Chapter21 - 1. Eq. 21-1 gives Coulomb's Law, F = k k | q1 |...

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1. Eq. 21-1 gives Coulomb’s Law, Fk qq r = 12 2 , which we solve for the distance: () ( ) ( ) 92 2 6 6 8.99 10 N m C 26.0 10 C 47.0 10 C || 1.39m. 5.70N kq q r F −− ×⋅ × × == =
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2. (a) With a understood to mean the magnitude of acceleration, Newton’s second and third laws lead to ma m 22 11 2 7 7 63 10 70 90 49 10 =⇒ = × .. . . kg m s ms kg. 2 2 ch c h (b) The magnitude of the (only) force on particle 1 is () 2 12 92 2 8.99 10 N m C . (0.0032 m) qq q Fm a k r == = × Inserting the values for m 1 and a 1 (see part (a)) we obtain | q | = 7.1 × 10 –11 C.
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3. The magnitude of the mutual force of attraction at r = 0.120 m is () ( )( ) 66 12 92 2 22 3.00 10 C 1.50 10 C 8.99 10 N m C 2.81N. (0.120 m) qq Fk r −− ×× == × =
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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 22 / Fk qr = . 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)(3 / 4) 3 3 ' 3 0.375. 88 8 qq q F k F rr F == = = =
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F qQ q r = 1 4 0 2 π ε bg 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. 5. The magnitude of the force of either of the charges on the other is given by
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6. The unit Ampere is discussed in §21-4. Using i for current, the charge transferred is () ( ) 46 2.5 10 A 20 10 s 0.50 C. qi t == × × =
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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 qq r k r a =− 1 4 0 12 2 2 π 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 r b == + ++ 1 44 0 22 2 2 2 π di bg . We solve the two force equations simultaneously for q 1 and q 2 . The first gives the product rF k a 2 2 9 12 0500 0108 899 10 300 10 ×⋅ × .. . ., mN Nm C C 2 and the second gives the sum qq r F k b 6 2 2 0500 0 0360 200 10 += = . . . m N 8.99 10 N m C C 92 2 where we have taken the positive root (which amounts to assuming q 1 + q 2 0). Thus, the product result provides the relation ( ) 12 2 2 1 3.00 10 C q q −× = which we substitute into the sum result, producing q q 1 12 1 6 × .
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Chapter21 - 1. Eq. 21-1 gives Coulomb's Law, F = k k | q1 |...

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