Two identical spheres are resting on a horizontal surface at a distance of 0.01 m from each other, each attached to an identical unstretched spring with a spring constant of 100 N/m. If we charge the spheres with a negative charge of -1.60 µC each, what is the compression of each spring? Coulomb's constant is k = 9.0 × 10⁹ N m²/C². A)0.14 m B)0.18 m C)0.32 m D)0.36 m

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Quizplus · Accepted Answer

The force between the two charged spheres can be calculated using Coulomb's law: $F = \frac{kq_1q_2}{r^2}$, where $k = 9.0 \times 10^9 \, \text{N m}^2/\text{C}^2$, $q_1 = q_2 = -1.60 \times 10^{-6} \, \text{C}$, and $r = 0.01 \, \text{m}$. The force due to the spring can be described by Hooke's law: $F = -kx$, where $k$ is the spring constant (100 N/m in this case) and $x$ is the compression of the spring. Setting the force from Coulomb's law equal to the force from Hooke's law and solving for $x$ gives the compression of the spring.

Two Identical Spheres Are Resting on a Horizontal Surface at a Distance