Step-by-step explanation:
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To determine how high above the base the charged medicine ball floats, we need to consider the electrostatic force between the two charged objects and the gravitational force acting on the ball.
The electrostatic force (Coulomb force) between two charges is given by Coulomb's law:
\(F = \dfracq_1 \cdot q_2{r^2}\)
where:
\(F\) = electrostatic force (in Newtons)
\(k\) = Coulomb's constant (\(8.99 \times 10^9 \, N \cdot m^2/C^2\))
\(q_1\) and \(q_2\) = magnitudes of the charges (in Coulombs)
\(r\) = distance between the charges (in meters)
The gravitational force acting on the medicine ball is given by Newton's law of gravitation:
\(F_g = m \cdot g\)
where:
\(F_g\) = gravitational force (in Newtons)
\(m\) = mass of the ball (in kilograms)
\(g\) = acceleration due to gravity (\(9.81 \, m/s^2\))
Since the ball is floating, the electrostatic force and the gravitational force are balanced:
\(F_{\text{electrostatic}} = F_{\text{gravitational}}\)
Now, let's calculate the electrostatic force:
\(F_{\text{electrostatic}} = \dfrac{(8.99 \times 10^9 \, N \cdot m^2/C^2) \cdot |(100 \times 10^{-6} \, C) \cdot (17 \times 10^{-6} \, C)|}{r^2}\)
\(F_{\text{electrostatic}} = \dfrac{(8.99 \times 10^9 \, N \cdot m^2/C^2) \cdot (1.7 \times 10^{-3} \, C^2)}{r^2}\)
Now, set the electrostatic force equal to the gravitational force:
\(\dfrac{(8.99 \times 10^9 \, N \cdot m^2/C^2) \cdot (1.7 \times 10^{-3} \, C^2)}{r^2} = 10 \, kg \cdot 9.81 \, m/s^2\)
Now, solve for the distance \(r\):
\(r^2 = \dfrac{(8.99 \times 10^9 \, N \cdot m^2/C^2) \cdot (1.7 \times 10^{-3} \, C^2)}{10 \, kg \cdot 9.81 \, m/s^2}\)
\(r^2 \approx 31.14 \, m^2\)
\(r \approx \sqrt{31.14} \approx 5.58 \, m\)
So, the medicine ball floats approximately 5.58 meters above the base.