A small metal sphere, carrying a net charge of q1 ? q2 get to m apart? B. How cl

Question

A small metal sphere, carrying a net charge of q1 ? q2 get to m apart? B. How close does q2 when the spheres are 0.420 A. What is the speed of m/s . Assume that the two spheres can be treated as point charges. You can ignore the force of gravity. q1 with speed 22.0 q2 is moving toward m apart, q1 . When the two spheres are 0.800 g , is projected toward MuC and mass 1.70 q2 = -7.50 MuC , is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of q1 = -2.70

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 = 1.80g = 0.00180kg and v2 = 22.0 m/s The potential energy of q2 in the electric field of q1 is: Pa = k*q1*q2/D, where k = 1/4*p*eo the Coulomb constant, q1 = -2.70 µC, q2 = -7.50 µC, and D = 0.800 m When the two spheres are 0.420 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 = 0.420 m. Applying energy conservation, we must have: Ka+Pa = Kb+Pb The only unknown in the equation above is V2. Solve the equation to get the answer to Question 1: V2. 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. Solve this equation you get the answer to Question 2

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