Eletromagnetismo halliday cap 21

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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 × 10–11C.

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 thespheres 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’scharge 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 eitherof 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 resultdoes not depend on this particular choice). (a) The x-component of the force experienced by q1 = Q is 1 4πε 0

F1x =



( Q )( Q ) cos 45° + (| q |) ( Q )

(

2a

)

2

a

2

=

Q|q| Q /|q| − +1 2 4 πε 0 a 2 2

which (upon requiring F1x = 0) leads to Q / | q |= 2 2 , or Q / q = −2 2 = −2.83.

(b) The y-component of the net force on q2 = q is 1 4πε 0 | q |2

F2 y =(

2a

)

sin 45° − 2

(| q |) ( Q )
a2

=

| q |2 1 Q − 2 4πε 0 a 2 2 | q |

which (if we demand F2y = 0) leads to Q / q = −1/ 2 2 . The result is inconsistent with that obtained in part (a). Thus, we are unable to construct an equilibrium configuration with this geometry, where the only forces present are given by Eq. 21-1.

7. The force experienced by q3 is
F3 = F31 + F32 +F34 = 1 4πε 0 − | q3 || q1 | ˆ | q3 || q2 | | q || q | j+ (cos45°ˆ + sin 45°ˆ + 3 2 4 ˆ i j) i 2 2 a a ( 2a )

(a) Therefore, the x-component of the resultant force on q3 is 2 (1.0 ×10−7 ) | q3 | | q2 | 9 F3 x = + | q4 | = ( 8.99 ×10 ) 4πε 0 a 2 2 2 (0.050) 2 (b) Similarly, the y-component of the net force on q3 is 2 (1.0 ×10−7 ) | q3 | | q2 | 9 F3 y = − | q1 | + = ( 8.99 ×10 ) 4πε 0 a 2 (0.050)2 2 2
2 2

1 2 2

+ 2 = 0.17N.

−1 +

1 2 2

= −0.046N.

8. (a) The individual force magnitudes (acting on Q) are, by Eq. 21-1, k q1 Q =k q2 Q

b−a − g

a 2 2

ba − g

a 2 2

which leads to |q1| = 9.0 |q2|. Since Q is located between q1 and q2, we conclude q1 and q2 are like-sign. Consequently, q1/q2 = 9.0. (b) Now we have k q1 Q =k q2 Q

b−a − g

3a 2 2

ba − g

3a2 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/q2 = –25.

9. 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 system so the force...
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