 (Parte 10 de 11)

tvxx∆= and tvyy∆=

Chapter 21 42

Eliminate ∆t from these equations to obtain: () x y x yθtan==

Substitute numerical values and evaluate y:

Substitute for y4 cm and y12 cm and

i.e., the electron will strike the fluorescent screen 4.47 cm below the horizontal axis.

57 • Picture the Problem We can use its definition to find the dipole moment of this pair of charges.

(a) Apply the definition of electric dipole moment to obtain: Lprrq=

(b) If we assume that the dipole is oriented as shown to the right, then is to the right; pointing from the negative charge toward the positive charge.

*58 • Picture the Problem The torque on an electric dipole in an electric field is given by and the potential energy of the dipole by Epτ rrr×=.Eprr⋅−=U

Using its definition, express the torque on a dipole moment in a uniform electric field:

Epτ rrr×= and θτsinpE= where θ is the angle between the electric dipole moment and the electric field.

The Electric Field 1: Discrete Charge Distributions 43

(d) Using its definition, express the potential energy of a dipole in an electric field:

*59 •• Picture the Problem We can combine the dimension of an electric field with the dimension of an electric dipole moment to prove that, in any direction, the dimension of the far field is proportional to []31Land, hence, the electric field far from the dipole falls off as 1/r3.

Express the dimension of an electric

Express the dimension an electric dipole moment:

Write the dimension of charge in terms of the dimension of an electric dipole moment:

L pkE==

This shows that the field E due to a dipole

Chapter 21 4 p falls off as 1/r3. 60 •• Picture the Problem We can use its definition to find the molecule’s dipole moment. From the symmetry of the system, it is evident that the x component of the dipole moment is zero.

Using its definition, express the molecule’s dipole moment:

From symmetry considerations we

The y component of the molecule’s

− eLqLpy

61 •• Picture the Problem We can express the net force on the dipole as the sum of the forces acting on the two charges that constitute the dipole and simplify this expression to show that We can show that, under the given conditions, is also given by

.net iF Cp=r netF

()ipdxdExby differentiating the dipole’s potential energy function with respect to x.

r net

Apply Coulomb’s law to express the

and

CpaqC where p = 2aq.

The Electric Field 1: Discrete Charge Distributions 45

(b) Express the net force acting on the dipole as the spatial derivative of

dx dEp

Epdxddx dU x x=

62 ••• Picture the Problem We can express the force exerted on the dipole by the electric field as −dU/dr and the potential energy of the dipole as −pE. Because the field is due to a point charge, we can use Coulomb’s law to express E. In the second part of the problem, we can use Newton’s 3rd law to show that the magnitude of the electric field of the dipole along the line of the dipole a distance r away is approximately 2kp/r3.

(a) Express the force exerted by the electric field of the point charge on the dipole:

where is a unit radial vector pointing from Q toward the dipole. rˆ

Express the potential energy of the

dipole in the electric field:

32 rkQpr kQpdr

(b) Using Newton’s 3rd law, express the force that the dipole exerts on the charge Q at the origin:

F r−=Qon or ronFFQ−= and

FFQ=on

Express in terms of the field in which Q finds itself: QFon

QEFQ=on kpE=

General Problems

*63 • Picture the Problem We can equate the gravitational force and the electric force acting on a proton to find the mass of the proton under the given condition.

(Parte 10 de 11)