IP A charge of 20.0 C is held fixed at the origin. Part A. If a -5.00 C charge w

Question

IP A charge of 20.0 C  is held fixed at the origin.

Part A. If a -5.00 C charge with a mass of 3.70 g is released from rest at the position (0.925 m, 1.17 m), what is its speed when it is halfway to the origin?

Part B. Suppose the -5.00 C charge is released from rest at the point x = 12(0.925m) and y= 12(1.17m). When it is halfway to the origin, is its speed greater than, less than, or equal to the speed found in part A?

Part C. Find the speed of the charge for the situation described in part B.

Explanation / Answer

(A) PE =k q1 q2 / r

initially , ri = sqrt(0.925^2 + 1.17^2) = 1.492 m

rf = ri/2 = 0.746 m

PEi + KEi = PEf + KEf

(9 x 10^9 x 20 x 10^-6 x -5 x 10^-6)/1.492 + 0 = (9 x 10^9 x 20 x 10^-6 x -5 x 10^-6)/0.746 + (0.0037) v^2 /2

v = 25.5 m/s

(B) distance has increased hence delta(PE) will be less.

hence speed will be less.

(C) (9 x 10^9 x 20 x 10^-6 x -5 x 10^-6)/12(1.492) + 0 = (9 x 10^9 x 20 x 10^-6 x -5 x 10^-6)/12(0.746) + (0.0037) v^2 /2

v = 7.4 m/s

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