**UFBA**

# Respostas e resolução Física Halliday 1 2 3 4 8 edição - ch22

(Parte **2** de 2)

38. (a) Fe = Ee = (3.0 × 106 N/C)(1.6 × 10–19 C) = 4.8 × 10 – 13 N.

(b) Fi = Eqion = Ee = 4.8 × 10 – 13 N.

39. (a) The magnitude of the force on the particle is given by F = qE, where q is the magnitude of the charge carried by the particle and E is the magnitude of the electric field at the location of the particle. Thus,

The force points downward and the charge is negative, so the field points upward. (b) The magnitude of the electrostatic force on a proton is

(c) A proton is positively charged, so the force is in the same direction as the field, upward.

(d) The magnitude of the gravitational force on the proton is

The force is downward.

(e) The ratio of the forces is

40. (a) The initial direction of motion is taken to be the +x direction (this is also the direction of G E). We use vvax aF m eE me ==− to solve for distance ∆x:

va mv eE c hc h

(b) Eq. 2-17 leads to

xv x v avg m ms e ie i i i e i mvKv a x eE x

Km v v v m v −−

Thus, the fraction of the initial kinetic energy lost in the region is 0.12 or 1.2%.

41. (a) The magnitude of the force acting on the proton is F = eE, where E is the magnitude of the electric field. According to Newton’s second law, the acceleration of the proton is a = F/m = eE/m, where m is the mass of the proton. Thus, ms 4 2c hc h

(b) We assume the proton starts from rest and use the kinematic equation vvax2022=+

(or else xat= 12 2 and v = at) to show that

42. When the drop is in equilibrium, the force of gravity is balanced by the force of the electric field: mg = −qE, where m is the mass of the drop, q is the charge on the drop, and E is the magnitude of the electric field. The mass of the drop is given by m = (4π/3)r3ρ, where r is its radius and ρ is its mass density. Thus,

mg r g q

43. (a) We use ∆x = vavgt = vt/2:

v x ms . c h

mae xm et c hc h c hc h

4. We assume there are no forces or force-components along the x direction. We combine Eq. 2-28 with Newton’s second law, then use Eq. 4-21 to determine time t followed by Eq. 4-23 to determine the final velocity (with –g replaced by the ay of this problem); for these purposes, the velocity components given in the problem statement are re-labeled as v0x and v0y respectively.

(b) Since vx = v0x in this problem (that is, ax = 0), we obtain t x

v a t x y y

.. di c h which leads to vy = –2.8 × 106 m/s. Therefore, the final velocity is

45. We take the positive direction to be to the right in the figure. The acceleration of the proton is ap = eE/mp and the acceleration of the electron is ae = –eE/me, where E is the magnitude of the electric field, mp is the mass of the proton, and me is the mass of the electron. We take the origin to be at the initial position of the proton. Then, the coordinate of the proton at time t is xatp =

They pass each other when their coordinates are the same, or 12 2 12 at L a tpe =+. This means t2 = 2L/(ap – ae) and p e pe e ppe ae E m m xL L L a m meE m eE m

46. Due to the fact that the electron is negatively charged, then (as a consequence of Eq.

2-28 and Newton’s second law) the field E → pointing in the +y direction (which we will call “upward”) leads to a downward acceleration. This is exactly like a projectile motion problem as treated in Chapter 4 (but with g replaced with a = eE/m = 8.78 × 1011 m/s2 ).

Thus, Eq. 4-21 gives t = x vo cos 40º

= 3.0 m

1.53 x 107 m/s = 1.96 × 10 − 6 s.

This leads (using Eq. 4-23) to vy = vo sin 40º − a t = − 4.34 × 105 m/s .

Since the x component of velocity does not change, then the final velocity is

C 3.0 10 N C i C N C j

Ni N j. 3c hc h c hb g b g b g

Therefore, the force has magnitude equal to

(b) The angle the force FG makes with the +x axis is

measured counterclockwise from the +x axis.

(c) With m = 0.010 kg, the (x, y) coordinates at t = 3.0 s can be found by combining Newton’s second law with the kinematics equations of Chapters 2–4. The x coordinate is

x x

(d) Similarly, the y coordinate is

y y

Ft ya t m

47. (a) Using Eq. 2-28, we find

48. We are given σ = 4.0 × 10 −6 C/m2 and various values of z (in the notation of Eq. 2-

26 which specifies the field E of the charged disk). Using this with F = eE (the magnitude of Eq. 2-28 applied to the electron) and F = ma, we obtain

(a) The magnitude of the acceleration at a distance R is a = e σ (2 − 2 )

4 m εo

= 1.16 × 1016 m/s2 .

(b) At a distance R/100, a= e σ (10001 − 10001 )

20002 m εo

= 3.94 × 1016 m/s2 .

(c) At a distance R/1000, a = e σ (1000001 − 1000001 )

2000002 m εo

= 3.97 × 1016 m/s2 .

(d) The field due to the disk becomes more uniform as the electron nears the center point. One way to view this is to consider the forces exerted on the electron by the charges near the edge of the disk; the net force on the electron caused by those charges will decrease due to the fact that their contributions come closer to canceling out as the electron approaches the middle of the disk.

Eq. 2-28 and Newton’s second law) the field E → pointing in the same direction as the velocity leads to deceleration. Thus, with t = 1.5 × 10 − 9 s, we find v = vo − |a| t = vo − eE

(b) The displacement is equal to the distance since the electron does not change its direction of motion. The field is uniform, which implies the acceleration is constant. Thus,

49. (a) Due to the fact that the electron is negatively charged, then (as a consequence of

(Parte **2** de 2)

- 1
- 2