636_PartUniversity Physics Solution

636_PartUniversity Physics Solution - Electric Charge and...

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Electric Charge and Electric Field 21-29 Since x a << we can use the binomial expansion for 22 ( 1 /) a n d ( 1 /) x ax a −+ and keep only the first two terms: (1 ) 1 . n zn z +≈ + For 2 / ) , xa / zx a =− and 2 n = − so 2 / ) 1 2 / . x a −≈ + For 2 + / a =+ and 2 n so 2 1 x a Then 23 00 11 . 4 qQ x x qQ Fx aa a a ππ ⎛⎞ ⎡⎤ ≈− + = ⎜⎟ ⎢⎥ ⎝⎠ ⎣⎦ PP For simple harmonic motion Fk x and the frequency of oscillation is () 1/2 / . f km π = The net force here is of this form, with 3 0 /. kq Q a = P Thus 3 0 1 . 2 qQ f ma = P (b) The forces and their components are shown in Figure 21.73c. Figure 21.73c The x -components of the forces exerted by the two charges cancel, the y -components add, and the net force is in the + y -direction when y > 0 and in the -direction y when y < 0. The charge moves away from the origin on the y -axis and never returns. EVALUATE: The directions of the forces and of the net force depend on where q is located relative to the other two charges. In part (a), 0 at 0 == and when the charge q is displaced in the + x - or – x -direction the net force is a restoring force, directed to return to 0. qx = The charge oscillates back and forth, similar to a mass on a spring. 21.74. IDENTIFY: Apply 0 x F = and 0 y F = to one of the spheres. SET UP: The free-body diagram is sketched in Figure 21.74. e F is the repulsive Coulomb force between the spheres. For small , θ sin tan . EXECUTE: e sin 0 x FT F ∑= −= and cos 0 y m g = . So 2 e 2 sin cos mg kq F d . But tan sin 2 d L θθ ≈= , so 2 3 2 kq L d mg = and 1/3 2 0 2 qL d mg = P . EVALUATE: d increases when q increases. Figure 21.74 21.75. IDENTIFY: Use Coulomb's law for the force that one sphere exerts on the other and apply the 1st condition of equilibrium to one of the spheres. (a) SET UP: The placement of the spheres is sketched in Figure 21.75a. Figure 21.75a
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21-30 Chapter 21 The free-body diagrams for each sphere are given in Figure 21.75b. Figure 21.75b F c is the repulsive Coulomb force exerted by one sphere on the other. (b) EXECUTE: From either force diagram in part (a): y y Fm a = cos25.0 0 and cos25.0 mg Tm g T °− = = ° x x a = cc sin25.0 0 and sin25.0 TF F T = = ° Use the first equation to eliminate T in the second: ( )( ) c /cos25.0 sin25.0 tan25.0 g m g = °° = ° 22 12 c 2 00 0 11 1 4 4 4 [2(1.20 m)sin25.0 ] qq F rr ππ π == = ° PP P Combine this with c tan25.0 g and get 2 2 0 1 tan25.0 4 [2(1.20 m)sin25.0 ] q mg °= ° P () 0 tan25.0 2.40 m sin 25.0 1/4 mg q ° P ( ) 32 6 92 2 15.0 10 kg 9.80 m/s tan25.0 2.40 m sin25.0 2.80 10 C 8.988 10 N m /C q ×° = × ×⋅ (c) The separation between the two spheres is given by 2 sin . 2.80 C Lq θμ = as found in part (b). ( ) 2 2 c0 c 1/ 4 / 2 sin and tan .
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This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

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636_PartUniversity Physics Solution - Electric Charge and...

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