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

speed

ratio of distance traveled to time interval

net force

vector sum of forces on object

object

source of diverging light rays either luminous or illuminated

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explanation: