(Parte 7 de 11)

For a << x:

and

*45 •• Picture the Problem The diagram shows the electric field vectors at the point of interest

P due to the two charges. We can use Coulomb’s law for Er due to point charges and the superposition principle for electric fields to findPEr

. We can apply EF rrq=to find the force on a proton at (−3 m, 1 m).

Chapter 21 30

(a) Express the electric field at

Evaluate : 1E r jirE kqr

Evaluate 2Er :

jirE kqr

Substitute and simplify to find PEr :

The magnitude of is: PE r

The Electric Field 1: Discrete Charge Distributions 31

The direction of is: PE r

Note that the angle returned by your

reference angle and must be increased by 180° to yield θE.

The magnitude of Fr is:

The direction of Fr

where, as noted above, the angle returned by your calculator for

16 1is the reference

angle and must be increased by 180° to yield θE.

46 •• Picture the Problem In Problem 4 it is shown that the electric field on the x axis, due

(a) Evaluate dx dEx:

Chapter 21 32

axaxxkq axxaxxkq axaxdx dxkq axxdx dkqaxkqxdxddx dEx

Solve for x to obtain: 2 ax±=

(b) The following graph was plotted using a spreadsheet program: 2kq = 1 and a = 1

E x

47 ••• Picture the Problem We can determine the stability of the equilibrium in Part (a) and Part (b) by considering the forces the equal charges q at y = +a and y = −a exert on the test charge when it is given a small displacement along either the x or y axis. The application of Coulomb’s law in Part (c) will lead to the magnitude and sign of the charge that must be placed at the origin in order that a net force of zero is experienced by each of the three charges.

(a) Because Ex is in the x direction, a positive test charge that is displaced from

The Electric Field 1: Discrete Charge Distributions 3

(0, 0) in either the +x direction or the −x direction will experience a force pointing away from the origin and accelerate in the direction of the force.

axis. the alongnt displaceme small afor unstable is (0,0)at mequilibriu thely,Consequent x

If the positive test charge is displaced in the direction of increasing y (the positive y direction), the charge at y = +a will exert a greater force than the charge at y = −a, and the net force is then in the −y direction; i.e., it is a restoring force. Similarly, if the positive test charge is displaced in the direction of decreasing y (the negative y direction), the charge at y = −a will exert a greater force than the charge at y = −a, and the net force is then in the −y direction; i.e., it is a restoring force.

axis. the alongnt displaceme small afor stable is (0,0)at mequilibriu thely,Consequent y

axisthealong ntsdisplacemefor unstable and

axis thealong ntsdisplacemefor (0,0)at stable is mequilibriu thecharge, testnegative afor that,finds one ),(Part in as arguments same theFollowing y xa

(c) Express the net force acting on the kqqFayq

Remarks: In Part (c), we could just as well have expressed the net force acting on the charge at y = −a. Due to the symmetric distribution of the charges at y = −a and y

= +a, summing the forces acting on q0 at the origin does not lead to a relationship between q0 and q.

*48 ••• Picture the Problem In Problem 4 it is shown that the electric field on the x axis, due to

(a) Express the force acting on the on the bead when its displacement from

Chapter 21 34

Factor a2 from the denominator to

a xa xkqFx

For x << a:

i.e., the bead experiences a linear restoring force.

(b) Express the period of a simple harmonic oscillator:

k' mTπ2=

Obtain k′ from our result in part (a):

kqk'=

(Parte 7 de 11)

Comentários