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OLEGan [10]
3 years ago
7

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
Physics
1 answer:
Umnica [9.8K]3 years ago
5 0
The answer is c. +2.0 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

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