Respuesta :
Answer:
(a) q1 = 0.4μC, q2 = 14.6μC
(b) q1 = 15.36μC, q2 = -0.36μC
Explanation:
Parameters given:
Charges q1 and q2, such that:
q1 + q2 = 15μC
Distance between them, r = 2.5m
Electrostatic force between them, F = 0.008N
Electrostatic force, F is given as
F = (k*q1*q2)/(r^2)
A. If they are both positive and q1 is smaller:
q1 + q2 = (15 * 10^-6)
q1 = (15 * 10^-6) - q2
=> F = kq2{(15 * 10^-6) - q2} / r^2
=> 0.008 * r^2 = k * (15 * 10^-6)q2 - k(q2)^2
Inputting the values of k = 9 * 10^9 and r = 2.5:
0.05 = 135000q2 - (9*10^9)(q2)^2
=> (9*10^9)(q2)^2 - 135000q2 + 0.05 = 0
This is a quadratic equation. Solving using the quadratic formula yields:
q2 = 14.6μC or 0.38μC
=> q1 = (15 * 10^-6) - (14.6 * 10^-6)
q1 = 0.4μC
Or
=> q1 = (15 * 10^-6) - (0.38 * 10^-6)
q1 = 14.62μC
Since q1 < q2, then:
q1 = 0.4μC and q2 = 14.6μC
B. q1 is positive and q2 is negative, then:
q1 + q2 = 15μC
q1 = 15μC - q2
=> F = -(k*q1*q2)/(r^2)
F = -k*q2{(15 * 10^-6) - q2} / r^2
0.008 * r^2 = -k * (15 * 10^-6)q2 + k(q2)^2
Inputting the values of k = 9 * 10^9 and r = 2.5:
0.05 = -135000q2 + (9*10^9)(q2)^2
=> (9*10^9)(q2)^2 - 135000q2 - 0.05 = 0
Solving using the quadratic formula yields:
q2 = 15.4μC or -0.3617μC
Since q2 must be negative,
q2 = -0.36μC
=> q1 = (15 * 10^-6) - (-0.36 * 10^-6)
q1 = (15 * 10^-6) + (0.36 * 10^-6)
q1 = 15.36μC
