Capitulo 22 halliday resolvido

1. Eq. 22-1 gives Coulomb’s Law, F = k r =
=

|q1 ||q2 | r2 ,

which we solve for the distance:

k |q1 ||q2 |
F
(8.99 × 109 N · m2 /C2 ) (26.0 × 10−6 C) (47.0 × 10−6 C)
= 1.39 m .
5.70 N

2. The magnitude of the mutual force of attraction at r = 0.120 m is
F =k

|q1 ||q2 |
= 8.99 × 109 r2 3.00 × 10−6 1.50 × 10−6
= 2.81 N .
0.1202

3. (a) With a understood to mean the magnitude of acceleration, Newton’s second and third laws lead to m2 a2 = m1 a1 =⇒ m2 =

6.3 × 10−7 kg 7.0 m/s2
= 4.9 × 10−7 kg .
9.0 m/s 2

(b) The magnitude of the (only) force on particle 1 is
F = m 1 a1 = k

|q |2
|q1 ||q2 |
= 8.99 × 109
.
r2
0.00322

Inserting the values for m1 and a1 (see part (a)) we obtain |q | = 7.1 × 10−11 C.

4. The fact that the spheres are identical allows us to conclude that when two spheres are in contact, they share equal charge. Therefore, when a charged sphere (q ) touches an uncharged one, they will
(fairly quickly) each attain half that charge (q/2). We start with spheres 1 and 2 each having charge q and experiencing a mutual repulsive force F = kq 2 /r2 . When the neutral sphere 3 touches sphere 1, sphere 1’s charge decreases to q/2. Then sphere 3 (now carrying charge q/2) is brought into contact with sphere 2, a total amount of q/2 + q becomes shared equally between them. Therefore, the charge of sphere 3 is 3q/4 in the final situation. The repulsive force between spheres 1 and 2 is finally
F =k

q
2

3q
4

r2

=

3 q2
3
k = F.
8 r2
8

5. We put the origin of a coordinate system at the lower left corner of the square and take +x rightward and +y upward. The force exerted by the charge +q on the charge +2q is
F1 = k

q (2q )
(−ˆ ) . j a2

The force exerted by the charge −q on the +2q charge is directed along the diagonal of the square and has magnitude q (2q )
F2 = k √
(a 2)2
√
which becomes, upon finding its components (and using the fact that cos 45◦ = 1/ 2), q (2q ) ˆ
q