1. A point charge q is placed midway between two identical positive point charge
ID: 1682934 • Letter: 1
Question
1. A point charge q is placed midway between two identical positive point charges of magnitude 5.14 10-9 C. The resultant force on each of the three charges is zero. Find q.2. Two point charges 0.564 m apart experience a repulsive force of 0.396 N. The sum of the two charges equals 1.10 10-5 C. Find the values of the two charges. (Enter your answers from smallest to largest.)
anything will help im stuck thinking i need the sum of the forces to equal zero but im left to 2q=0 which is 0 but i know thats not right! thanks
Explanation / Answer
(a) q1 = q2 = 5.14*10-9 C q = ? Let ' r ' be the distance between q1 & q2 force acting on q1 due to q, F1 = Kq1*q/(r/2)2 where k = electrostatic constant = 9*109 Nm2/C2 = (k *5.14*10-9 *q *4)/r2 (from q to q1) force acting on q1 due to q2, F2 = Kq1*q2/r2 where k = electrostatic constant = 9*109 Nm2/C2 = k*(5.14*10-9 )2/r2 (from q2 to q1) Given that the resultant force acting on each of three particles is 0. so F1 + F2 = 0 k *5.14*10-9 *q *4)/r2 + k*(5.14*10-9 )2/r2 = 0 (k *5.14*10-9 *q *4)/r2 = - k*(5.14*10-9 )2/r2 4q = - 5.14*10-9 q = -1.285*10-9 C (b) Given q1 + q2 = 1.10*10-5 C -----------------------(1) distance between q1 & q2, r = 0.564 m repulsive force F = 0.396 N kq1q2/r2 = 0.396 q1*q2 = 0.396*(0.564)2 / 9*109 = 0.1399*10-10 C2 -----------------------------(2) (q1 - q2)2 = (q1 + q2)2 -4*q1*q2 = (1.10*10-5)2 - 4 *0.1399*10-10 = 1.21 * 10-10 - 0.5596 *10-10 = 0.6504*10-10 so q1 - q2 = 0.8064*10-5 C -----------------------------(3) Add equations (1) & (3) we get 2 q1 = 1.9064*10-5 C q1 = 0.9532*10-5 C substitute the q1 value in (1) then q2 = 0.1468*10-5 C q1 = 0.9532*10-5 C & q2 = 0.1468*10-5 C force acting on q1 due to q2, F2 = Kq1*q2/r2 where k = electrostatic constant = 9*109 Nm2/C2 = k*(5.14*10-9 )2/r2 (from q2 to q1) Given that the resultant force acting on each of three particles is 0. so F1 + F2 = 0 k *5.14*10-9 *q *4)/r2 + k*(5.14*10-9 )2/r2 = 0 (k *5.14*10-9 *q *4)/r2 = - k*(5.14*10-9 )2/r2 4q = - 5.14*10-9 q = -1.285*10-9 C (b) Given q1 + q2 = 1.10*10-5 C -----------------------(1) distance between q1 & q2, r = 0.564 m repulsive force F = 0.396 N kq1q2/r2 = 0.396 q1*q2 = 0.396*(0.564)2 / 9*109 = 0.1399*10-10 C2 -----------------------------(2) (q1 - q2)2 = (q1 + q2)2 -4*q1*q2 = (1.10*10-5)2 - 4 *0.1399*10-10 = 1.21 * 10-10 - 0.5596 *10-10 = 0.6504*10-10 so q1 - q2 = 0.8064*10-5 C -----------------------------(3) Add equations (1) & (3) we get 2 q1 = 1.9064*10-5 C q1 = 0.9532*10-5 C substitute the q1 value in (1) then q2 = 0.1468*10-5 C q1 = 0.9532*10-5 C & q2 = 0.1468*10-5 C force acting on q1 due to q2, F2 = Kq1*q2/r2 where k = electrostatic constant = 9*109 Nm2/C2 = k*(5.14*10-9 )2/r2 (from q2 to q1) Given that the resultant force acting on each of three particles is 0. so F1 + F2 = 0 k *5.14*10-9 *q *4)/r2 + k*(5.14*10-9 )2/r2 = 0 (k *5.14*10-9 *q *4)/r2 = - k*(5.14*10-9 )2/r2 4q = - 5.14*10-9 q = -1.285*10-9 C (b) Given q1 + q2 = 1.10*10-5 C -----------------------(1) distance between q1 & q2, r = 0.564 m repulsive force F = 0.396 N kq1q2/r2 = 0.396 q1*q2 = 0.396*(0.564)2 / 9*109 = 0.1399*10-10 C2 -----------------------------(2) (q1 - q2)2 = (q1 + q2)2 -4*q1*q2 = (1.10*10-5)2 - 4 *0.1399*10-10 = 1.21 * 10-10 - 0.5596 *10-10 = 0.6504*10-10 so q1 - q2 = 0.8064*10-5 C -----------------------------(3) Add equations (1) & (3) we get 2 q1 = 1.9064*10-5 C q1 = 0.9532*10-5 C substitute the q1 value in (1) then q2 = 0.1468*10-5 C q1 = 0.9532*10-5 C & q2 = 0.1468*10-5 C q = -1.285*10-9 C (b) Given q1 + q2 = 1.10*10-5 C -----------------------(1) distance between q1 & q2, r = 0.564 m repulsive force F = 0.396 N kq1q2/r2 = 0.396 q1*q2 = 0.396*(0.564)2 / 9*109 = 0.1399*10-10 C2 -----------------------------(2) (q1 - q2)2 = (q1 + q2)2 -4*q1*q2 = (1.10*10-5)2 - 4 *0.1399*10-10 = 1.21 * 10-10 - 0.5596 *10-10 = 0.6504*10-10 so q1 - q2 = 0.8064*10-5 C -----------------------------(3) Add equations (1) & (3) we get 2 q1 = 1.9064*10-5 C q1 = 0.9532*10-5 C substitute the q1 value in (1) then q2 = 0.1468*10-5 C q1 = 0.9532*10-5 C & q2 = 0.1468*10-5 CRelated Questions
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