Rating based on step by step help A small metal sphere, carrying a net charge of
ID: 2233415 • Letter: R
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Rating based on step by step help
A small metal sphere, carrying a net charge of q1 =-2.80 muC, is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of q2 = -7.80 muC and mass 1.50 g, is projected toward q1.-. When the two spheres arc 0.800 apart, q2 is moving toward q1 with speed 22.0 m/s (see figure). Assume that the two spheres can be treated as point charges. You can ignore the force of gravity, (a) What is the speed of q2 when the spheres are 0.400 m apart? (b) How close does q2 get to q1?Explanation / Answer
You can easily solve this problem from energy conservation point of view. When the two spheres are 0.800 m apart, the kinetic energy is: Ka = 0.5*m2*v2^2, where m2 = 1.50g = 0.00150kg and v2 = 22.0 m/s Ka=0.363 The potential energy of q2 in the electric field of q1 is: Pa = k*q1*q2/D, where k = 1/4*?*eo the Coulomb constant, q1 = -2.80 ?C, q2 = -7.80 ?C, and D = 0.800 m4 Pa=0.2457 When the two spheres are 0.400 m apart, the kinetic energy is: Kb = 0.5*m2*V2^2, where V2 is the unknown speed of q2. The potential energy of q2 in the electric field of q1 is: Pb = k*q1*q2/d, where d = 0.040 m.= 0.491 Applying energy conservation, we must have: Ka+Pa = Kb+Pb ==>0.363+0.2457=0.491+ 1/2*0.00150*v^2==> v2=12.52 Applying energy conservation again, we must have: Ka+Pa = Kc+Pc, where Kc =0 is the kinetic energy when q2 is closest to q1 Pc = k*q1*q2/S and S is the unknown closest distance between q1 and q2. 0.363+0.2457= 0+ 9*10^9* 2.8*7.8/s==>0.3229
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