25 \times 10^{2} \mathrm{N} / \mathrm{C}\) and points toward the ball. What charge resides on the ball?

Neutral hydrogen can be modeled as a positive point charge \(+1.6 \times 10^{-19} \mathrm{C}\) surrounded by a distribution of negative charge with volume density given by \(\rho_{\mathrm{E}}(r)=-A e^{-2 r / a_{0}}\) where \(a_{0}=0.53 \times 10^{-10} \mathrm{~m}\) is called the Bohr radius, \(A\) is a constant such that the total amount of negative charge is \(-1.6 \times 10^{-19} \mathrm{C},\) and \(e=2.718 \cdots\) is the base of the natural log. (a) What is the net charge inside a sphere of radius \(a_{0}\) ? (b) What is the strength of the electric field at a distance \(a_{0}\) from the nucleus? [Hint: Do not confuse the exponential number \(e\) with the elementary charge \(e\) which uses the same symbol but has a completely different meaning and value \(\left(e=1.6 \times 10^{-19} \mathrm{C}\right)\)

(III) A flat slab of nonconducting material (Fig. \(22-40\) ) carries a uniform charge per unit volume, \(\rho_{\mathrm{E}}\). The slab has thickness \(d\) which is small compared to the height and breadth of the slab. Determine the electric field as a function of \(x(a)\) inside the slab and ( \(b\) ) outside the slab (at distances much less than the slab's height or breadth). Take the origin at the center of the slab.

(I) The total electric flux from a cubical box \(28.0 \mathrm{~cm}\) on a side is \(1.84 \times 10^{3} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}\). What charge is enclosed by the box?

A conducting spherical shell (Fig. 22-49) has inner radius \(=10.0 \mathrm{~cm},\) outer radius \(=15.0 \mathrm{~cm},\) and has a \(+3.0 \mu \mathrm{C}\) point charge at the center. A charge of \(-3.0 \mu \mathrm{C}\) is put on the conductor. ( \(a\) ) Where on the conductor does the \(-3.0 \mu \mathrm{C}\) end up? \((b)\) What is the electric field both inside and outside the shell?
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