In copper at room temperature, the mobility of mobile electrons is about 4.5×10-3(m/s)V/m,and there are about8×1028mobile electrons perm3Calculate the conductivityσand include the correct units. In actual practice, it is usually easier to measure the conductivityσand deduce the mobilityufrom this measurement.

Short Answer

Expert verified

57.6×106C·m2/V·s

Step by step solution

01

Given Data

Mobility of electrons, μ=4.5×10-3m/sV/m

Number of electrons perm3,n=8×108

02

Concept

The conductivity is defined as the current flowing through a unit cross-sectional area.

03

Calculate the conductivity

Conductivity,

σ=neμwhere,eischargeofelectrons(1.6×10-19C=8×10281.6×10-19C4.5×10-3=57.6×106C·m2/V·s

Hence, the conductivity is 57.6×106C·m2/V·s

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The capacitor in Figure 19.68 is initially uncharged, then the circuit is connected. Which graph in Figure 19.66 best describes the magnitude of the fringe field of the capacitor at location A(inside the connecting wire) as a function of time?

Why can birds perch on the bare wire of a power line without being electrocuted?

A circuit consists of two batteries (with negligible resistance), six ohmic resistors and connecting wires that have negligible resistance. The resistance R1is 10Ω, R2 is 20Ω, R3 is 30Ω, R4is 12Ω, R5is 15Ω and R6 is 20Ω. Unknown currents I1,I2 ,I3 ,I4 , I5 and I6 have their directions marked on the circuit diagram in figure 19.87.

(a) Write down a set of equations that could be solved for the six unknown currents. Make sure you can explain how to you got these equations. (b) When a correct set of equations is solved the currents are as follows (to the nearest miiampeares) I1=0.4394A, I2=0.3312A, I3=0.0065A, I4=0.1082A, I5=0.3247Aand I6=0.4329A. Check your equations by substituting in these numbers. (c) Suppose that you connect the negative lead of a voltmeter to location C. What does the voltmeter read, including both magnitude and sign? (d) What does the power output of the 5 V battery? (e) Resistor is made of a very thin metal wire that is 3 mm long, with a diameter of 0.1 mm. What is the electric field inside the metal resistor.

The insulating layer between the plates of a capacitor not only holds the plates apart to prevent conducting contact but also has a big effect on charging. Consider two capacitors whose only difference is that capacitor number has nothing between the plates, while capacitor number has a layer of plastic in the gap (Figure 19.57). They are placed in two different circuits having similar batteries and bulbs in series with the capacitor.

Show that in the first fraction of a second the current stays more nearly constant (decreases less rapidly) in the circuit with capacitor number . Explain your reasoning in detail. Hint: Consider the electric fields produced in the nearby wires by this plastic-filled capacitor. Suppose that the plastic is replaced by a different plastic that polarizes more easily. In the same circuit, would this capacitor keep the current more nearly constant or less so than capacitor ?

A more extensive analysis shows that this trend holds true for the entire charging process: the capacitor containing an easily polarized insulator ends up with more charge on its plates. The capacitor you have been using is filled with an insulator that polarizes extremely easily.

Consider a copper wire with a cross-sectional area of 1 mm2 (similar to your connecting wires ) and carrying 0.3 A of current, which is about what you get in a circuit with a thick-filament bulb and two batteries in series. Calculate the strength of the very small electric field required to drive this current through the wire.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free