A mass with a charge of 4.60 x 10-7 C rests on a frictionless surface. A compressed spring exerts a force on the mass on the left side. Three centimeters to the right of the mass is a 7.50 x 10-7 C fixed charge. How much is the spring compressed if its spring constant is 14 N/m

The compression of the spring is 24.6 cm

Explanation:

magnitude of the charge on the left, q₁ = 4.6 x 10⁻⁷ C

magnitude of the charge on the right, q₂ = 7.5 x 10⁻⁷ C

distance between the two charges, r = 3 cm = 0.03 m

spring constant, k = 14 N/m

The attractive force between the two charges is calculated using Coulomb's law;

The extension of the spring is calculated as follows;

F = kx

x = F/k

x = 3.45 / 14

x = 0.246 m

x = 24.6 cm

The compression of the spring is 24.6 cm

a. is the answer because they are not shared

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