(Parte 2 de 11)

20 •• Picture the Problem Because it is both very small and repulsive, we can ignore the gravitational force between the spheres. It is also true that we are given no information about the masses of these spheres. We can find the largest possible value of Q by equating the electrostatic force between the charged spheres and the maximum force the cable can withstand.

Using Coulomb’s law, express the electrostatic force between the two

Express the tensile strength Stensile of steel in terms of the maximum force

Fmax in the cable and the crosssectional area of the cable and solve for F:

A FSmaxtensile=⇒ tensilemaxASF=

Solve for Q: k ASQtensilel=

Substitute numerical values and evaluate Q:

21 •• Picture the Problem We can use Coulomb’s law to express the net force acting on the copper cube in terms of the unbalanced charge resulting from the assumed migration of half the charges to opposite sides of the cube. We can, in turn, find the unbalanced charge

Qunbalanced from the number of copper atoms N and the number of electrons per atom.

The Electric Field 1: Discrete Charge Distributions 7

(a) Using Coulomb’s law, express the net force acting on the copper rod due to the imbalance in the positive and negative charges:

kQF=(1)

unbalancedr

Relate the number of copper atoms N to the mass m of the rod, the molar mass M of copper, and

Avogadro’s number NA:

MVMmN N rodCuA

Solve for N to obtain: A rodCu NM

Because each atom has 29 electrons and protons, we can express

Qunbalanced as:

Substitute numerical values and evaluate Qunbalanced:

Substitute for Qunbalanced in equation (1) to obtain:

(b) Using Coulomb’s law, express the maximum force of repulsion

Fmax in terms of the maximum possible charge Qmax:

2 max max r kQF=

Solve for Qmax: k FrQ max

Express Fmax in terms of the tensile strength Stensile of copper: ASFtensilemax= where A is the cross sectional area of the cube.

Chapter 21 8

Substitute to obtain:

k ASrQ tensile

Substitute numerical values and evaluate Qmax:

Remarks: A net charge of −32 µC means an excess of 2.00×1014 electrons, so the net imbalance as a percentage of the total number of charges is 4.06×10−1 = 4×10−9 %.

2 ••• Picture the Problem We can use the definition of electric field to express E in terms of the work done on the ionizing electrons and the distance they travel λ between collisions. We can use the ideal-gas law to relate the number density of molecules in the gas ρ and the scattering cross-section σ to the mean free path and, hence, to the electric field.

(a) Apply conservation of energy to relate the work done on the electrons by the electric field to the change in their kinetic energy:

From the definition of electric field we have: qEF=

Substitute for F and ∆s to obtain: λqEW=, where the mean free path λ is the distance traveled by the electrons between ionizing collisions with nitrogen atoms.

Referring to pages 545-546 for a discussion on the mean-free path,

use its definition to relate λ to the scattering cross-section σ and the number density for air molecules n:

Substitute for λ and solve for E to obtain: q nWEσ=

Use the ideal-gas law to obtain:

kTPV Nn==

The Electric Field 1: Discrete Charge Distributions 9

Substitute for n to obtain:

PWEσ=(1)

qkT Substitute numerical values and evaluate E:

i.e., E increases linearly with pressure and varies inversely with temperature.

*23 •• Picture the Problem We can use Coulomb’s law to express the charge on the rod in terms of the force exerted on it by the soda can and its distance from the can. We can apply Newton’s 2nd law in rotational form to the can to relate its acceleration to the electric force exerted on it by the rod. Combining these equations will yield an expression for Q as a function of the mass of the can, its distance from the rod, and its acceleration.

Use Coulomb’s law to relate the force on the rod to its charge Q and distance r from the soda can: 22r kQF=

Solve for Q to obtain:

=(1)

FrQ 2

Apply to the can:

ατI=∑mass ofcenter αIFR=

Because the can rolls without slipping, we know that its linear acceleration a and angular acceleration α are related according to:

(Parte 2 de 11)

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