ch21 - 1(a With a understood to mean the magnitude of...

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1. (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 F m a k q q r q = = = × 1 1 1 2 2 9 2 2 8 99 10 0 0032 . . . c h Inserting the values for m 1 and a 1 (see part (a)) we obtain | q | = 7.1 × 10 –11 C.
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2. The magnitude of the mutual force of attraction at r = 0.120 m is F k q q r = = × × × = 1 2 2 9 6 6 2 8 99 10 300 10 150 10 0120 2 81 . . . . . c h c hc h N.
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3. Eq. 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 | || | 1.39m. 5.70N k q q r F × × × = = =
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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 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 ′= = = ¡ = =
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5. 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 π 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.
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6. For ease of presentation (of the computations below) we assume Q > 0 and q < 0 (although the final result does not depend on this particular choice). (a) The x -component of the force experienced by q 1 = Q is ( )( ) ( ) ( ) ( ) 1 2 2 2 0 0 | | 1 | | /| | cos45 1 4 4 2 2 2 x Q Q q Q Q q Q q F a a a § · § · ¨ ¸ = °+ = + ¨ ¸ ¨ ¸ © ¹ ¨ ¸ © ¹ πε ε π which (upon requiring F 1 x = 0) leads to /| | 2 2 Q q = , or / 2 2 2.83. Q q = − = − (b) The y -component of the net force on q 2 = q is ( ) ( ) ( ) 2 2 2 2 2 2 0 0 | | 1 | | | | 1 sin 45 4 4 | | 2 2 2 y q Q q q Q F a a q a § · § · ¨ ¸ = °− = ¨ ¸ ¨ ¸ © ¹ ¨ ¸ © ¹ πε πε πε πε πε πε πε πε which (if we demand F 2 y = 0) leads to / 1/ 2 2 Q q = − . 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.
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7. The force experienced by q 3 is 3 1 3 2 3 4 3 31 32 34 2 2 2 0 | || | | || | | || | 1 ˆ ˆ ˆ ˆ j (cos45 i sin 45 j) i 4 ( 2 ) q q q q q q F F F F a a a πε § · = + + = + ° + ° + ¨ ¸ © ¹ G G G G (a) Therefore, the x -component of the resultant force on q 3 is ( ) ( ) 2 7 9 3 2 3 4 2 2 0 2 1.0 10 | | | | 1 | | 8.99 10 2 0.17N.
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