Question

 A negative charge of -0.590 μC exerts an upward 0.900-N force on an unknown charge that is located 0.500 m directly below the first charge.

 Part A

 What is the value of the unknown charge (magnitude and sign)?

 Part B

 What is the magnitude of the force that the unknown charge exerts on the -0.590 μC charge?

Answer 1

Part A

The negative charge -0.590 C attracts( upwards force on a below charge) an unknown charge that is 0.5 m below it. Since the force is attraction, hence it has to be a positive charge.
Let the unknown charge be q
Now the force of attraction given by Columb's Law is
F_e = k (0.590 * 10^-6) (q) / (0.5)²
F_e = 0.9 N

0.9 = 9*10⁹ * (0.590 * 10^-6) (q) / (0.25) [ here k = 9*10⁹ ]
q = (0.9 * 0.25) / (9*10⁹)* (0.590 * 10^-6)
q = 4.23 * 10^-5 = 42.43 C

Part B

The force by which the +42.43 C charge acts on the -0.590 C is equal and opposite by Newtons Law.
The magnitude is F_e = 0.900N