Answer:
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;
[tex]F = \frac{kq_1q_2}{r^2} \\\\F = \frac{(9\times 10^9)(4.6\times 10^{-7})(7.5\times 10^{-7})}{(0.03)^2} \\\\F= 3.45 \ N[/tex]
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
