Question

Two charges q1 and q2 have a total charge of 15 ?C. When they are separated by 2.5 m, the force exerted by one charge on the other has a magnitude of 8 mN. Find q1 and q2 if
(a) both are positive so that they repel each other, and q1 is the smaller of the two.
q1=_______?C , q2= _______?C

(b) one, say q1, is positive and the other is negative so that they attract each other.
q1=_______?C , q2= _______?C

Answer 1

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