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691_PartUniversity Physics Solution

691_PartUniversity Physics Solution - 23-14 Chapter 23...

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23-14 Chapter 23 23.36. I DENTIFY : The voltmeter reads the potential difference between the two points where the probes are placed. Therefore we must relate the potential difference to the distances of these points from the center of the cylinder. For points outside the cylinder, its electric field behaves like that of a line of charge. S ET U P : Using ( ) 0 ln / 2 b a V r r λ π Δ = P and solving for r b , we have 0 2 / V b a r r e π λ Δ = P . E XECUTE : The exponent is 9 2 2 9 1 (175 V) 2 9.00 10 N m /C 0.648 15.0 10 C/m × × = × , which gives r b = (2.50 cm) e 0.648 = 4.78 cm. The distance above the surface is 4.78 cm – 2.50 cm = 2.28 cm. E VALUATE : Since a voltmeter measures potential difference, we are actually given Δ V , even though that is not stated explicitly in the problem. We must also be careful when using the formula for the potential difference because each r is the distance from the center of the cylinder, not from the surface. 23.37. I DENTIFY : For points outside the cylinder, its electric field behaves like that of a line of charge. Since a voltmeter reads potential difference, that is what we need to calculate. S ET U P : The potential difference is ( ) 0 ln / 2 b a V r r λ π Δ = P . E XECUTE : (a) Substituting numbers gives ( ) 0 ln / 2 b a V r r λ π Δ = P = ( )( ) 6 9 2 2 10.0 cm 8.50 10 C/m 2 9.00 10 N m /C ln 6.00 cm × × × V Δ = 7.82 × 10 4 V = 78,200 V = 78.2 kV (b) E = 0 inside the cylinder, so the potential is constant there, meaning that the voltmeter reads zero. E VALUATE : Caution! The fact that the voltmeter reads zero in part (b) does not mean that V = 0 inside the cylinder. The electric field is zero, but the potential is constant and equal to the potential at the surface. 23.38. I DENTIFY : The work required is equal to the change in the electrical potential energy of the charge-ring system. We need only look at the beginning and ending points, since the potential difference is independent of path for a conservative field. S ET U P : (a) W = ( ) center 0 1 0 4 Q U q V q V V q a πε Δ = Δ = = E XECUTE : Substituting numbers gives Δ U = (3.00 × 10 -6 C)(9.00 × 10 9 N m 2 /C 2 )(5.00 × 10 –6 C)/(0.0400 m) = 3.38 J (b) We can take any path since the potential is independent of path. (c) S ET U P : The net force is away from the ring, so the ball will accelerate away. Energy conservation gives 2 1 0 max 2 . U K mv = = E XECUTE : Solving for v gives 0 2 2(3.38 J) 0.00150 kg U v m = = = 67.1 m/s E VALUATE : Direct calculation of the work from the electric field would be extremely difficult, and we would need to know the path followed by the charge. But, since the electric field is conservative, we can bypass all this calculation just by looking at the end points (infinity and the center of the ring) using the potential. 23.39. I DENTIFY : The electric field is zero everywhere except between the plates, and in this region it is uniform and points from the positive to the negative plate (to the left in Figure 23.32).
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