An object with mass \(m=1.0 \mathrm{~g}\) and charge \(q\) is placed at point \(A\), which is \(0.05 \mathrm{~m}\) above an infinitely large, uniformly charged, nonconducting sheet \(\left(\sigma=-3.5 \cdot 10^{-5} \mathrm{C} / \mathrm{m}^{2}\right)\), as shown in the figure. Gravity is acting downward \(\left(g=9.81 \mathrm{~m} / \mathrm{s}^{2}\right)\). Determine the number, \(N\), of electrons that must be added to or removed from the object for the object to remain motionless above the charged plane.

We know that the gravitational force (\(\textit{F}_g\)) acts vertically downward on the object, and due to the charged sheet, an electric force (\(\textit{F}_e\)) acts on it as well. We want to find the number of electrons, \(N\), which needs to be added/removed, so we need to know the electric force on this object and the gravitational force to balance them.

The gravitational force acting on the object is given by the equation: \(\textit{F}_g = mg\). Here, \(m\) is the mass of the object and \(g\) is the acceleration due to gravity. Using the given mass, \(m = 1.0\ \text{g} = 0.001\ \text{kg}\), and \(g = 9.81\ \text{m/s}^2\), we can find the gravitational force acting on the object. \(\textit{F}_g = (0.001\ \text{kg})(9.81\ \text{m/s}^2) = 9.81 \cdot 10^{-3} \mathrm{N}\).

To find the electric force acting on the object, we first need to determine the electric field generated by the charged sheet. The formula to calculate the electric field (\(E\)) for an infinitely large, uniformly charged, nonconducting sheet is given by: \(E = \dfrac{\sigma}{2 \epsilon_0}\), where \(\sigma\) is the surface charge density and \(\epsilon_0\) is the vacuum permittivity (\(\epsilon_0 = 8.85 \cdot 10^{-12} \mathrm{C} / \mathrm{N} \cdot \mathrm{m}^{2}\)). Using the given surface charge density, \(\sigma = -3.5 \cdot 10^{-5} \mathrm{C} / \mathrm{m}^{2}\), we can find the electric field due to the charged sheet: \(E = \dfrac{-3.5 \cdot 10^{-5} \mathrm{C} / \mathrm{m}^{2}}{2(8.85 \cdot 10^{-12} \mathrm{C} / \mathrm{N} \cdot \mathrm{m}^{2})} = -1.98 \cdot 10^{6} \mathrm{N} / \mathrm{C}\)

The electric force acting on the object can be calculated using the formula: \(\textit{F}_e = qE\), where \(q\) is the charge of the object and \(E\) is the electric field. To keep the object motionless, we want the electric force to balance the gravitational force. So, we can set up the equation: \(\textit{F}_e = \textit{F}_g\) and solve for the charge \(q\): \(qE = \textit{F}_g \Rightarrow q = \dfrac{\textit{F}_g}{E} = \dfrac{9.81 \cdot 10^{-3} \mathrm{N}}{-1.98 \cdot 10^{6} \mathrm{N} / \mathrm{C}} \approx -4.95 \cdot 10^{-9}\ \mathrm{C}\)

The charge of an electron is \(e = 1.6 \cdot 10^{-19}\ \mathrm{C}\) (negative sign indicates that the electron is negatively charged). To find the number of electrons (\(N\)) that need to be added or removed from the object, we can use the following equation: \(q = Ne\), where \(N\) is the number of electrons and \(e\) is the elementary charge. Solving for \(N\), we get: \(N = \dfrac{q}{e} = \dfrac{-4.95 \cdot 10^{-9}\ \mathrm{C}}{-1.6 \cdot 10^{-19}\ \mathrm{C}} \approx 3.09 \cdot 10^{10}\) Since the object's charge is negative and we want to keep it motionless, we need to add electrons to have a balanced electric force. Hence, approximately \(3.09 \cdot 10^{10}\) electrons need to be added to the object to keep it motionless above the charged plane.