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# ch21 - Chapter 21 1 The magnitude of the force of either of...

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875 Chapter 21 1. The magnitude of the force of either of the charges on the other is given by F q Q q r 1 4 0 2 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. 2. 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  3. Equation 21-1 gives Coulomb’s law, F k q q r 1 2 2 , which we solve for the distance:     9 2 2 6 6 1 2 8.99 10 N m C 26.0 10 C 47.0 10 C | || | 1.39 m. 5.70N k q q r F 4. The unit ampere is discussed in Section 21-4. Using i for current, the charge transferred is  4 6 2.8 10 A 20 10 s 0.56 C. q it 5. The magnitude of the mutual force of attraction at r = 0.120 m is  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

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CHAPTER 21 876 6. (a) With a understood to mean the magnitude of acceleration, Newton’s second and third laws lead to 7 2 7 2 2 1 1 2 2 6.3 10 kg 6.0m s 4.2 10 kg. 9.0m s m a m a m (b) The magnitude of the (only) force on particle 1 is 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 | = 6.6 10 –11 C. 7. With rightward positive, the net force on q 3 is 1 3 2 3 3 13 23 2 2 23 12 23 . q q q q F F F k k L L L We note that each term exhibits the proper sign (positive for rightward, negative for leftward) for all possible signs of the charges. For example, the first term (the force exerted on q 3 by q 1 ) is negative if they are unlike charges, indicating that q 3 is being pulled toward q 1 , and it is positive if they are like charges (so q 3 would be repelled from q 1 ). Setting the net force equal to zero L 23 = L 12 and canceling k , q 3 , and L 12 leads to 1 1 2 2 0 4.00. 4.00 q q q q   8. In experiment 1, sphere C first touches sphere A , and they divided up their total charge ( Q /2 plus Q ) equally between them. Thus, sphere A and sphere C each acquired charge 3 Q /4. Then, sphere C touches B and those spheres split up their total charge (3 Q /4 plus – Q /4) so that B ends up with charge equal to Q /4. The force of repulsion between A and B is therefore
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