Capitulo 22 halliday resolvido

Disponível somente no TrabalhosFeitos
  • Páginas : 28 (6768 palavras )
  • Download(s) : 0
  • Publicado : 13 de setembro de 2012
Ler documento completo
Amostra do texto
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 meanthe 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 areidentical 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 decreasesto 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 leftcorner 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(2q ) ˆ
i+k √
j.
F2 = k √
2
2 2a
2 2 a2
Finally, the force exerted by the charge −2q on +2q is
F3 = k

(2q )(2q ) ˆ
i.
a2

(a) Therefore, the horizontal component of the resultant force on +2q is
Fx

=
=

F 1 x + F 2 x + F3 x = k
8.99 × 109

q2
a2

1
√ +4
2

1.0 × 10−7
0.0502

2

1
√ +4
2

= 0.17 N .

(b) The vertical component of the net force is
F y = F1y + F2 y + F 3 y = k

q2
a2

1
−2 + √
2

= −0.046 N .

6. (a) The individual force magnitudes (acting on Q) are, by Eq. 22-1,
k

|q1 | Q
−a −

a2
2

=k

|q2 | Q
a−

a2
2

which leads to |q1 | = 9 |q2 |. Since Q is located between q1 and q2 , we conclude q1 and q2 are
like-sign. Consequently, q1 = 9q2 .
(b) Now we have
k

|q1 | Q
−a −

3a 2
2

=k

|q2 | Qa−

3a 2
2

which yields |q1 | = 25 |q2 | . Now, Q is not located between q1 and q2 , one of them must push and
the other must pull. Thus, they are unlike-sign, so q1 = −25q2 .

7. We assume the spheres are far apart. Then the charge distribution on each of them is spherically
symmetric and Coulomb’s law can be used. Let q1 and q2 be the original charges. We choose the
coordinate systemso the force on q2 is positive if it is repelled by q1 . Then, the force on q2 is
Fa = −

1 q1 q2
q1 q2
= −k 2
4πε0 r2
r

where r = 0.500 m. The negative sign indicates that the spheres attract each other. After the wire is
connected, the spheres, being identical, acquire the same charge. Since charge is conserved, the total
charge is the same as it was originally. This means thecharge on each sphere is (q1 + q2 )/2. The force
is now one of repulsion and is given by
Fb =

1
4πε0

q1 + q2
2

q1 + q2
2

r2

=k

(q1 + q2 )2
.
4r 2

We solve the two force equations simultaneously for q1 and q2 . The first gives the product
q1 q2 = −

r 2 Fa
(0.500 m)2 (0.108 N)
= −3.00 × 10−12 C2 ,
=−
k
8.99 × 109 N · m2 /C2

and the second gives the sum
q1 + q2 =...
tracking img