Eletromagnetismo halliday cap 21

1. (a) With a understood to mean the magnitude of acceleration, Newton’s second and third laws lead to m2 a2 = m1a1

c6.3 × 10 kghc7.0 m s h = 4.9 × 10 m =
−7 2 2

−7

9.0 m s

2

kg.

(b) The magnitude of the (only) force on particle 1 is q q q F = m1a1 = k 1 2 2 = 8.99 × 109 . r 0.0032 2
2

c

h

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

2. The magnitude of the mutual force of attraction at r = 0.120 m is 3.00 × 10−6 150 × 10−6 . q1 q2 9 F=k = 8.99 × 10 = 2.81 N . 2 r 01202 .

c

hc

hc

h

3. Eq. 21-1 gives Coulomb’s Law, F = k k | q1 || q2 | r= = F

q1 q2 r2

, which we solve for the distance:

(8.99 ×10 N ⋅ m
9

2

C2 ) ( 26.0 ×10−6 C ) ( 47.0 ×10−6 C ) 5.70N

= 1.39m.

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 / r 2 . 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 (q / 2)(3q / 4) 3 q 2 3 F′ = k = k 2= F r2 8 r 8

F' 3 = = 0.375. F 8

5. The magnitude of the force of either of the charges on the other is given by

F=

1 q Q−q r2 4 πε 0

b

g

where r is the distance between the charges. We want the value of q that maximizes the function f(q) = q(Q – q). Setting the derivative df/dq equal to zero leads to Q – 2q = 0, or q = Q/2. Thus, q/Q = 0.500.

6. For ease of presentation (of the computations below) we assume Q > 0 and q < 0 (although the final result