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Chapter 23: Problem 31

A very long wire carries a uniform linear charge density \(\lambda\). Using a voltmeter to measure potential difference, you find that when one probe of the meter is placed 2.50 \(\mathrm{cm}\) from the wire and the other probe is 1.00 \(\mathrm{cm}\) farther from the wire, the meter reads 575 \(\mathrm{V}\) (a) What is \(\lambda ?\) (b) If you now place one probe at 3.50 \(\mathrm{cm}\) from the wire and the other probe 1.00 \(\mathrm{cm}\) farther away, will the voltmeter read 575 \(\mathrm{V}\) ? If not, will it read more or less than 575 \(\mathrm{V}\) ? Why? (c) If you place both probes 3.50 \(\mathrm{cm}\) from the wire but 17.0 \(\mathrm{cm}\) from each other, what will the voltmeter read?

Short Answer

Expert verified
(a) \( \lambda \approx 2.38 \times 10^{-9} \ \mathrm{C/m} \), (b) less than 575 V, (c) 0 V.

Step by step solution

01

Understanding Potential Difference

The potential difference between two points near a long charged line can be given by the equation: \[ V = \frac{\lambda}{2\pi \varepsilon_0} \ln \left( \frac{r_2}{r_1} \right) \]where \( \lambda \) is the linear charge density, \( r_1 \) and \( r_2 \) are the distances from the wire to the points, and \( \varepsilon_0 \) is the vacuum permittivity \( (8.85 \times 10^{-12} \ \mathrm{C^2/N \, . \, m^2}) \).
02

Solving for 位

We are given that the voltmeter reads 575 V when one probe is at 2.50 cm (\(0.025 \ \mathrm{m}\)) and the other is at 3.50 cm (\(0.035 \ \mathrm{m}\)). Substituting these into the potential difference formula, we have:\[ 575 = \frac{\lambda}{2\pi \times 8.85 \times 10^{-12}} \ln \left( \frac{0.035}{0.025} \right) \]We need to solve for \( \lambda \).
03

Calculating the Natural Logarithm

Calculate the natural logarithm: \[ \ln \left( \frac{0.035}{0.025} \right) = \ln(1.4) \approx 0.3365 \]
04

Solving for 位

Rearranging the formula from step 2 to solve for \( \lambda \):\[ \lambda = \frac{575 \times 2\pi \times 8.85 \times 10^{-12}}{0.3365} \]Calculating this, we find:\[ \lambda \approx 2.38 \times 10^{-9} \ \mathrm{C/m} \]
05

Analyzing New Probe Placement (Part b)

Now, place one probe at 3.50 cm (0.035 m) and the other probe 1.00 cm farther, or 4.50 cm (0.045 m). Using the same formula with the previous 位 value:\[ V = \frac{\lambda}{2\pi \varepsilon_0} \ln \left( \frac{0.045}{0.035} \right) \]Calculate the natural log:\[ \ln \left( \frac{0.045}{0.035} \right) = \ln(1.286) \approx 0.2513 \]Substituting to find the voltage, the voltmeter will not read 575 V; it will be less.
06

Placing Probes 17 cm Apart at Constant Distance (Part c)

If both probes are placed at 3.50 cm but 17.0 cm apart, this means there is no potential difference due to their equal radial distance from the wire. Hence, the wire will read 0 V because the potential depends only on the radial distance from the wire, not the distance between probes.

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Key Concepts

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

Linear Charge Density
Linear charge density is a fundamental concept in electrostatics. It represents how much electric charge is distributed along a line, such as a wire. The symbol for linear charge density is typically \(\lambda\), and it has units of coulombs per meter (C/m).

Linear charge density is crucial when dealing with infinitely long charged wires. Such wires create electric fields that vary with distance from the wire. By knowing \(\lambda\), you can predict how these fields behave, which helps solve problems involving potential differences around charged wires.

Understanding \(\lambda\) allows you to solve equations like the one used in the exercise, where the potential difference between two points is calculated by integrating the electric field produced by the charged wire.
Potential Difference
The potential difference, often referred to as voltage, measures the electric potential between two points. This difference tells you how much work would be done by or against the electric field to move a test charge between those points. Mathematically, for a long charged wire, it can be expressed as: \[ V = \frac{\lambda}{2\pi \varepsilon_0} \ln \left( \frac{r_2}{r_1} \right) \]where \(\lambda\) is the linear charge density, \(r_1\) and \(r_2\) are distances from the wire, and \(\varepsilon_0\) is the vacuum permittivity.

In the exercise, we used this equation to find the potential difference when probes were placed at different distances from the wire. By calculating the logarithm of the radius ratio, you determine how much potential difference exists based on the distribution of charge along the wire.
  • The logarithmic term \(\ln\left( \frac{r_2}{r_1} \right)\) emphasizes how the potential difference changes with distance.
  • As you move the probes further from each other, the potential difference reflects the strength and influence of the electric field.
Vacuum Permittivity
Vacuum permittivity, denoted as \(\varepsilon_0\), is a constant that appears in many electrostatic equations. Its value is approximately \(8.85 \times 10^{-12} \ \mathrm{C^2/N \, . \, m^2}\). This constant plays a crucial role in describing how electric fields interact in a vacuum.

In the exercise, \(\varepsilon_0\) is part of the formula used to calculate the potential difference. It influences how electric charge in free space affects other charges.
  • \(\varepsilon_0\) helps define how strongly electric fields are set up between charges.
  • It's an essential part of Coulomb's law, which describes the force between two point charges.
  • Permittivity affects the strength and propagation of electric fields in various materials. For a vacuum, \(\varepsilon_0\) provides the baseline from which other materials' permittivities are compared.

Understanding vacuum permittivity helps make sense of how various dielectric materials would alter these fields when placed in them. This understanding can further inform us about materials' effectiveness in insulating electric charges.

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Most popular questions from this chapter

Charge \(Q=+4.00 \mu \mathrm{C}\) is distributed uniformly over the volume of an insulating sphere that has radius \(R=5.00 \mathrm{cm} .\) What is the potential difference between the center of the sphere and the surface of the sphere?

Axons. Neurons are the basic units of the nervous system. They contain long tubular structures called axons that propagate electrical signals away from the ends of the neurons. The axon contains a solution of potassium \(\left(\mathrm{K}^{+}\right)\) ions and large negative organic ions. The axon membrane prevents the large ions from leaking out, but the smaller \(\mathrm{K}^{+}\) ions are able to penetrate the membrane to some degree (Fig. E23.37). This leaves an excess negative charge on the inner surface of the axon membrane and an excess positive charge on the outer surface, resulting in a potential difference across the membrane that prevents further \(K^{+}\) ions from leaking out. Measurements show that this potential difference is typically about 70 \(\mathrm{mV}\) . The thickness of the axon membrane itself varies from about 5 to \(10 \mathrm{nm},\) so we'll use an average of 7.5 \(\mathrm{nm}\) . We can model the membrane as a large sheet having equal and opposite charge densities on its faces. (a) Find the electric field inside the axon membrane, assuming (not too realistically) that it is filled with air. Which way does it point: into or out of the axon? (b) Which is at a higher potential: the inside surface or the outside surface of the axon membrane?

The Millikan Oil-Drop Experiment. The charge of an electron was first measured by the American physicist Robert Millikan during \(1909-1913 .\) In his experiment, oil is sprayed in very fine drops (around \(10^{-4} \mathrm{mm}\) in diameter) into the space between two parallel horizontal plates separated by a distance \(d . \mathrm{A}\) potential difference \(V_{A B}\) is maintained between the parallel plates, causing a downward electric field between them. Some of the oil drops acquire a negative charge because of frictional effects or because of ionization of the surrounding air by \(x\) rays or radioactivity. The drops are observed through a microscope. (a) Show that an oil drop of radius \(r\) at rest between the plates will remain at rest if the magnitude of its charge is $$q=\frac{4 \pi}{3} \frac{\rho r^{3} g d}{V_{A B}}$$ where \(\rho\) is the density of the oil. (Ignore the buoyant force of the air.) By adjusting \(V_{A B}\) to keep a given drop at rest, the charge on that drop can be determined, provided its radius is known. (b) Millikan's ooil drops were much too small to measure their radii directly. Instead, Millikan determined \(r\) by cutting off the electric field and measuring the terminal speed \(v_{t}\) of the drop as it fell. (We discussed the concept of terminal speed in Section \(5.3 . )\) The viscous force \(F\) on a sphere of radius \(r\) moving with speed \(v\) through a fluid with viscosity \(\eta\) is given by Stokes's law: \(F=6 \pi \eta r v .\) When the drop is falling at \(v_{1}\) , the viscous force just balances the weight \(w=m g\) of the drop. Show that the magnitude of the charge on the drop is $$q=18 \pi \frac{d}{V_{A B}} \sqrt{\frac{\eta^{3} v_{t}^{3}}{2 \rho g}}$$ Within the limits of their experimental error, every one of the thousands of drops that Millikan and his coworkers measured had a charge equal to some small integer multiple of a basic charge \(e\) . That is, they found drops with charges of \(\pm 2 e, \quad \pm 5 e,\) and so on, but none with values such as 0.76\(e\) or 2.49\(e .\) A drop with charge \(-e\) has acquired one extra electron; if its charge is \(-2 e,\) it has acquired two extra electrons, and so on. (c) A charged oil drop in a Millikan oil-drop apparatus is observed to fall 1.00 mm at constant speed in 39.3 s if \(V_{A B}=0 .\) The same drop can be held at rest between two plates separated by 1.00 \(\mathrm{mm}\) if \(V_{A B}=9.16 \mathrm{V} .\) How many excess electrons has the drop acquired, and what is the radius of the drop? The viscosity of air is \(1.81 \times 10^{-5} \mathrm{N} \cdot \mathrm{s} / \mathrm{m}^{2},\) and the density of the oil is 824 \(\mathrm{kg} / \mathrm{m}^{3} .\)

A small metal sphere, carrying a net charge of \(q_{1}=\) \(-2.80 \mu \mathrm{C},\) is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of \(q_{2}=-7.80 \mu \mathrm{C} \quad\) and \(\operatorname{mass}\) \(1.50 \mathrm{g},\) is projected toward \(q_{1}\). When the two spheres are 0.800 \(\mathrm{m}\) apart, \(q_{2}\) is moving toward \(q_{1}\) with speed 22.0 \(\mathrm{m} / \mathrm{s}\) (Fig. E23.5). Assume that the two spheres can be treated as point charges. You can ignore the force of gravity. (a) What is the speed of \(q_{2}\) when the spheres are 0.400 \(\mathrm{m}\) apart? (b) How close does \(q_{2}\) get to \(q_{1}\) ?

The electric potential \(V\) in a region of space is given by $$V(x, y, z)=A\left(x^{2}-3 y^{2}+z^{2}\right)$$ where \(A\) is a constant. (a) Derive an expression for the electric field \(\vec{E}\) at any point in this region. (b) The work done by the field when a \(1.50-\mu \mathrm{C} \quad\) test charge moves from the point \((x, y, z)=\) \((0,0,0.250 \mathrm{m})\) to the origin is measured to be \(6.00 \times 10^{-5} \mathrm{J}\). Determine \(A .\) (c) Determine the electric field at the point \((0,0,\) 0.250 \(\mathrm{m} ) .\) (d) Show that in every plane parallel to the \(x z\) -plane the equipotential contours are circles. (e) What is the radius of the equipotential contour corresponding to \(V=1280 \mathrm{V}\) and \(y=2.00 \mathrm{m} ?\)
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