hw1 - Problem 4. A 2-g ping-pong ball rubbed again st a...

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Problem 4. A 2-g ping-pong ball rubbed against a wool jacket acquires a net positive charge of 1 μ C . Estimate the fraction of the ball’s electrons that have been removed. Solution If half the ball’s mass is protons, their number (equal to the original number of electrons) is 1 g = m p . The number of electrons removed is C = e , so the fraction removed is ( C = e ) ( g = m p ) = 10 ! 6 C " 1 . 67 " 10 ! 24 g 1 . 6 " 10 ! 19 C " 1 g = 1 . 04 " 10 ! 11 (a hundred billionth). Problem 6. Find the ratio of the electrical force between a proton and an electron to the gravitational force between the two. Why doesn’t it matter that you aren’t told the distance between them? Solution At all distances (for which the particles can be regarded as classical point charges), the Coulomb force is stronger than the gravitational force by a factor of: F elec F grav = ke 2 r 2 ! " # \$ % & r 2 Gm p m e ! " # \$ % & = (9 ! 10 9 N " m 2 /C 2 )(1.6 ! 10 # C) 2 (6.67 ! 10 # N " m 2 /kg 2 )(1.67 ! 10 # 27 kg)(9.11 ! 10 # 31 kg) º 2.3 ! 10 39 . The spacial dependence of both forces is the same, and cancels out. Problem 9. Two charges, one twice as large as the other, are located 15 cm apart and experience a repulsive force of 95 N. What is the magnitude of the larger charge? Solution The product of the charges is q 1 q 2 = r 2 F Coulomb = k = (0.15 m) 2 (95 N) = (9 ! 10 9 N " m 2 2 ) = 2.38 ! 10 # 10 C 2 . If one charge is twice the other, q 1 = 2 q 2 , then 1 2 q 1 2 = 2.38 ! 10 " C and q 1 = ± 21.8 C. Problem 11. A proton is on the x- axis at x = 1.6 nm. An electron is on the y -axis at y = 0.85 nm. Find the net force the two exert on a helium nucleus (charge + 2 e ) at the origin. Solution A unit vector from the proton’s position to the origin is ! î , so the Coulomb force of the proton on the helium nucleus is F P,He = k ( e )(2 e )( ! î ) = (1.6 nm) 2 = ! 0.180 î nN. (Use Equation 23-1, with q 1 for the proton, q 2 for the helium nucleus, and the approximate values of k and e given.) A unit vector from the electron’s position to the origin is ! ˆ j , so its force on the helium nucleus is F e,He = k ( ! e )( 2 e )( ! ˆ j ) = ( 0 . 85 nm ) 2 = 0 . 638 ˆ j nN . The net Coulomb force on the helium nucleus is the sum

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