A point charge q1 = 88.5 uC is held fixedat the origin. A second point charge, w
ID: 1665716 • Letter: A
Question
A point charge q1 = 88.5 uC is held fixedat the origin. A second point charge, withmass m = 5.40×102 kgand charge q2 = -2.46 uC, is placed at thelocation (0.323 m, 0). Find the electric potential energy of this system ofcharges. If the second charge is released from rest, what is its speedwhen it reaches the point (0.121 m, 0)? A point charge q1 = 88.5 uC is held fixedat the origin. A second point charge, withmass m = 5.40×102 kgand charge q2 = -2.46 uC, is placed at thelocation (0.323 m, 0). Find the electric potential energy of this system ofcharges. If the second charge is released from rest, what is its speedwhen it reaches the point (0.121 m, 0)?Explanation / Answer
a. Potential energy ofcharges q1 and q2 separated by distance'r' U = - (1/40)* q1 * q2 / r here r = (x2+ y2) r1 = (0.3232+ 02) = 0.323 m U1 = 9* 109 * 88.5 * 10-6 * ( - 2.46 *10-6) / 0.323 = -6.066 J b. P.E. of chargesat r2 = (0.1212+02) = 0.121 m U2 = 9* 109 * 88.5 * 10-6 * ( - 2.46 *10-6) / 0.121 = -16.193 J ChangeinP.E. U = U2 - U1 = -16.193 - ( - 6.066) = -10.127 J - ve sign could be discarded as it onlyindicates the bound state. According to law of conservation ofmechanical energy, K.E. = U (1/2)* m *v2 = 10.127 J 0.5 * 5.40 * 10-2 *v2 = 10.127 v = (10.127/ 2.7 * 10-2) = 19.37 m/s = -16.193 - ( - 6.066) = -10.127 J - ve sign could be discarded as it onlyindicates the bound state. According to law of conservation ofmechanical energy, K.E. = U (1/2)* m *v2 = 10.127 J 0.5 * 5.40 * 10-2 *v2 = 10.127 v = (10.127/ 2.7 * 10-2) = 19.37 m/sRelated Questions
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