An attractive force of 7.2 n occurs between two point charges that are 0.10 m apart. if one charge is -4.0 µc, what is the other charge? (µc = 1.0 × 10^-6 c) a. -4.0 µc b. -2.0 µc c. +2.0 µc d. +4.0 µc

To calculate this, we will use Coulomb's Law:

F = k*Q1*Q2/r²

where F is force, k is constant, Q is a charge, r is a distance between charges.

k = 9.0 × 10⁹  N*m/C²

It is given:

F = 7.2 N

d = 0.1 m =  10⁻¹ m

Q1 = -4.0 µC = 4 * 1.0 × 10⁻⁶ = 4.0 × 10⁻⁶

Q2 = ?

Thus, let's replace this in the formula for the force:

7.2 = 9.0 × 10⁹ * 4.0 × 10⁻⁶ * Q2/(10⁻¹)²

7.2 = 9 * 4 * 10⁹⁻⁶ * Q2/10⁻¹°²

7.2 = 36 × 10³ * Q2 / 10⁻²

Multiply both sides of the equation by 10⁻²:

7.2 × 10⁻² = 36 × 10³ * Q2

⇒ Q2 = 7.2 × 10⁻² / 36 × 10³ = 7.2/36 × 10⁻²⁻³ = 0.2 × 10⁻⁵ = 2 × 10⁻⁶ 

Since µC = 1.0 × 10^-6:

Q2 = 2 * 1.0 × 10^-6 = 2 µC

he . if i may ask what i your question exactly?

explanation:

0.863333

86%

hope this