Question

There are two objects (A and B) each with its own mass and charge. The electrostatic force between them has a value of 23.2N and is attractive. The two objects are separated by a distance of 2.3cm with object A being to the left of B. Assume one charge has 4 times the charge of the other.

a. Do the two charges have the same signs or opposite signs?
b. IF the two charges have opposite signs, is charge A or B negative?
c. Which charge (A or B) has the higher value?
d. Find the value of each charge.

Answer 1

Step-by-step explanation:

a. Since the electrostatic force is attractive, the charges must have opposite signs.

b. Let's assume that charge A is positive and charge B is negative. Then, the electrostatic force would be given by:

F = (k * |qA| * |qB|) / r^2

where k is Coulomb's constant, r is the distance between the charges, and |qA| and |qB| are the magnitudes of the charges. Since the force is attractive, |qA| > |qB|.

c. From the given information, we know that:

F = 23.2 N

r = 2.3 cm = 0.023 m

Let |qB| = q, then |qA| = 4q. Substituting these values into the equation for the electrostatic force, we get:

23.2 = (k * 4q * q) / (0.023)^2

Solving for q, we get:

q = 3.38 x 10^-7 C

Then, |qA| = 4q = 1.35 x 10^-6 C.

d. Charge A has a value of 1.35 x 10^-6 C and charge B has a value of -3.38 x 10^-7 C.