IPA charge of 22.0 C is held fixed at the origin. Part A If a-6.50 C charge with

Question

IPA charge of 22.0 C is held fixed at the origin. Part A If a-6.50 C charge with a mass of 4.00 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? m/s Submit My Answers Give Up Part B Suppose the-650 C charge is released from rest at the point z = 0.925m) and y-, 1.17 m). When it is halfway to the origin, is its speed greater than, less than, or equal to the speed found in part A? O Greater than the speed found in part A O Lee than the speed found part in A O Equal to the speed found in part A Submit My Answers Give Up Part C Find the speed of the charge for the situation described in part B. m/s

Explanation / Answer

Since F = K * q1 * q2 / r2   = m v dV / dr

K * q1 * q2 * dr / r2 = m * v dv

9 * 10 9 * 22* 6.50 * 10 -20 * [-1 / r] = 4 * 10 -3 * [v2 /2]

Taking limit in Y -direction 0.925 to 0.925/2 ,

9 * 10 9 * 22* 6.50 * 10 -20 * [-1 / r]0.925/20.925 = 4 * 10 -3 * [vy2 /2]v0

Vy = 26.38 * 10 -4

similarly , applying the limit in x coordinate from 1.17 to 1.17/2

Vx = 23.45 * 10-4 m/s

so , net velocity = sqrt ( Vy2 + Vx2 )

= 35.30 m/s

b ) Speed will be greater than the speed found in part A because as charge particles go closer , greater force acts on it which will increse its speed

c) From above , 9 * 10 9 * 22* 6.50 * 10 -20 * [-1 / r] = 4 * 10 -3 * [v2 /2]

Taking limit in Y -coordinate from 0.925/2 to 0.925/4 ,

we get Vy = 37.30 * 10 -4 m/s

and Vx = 23.45 * 1*414 * 10 -4 m/s = 33.15 *10 -4 m/s

Now net velocity V = 49.90 *10 -4 m/s ( formula is given in part a , answer )

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