(Parte 5 de 11)

The Electric Field 1: Discrete Charge Distributions 19

Substitute numerical values to obtain:

Remarks: An alternative solution is to equate the electrostatic forces acting on q2 when it is at (17.75 cm, 0).

35 •• Picture the Problem By considering the symmetry of the array of charges we can see that the y component of the force on q is zero. We can apply Coulomb’s law and the principle of superposition of forces to find the net force acting on q.

Express the net force acting on q: qQqxQq,45ataxis,on2°+=F r

Express the force on q due to the charge Q on the x axis:

2axis,on RkqQ qxQ =r

Express the net force on q due to the charges at 45°:

kqQ RkqQ

Substitute to obtain:

i iiF

R kqQ

RkqQRkqQq r

36 ••• Picture the Poblem Let the Hr+ ions be in the x-y plane with H1 at (0, 0, 0), H2 at (a, 0,

Chapter 21 20

Express the net force acting on q1:

ji rF

r qkqr

Noting that the magnitude of q4 is three times that of the other charges and that it is

kji kji kji rF

C r

The Electric Field 1: Discrete Charge Distributions 21

Substitute to find : 1F r

k kji jiiF

Express the condition that molecule is

Solve for and evaluate : 4Fr()k F

The Electric Field *37 •

Picture the Problem Let q represent the charge at the origin and use Coulomb’s law for Er due to a point charge to find the electric field at x = 6 m and −10 m.

(a) Express the electric field at a point P located a distance x from a charge q:

Evaluate this expression for

(b) Evaluate Er at x = −10 m:

(c) The following graph was plotted using a spreadsheet program:

Chapter 21 2

Ex (N

Picture the Problem Let q represent the charges of +4 µC and use Coulomb’s law for Er due to a point charge and the principle of superposition for fields to find the electric field at the locations specified.

Noting that q1 = q2, use Coulomb’s law and the principle of superposition to express the electric field due to the given charges at a point P a distance x from the origin:

q q xxkqxkqx kqxxx rE r

(a) Apply this equation to the point at x = −2 m:

(b) EvaluateEr at x = 2 m:

The Electric Field 1: Discrete Charge Distributions 23

(c) EvaluateEr at x = 6 m:

(d) EvaluateEr at x = 10 m:

Ex ( k

N m 2 /C)

39 • Picture the Problem We can find the electric field at the origin from its definition and the force on a charge placed there fromEF rrq=. We can apply Coulomb’s law to find the value of the charge placed at y = 3 cm. (a) Apply the definition of electric

nC2

(b) Express and evaluate the force on a charged body in an electric field:

Chapter 21 24

40 • Picture the Problem We can compare the electric and gravitational forces acting on an electron by expressing their ratio. We can equate these forces to find the charge that would have to be placed on a penny in order to balance the earth’s gravitational force on it.

electric force acting on the electron: eEFe=

(Parte 5 de 11)

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