This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 881 Chapter 21 1. Eq. 211 gives Coulomb’s Law, F k q q r = 1 2 2 , which we solve for the distance: ( 29 ( 29 ( 29 9 2 2 6 6 1 2 8.99 10 N m C 26.0 10 C 47.0 10 C    1.39m. 5.70N k q q r F × ⋅ × × = = = 2. (a) With a understood to mean the magnitude of acceleration, Newton’s second and third laws lead to m a m a m 2 2 1 1 2 7 7 6 3 10 7 0 9 0 4 9 10 = ⇒ = × = × . . . . kg m s m s kg. 2 2 c hc h (b) The magnitude of the (only) force on particle 1 is ( 29 2 1 2 9 2 2 1 1 2 2 8.99 10 N m C . (0.0032 m) q q q F m a k r = = = × ⋅ Inserting the values for m 1 and a 1 (see part (a)) we obtain  q  = 7.1 × 10 –11 C. 3. The magnitude of the mutual force of attraction at r = 0.120 m is ( 29 ( 29 ( 29 6 6 1 2 9 2 2 2 2 3.00 10 C 1.50 10 C 8.99 10 N m C 2.81N. (0.120 m) q q F k r × × = = × ⋅ = 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 2 2 / F kq r = . 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 3 q /4 in the final situation. The repulsive force between spheres 1 and 2 is finally 2 2 2 ( / 2)(3 / 4) 3 3 ' 3 0.375. 8 8 8 q q q F F k k F r r F ′ = = = ⇒ = = 5. The magnitude of the force of either of the charges on the other is given by CHAPTER 21 882 F q Q q r = 1 4 2 b ε 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 – 2 q = 0, or q = Q /2. Thus, q / Q = 0.500. 6. The unit Ampere is discussed in §214. Using i for current, the charge transferred is ( 29 ( 29 4 6 2.5 10 A 20 10 s 0.50 C. q it = = × × = 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 q 1 and q 2 be the original charges. We choose the coordinate system so the force on q 2 is positive if it is repelled by q 1 . Then, the force on q 2 is F q q r k q q r a =  =  1 4 1 2 2 1 2 2 b ε 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 the charge on each sphere is ( q 1 + q 2 )/2. The force is now one of repulsion and is given by F r k q q r b q q q q = = + + + 1 4 4 2 2 2 1 2 2 2 1 2 1 2 b ε d id i b g ....
View
Full
Document
This note was uploaded on 08/09/2010 for the course PHY 202 101A2 taught by Professor Prof.yang during the Summer '10 term at National Taiwan University.
 Summer '10
 Prof.Yang
 Physics, Acceleration

Click to edit the document details