5. Two negative charges that are both - 3.8 C push each other apart with a force of 19.0 N. How far apart

are the two charges?

82700 m

Explanation:

Applying Coulomb's law of electrostatic

F = kq²/r² ... Equation 1

Note: Both charges are the same

Where F = force of attraction, q = negative charge, r = distance between the charges, k = coulomb's constant

make r the subject of the equation

r = √ (kq²/F) ... Equation 1

Given: F = 19 N, q = 3.8 C, k = 9*10⁹ Nm²/C²

Substitute these values into equation 2

r = √ (3.8²*9*10⁹/19)

r = √ (68.4*10⁸)

r = 8.27*10⁴ m

r = 82700 m

Hence the two charges are 82700 m apart