(Parte 6 de 11)

Express the magnitude of the gravitational force acting on the electron:

gmFeg=

Express the ratio of these forces to obtain:

mgeEFFg e=

Substitute numerical values and geFF

(b) Equate the electric and gravitational forces acting on the penny and solve for q to obtain:

E mgq=

Substitute numerical values and

N/C150

The Electric Field 1: Discrete Charge Distributions 25

41 •• Picture the Problem The diagram shows the locations of the charges q1 and q2 and the point on the x axis at which we are to find .Er From symmetry considerations we can conclude that the y component of Er at

any point on the x axis is zero. We can use Coulomb’s law for the electric field due to point charges to find the field at any point on the x axis and to find the force on a charge q

EF r q=

0 placed on the x axis at x = 4 cm.

(a) Letting q = q1 = q2, express the xcomponent of the electric field due to one charge as a function of the distance r from either charge to the point of interest:

iE cos2 θrkqx =r

Express for both charges: xEr iE cos2 2 θrkqx =r

Substitute for cosθ and r, substitute numerical values, and evaluate to obtain:

(b) Apply to find the force on a charge q

EF r q=

0 placed on the x axis at x = 4 cm:

*42 •• Picture the Problem If the electric field at x = 0 is zero, both its x and y components must be zero. The only way this condition can be satisfied with the point charges of +5.0 µC and −8.0 µC are on the x axis is if the point charge of +6.0 µC is also on the x axis. Let the subscripts 5, −8, and 6 identify the point charges and their fields. We can use

Coulomb’s law for Er due to a point charge and the principle of superposition for fields to determine where the +6.0 µC charge should be located so that the electric field at x = 0 is zero.

Chapter 21 26

Express the electric field at x = 0 in terms of the fields due to the charges of +5.0 µC, −8.0 µC, and +6.0 µC:

++= − µ E r

− − r rkqrkqr

− i rkqrkqr

Divide out the unit vector to obtain:

43 •• 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 an electron at (−1 m, 0).

(a) Express the electric field at

The Electric Field 1: Discrete Charge Distributions 27

Evaluate : 1E r jirE kqr

Evaluate : 2E r jirE kqr

Substitute for and and simplify to find 1Er 2Er PEr

The magnitude of is: PE r

The direction of is: PE r

Note that the angle returned by your

kN/C8.10 kN/C10.1tan1is the reference angle and must be increased by 180° to yield θE.

Chapter 21 28

Picture the Problem The diagram shows the locations of the charges q1 and q2 and the point on the x axis at which we are to findEr . From symmetry considerations we can conclude that the y component of Er at any point on the x axis is zero. We can use

Coulomb’s law for the electric field due to point charges to find the field at any point on the x axis. We can establish the results called for in parts (b) and (c) by factoring the radicand and using the approximation 1≈+α whenever α << 1.

(a) Express the x-component of the electric field due to the charges at y = a and y = −a as a function of the distance r from either charge to point P:

Substitute for cosθ and r to obtain: () () i iiiE kqx ax kqxrkqxrxrkq x

The Electric Field 1: Discrete Charge Distributions 29 and

(c) .2by given be wouldfield Its .2 magnitude of charge single a be appear to wouldby separated charges the, For

2x kqEq aax

axkqxEx

(Parte 6 de 11)

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