A charge of 19.0 C is held fixed at the origin. 1. If a -5.00 C charge with a ma

Question

A charge of 19.0 C is held fixed at the origin.

1. If a -5.00 C charge with a mass of 3.10 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? v =  m/s

2. Suppose the -5.00 C charge is released from rest at the point x = 1/2(0.925m) and y = 1/2(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?

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

a. Greater than the speed found in part A b. Lee than the speed found part in A c. Equal to the speed found in part A

Explanation / Answer

1. PE = k q1 q2 / r

r = sqrt(0.925^2 + 1.17^2) = 1.49 m

Applying energy conservation,

PEi + KEi = PEf + KEf

(9 x 10^9 x -5 x 10^-06 x 19 x 10^-6 / 1.49) + 0 = (9 x 10^9 x -5 x 10^-06 x 19 x 10^-6 / (1.49/2) ) + (3.10 x 10^-3)v^2 / 2

- 0.574 + 0 = - 1.1476 + (1.55 x 10^-3)v^2

v = 19.24 m/s

2. now r' = 0.746 m

now diffrence in Pe is greater so speed will be greater.

Ans: (a)

3. - 1.146 + 0 = 2.293 + (1.55 x 10^-3 ) v^2

v = 27.2 m/s

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