From collisions with cosmic rays and from the solar wind, the Earth has a net electric charge of approximately \(-6.8 \cdot 10^{5} \mathrm{C} .\) Find the charge that must be given to a \(1.0-\mathrm{g}\) object for it to be electrostatically levitated close to the Earth's surface.

First, we need to calculate the gravitational force acting on the object. We are given the mass of the object as 1.0 g, which we need to convert to kg: \(m = 1.0 \times 10^{-3} \mathrm{kg}\) To find the gravitational force, we use the formula \(F_g = m \cdot g\), where g is the acceleration due to gravity, which is approximately \(9.81\,\mathrm{m/s^2}\). \(F_g = (1.0 \times 10^{-3}\,\mathrm{kg})\cdot (9.81\,\mathrm{m/s^2}) = 9.81 \times 10^{-3}\,\mathrm{N}\)

Now, we will set up the equation for the electrostatic force between the Earth and the object. We will use the following formula: \(F_e = k \frac{q_1 \cdot q_2}{r^2}\). Since we want electrostatic levitation, we can set \(F_g = F_e\): \(9.81 \times 10^{-3}\,\mathrm{N} = k \frac{q_1 \cdot q_2}{r^2}\)

To solve for the charge of the object, we need to plug in the values that we know. We know that the Earth's charge is approximately \(-6.8 \times 10^{5}\,\mathrm{C}\), and the distance between the object and the Earth's surface can be assumed to be the Earth's radius, which is approximately \(6.37 \times 10^6\,\mathrm{m}\). Also, the Coulomb's constant, k, is \(8.988 \times 10^9\,\mathrm{N\,m^2/C^2}\) Substituting these values into the equation, we get: \(9.81 \times 10^{-3}\,\mathrm{N} = (8.988 \times 10^9\,\mathrm{N\,m^2/C^2})\frac{(-6.8 \times 10^5\,\mathrm{C}) \cdot q_2}{(6.37 \times 10^6\,\mathrm{m})^2}\) Now, we can solve for \(q_2\): \(q_2 = \frac{(9.81 \times 10^{-3}\,\mathrm{N})\cdot(6.37 \times 10^6\,\mathrm{m})^2}{(8.988 \times 10^9\,\mathrm{N\,m^2/C^2})\cdot(-6.8 \times 10^5\,\mathrm{C})} = -2.44 \times 10^{-8}\,\mathrm{C}\) So, the charge that must be given to the 1.0-g object for it to be electrostatically levitated close to the Earth's surface is approximately \(-2.44 \times 10^{-8}\,\mathrm{C}\).