A negative charge of -0.550 μC exerts an upward 0.900-N force on an unknown charge that is located 0.300 m directly below the first charge.

Required:

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

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

c. What is the direction of this force?

Answer:a. q2 = 16.4μC, positive charge

b. F = 0.900N

c. downward

Explanation:a.In order to calculate the charge of the unknown charge you use the following formula, for the electric force between two charges:[tex]F_e=k\frac{q_1q_2}{r^2}[/tex]

(1)k: Coulomb’s constant = 8.98*10^9Nm^2/C^2

r: distance between the charges = 0.300m

q1: charge 1 = -0.550 μC = 0.550*10^-6C

q2: charge 2 = ?

Fe: electric force = 0.900N

The force exerted in the second charge points upward, then, the sign of the second charge is positive because this charge is getting closer to the first one.

You solve the equation (1) for the second charge ans replace the values of the other parameters:

[tex]q_2=\frac{r^2F_e}{kq_1}=\frac{(0.300m)^2(0.900N)}{(8.98*10^9Nm^2/C^2)(0.550*10^{-6}C)}\\\\q_2=1.64*10^{-5}C\\\\q_2=16.4*10^{-6}C=16.4*10\mu C[/tex]

The values of the second charge is 1.64 μCb.By the third Newton Law, you have that the force exerted in the second charge is equal to the force exerted by the first charge on the second one.The force exerted on the first charge is 0.900Nc. The charges are attracting between them, then, the force exerted on the first charge is pointing downward.