Question

An object with mass \(m=1.00 \mathrm{~g}\) and charge \(q\) is placed at point \(A\), which is \(0.0500 \mathrm{~m}\) above an infinitely large, uniformly charged, nonconducting sheet \(\left(\sigma=-3.50 \cdot 10^{-5} \mathrm{C} / \mathrm{m}^{2}\right), \mathrm{a}\) 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.

Short answer

Answer: \(6.15 \times 10^{10}\) electrons.

Step-by-step solution

Calculate gravitational force

The gravitational force acting on the object is given by \(F_g = mg\), where \(m\) is the mass of the object and \(g\) is the acceleration due to gravity. In this case, \(m = 1.00 \, \text{g} = 1.00 \times 10^{-3} \, \text{kg}\) and \(g = 9.81 \, \text{m/s}^2\). Therefore, \(F_g = (1.00 \times 10^{-3} \, \text{kg})(9.81 \, \text{m/s}^2) = 9.81 \times 10^{-3} \, \text{N}\).

Calculate electric field due to charged sheet

The electric field due to an infinitely large charged sheet is given by \(E = \frac{\sigma}{2 \epsilon_0}\), where \(\sigma\) is the surface charge density and \(\epsilon_0\) is the vacuum permittivity. In this case, \(\sigma = -3.50 \times 10^{-5} \, \text{C/m}^2\) and \(\epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2\). Therefore, \(E = \frac{-3.50 \times 10^{-5} \, \text{C/m}^2}{2(8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2)} = -9.97 \times 10^5 \, \text{N/C}\).

Calculate electric force on the object

The electric force acting on the object is given by \(F_e = qE\), where \(q\) is the charge on the object and \(E\) is the electric field at its position. Since the object is motionless, it means that \(F_e = F_g\). Therefore, \(qE = 9.81 \times 10^{-3} \, \text{N}\). Now we can solve for the charge \(q\).

Solve for the charge on the object

We have the equation \(qE = 9.81 \times 10^{-3} \, \text{N}\). We can substitute the value of \(E\) to find the charge \(q\). Therefore, \(q(-9.97 \times 10^5 \, \text{N/C}) = 9.81 \times 10^{-3} \, \text{N}\). Therefore, \(q = 9.81 \times 10^{-3} \, \text{N} / (-9.97 \times 10^5 \, \text{N/C})=-9.84 \times 10^{-9} \, \text{C}\).

Convert charge to the number of electrons

Since the charge of an electron is \(e = -1.60 \times 10^{-19} \, \text{C}\), we can find the number of electrons added or removed from the object using the equation \(N = \frac{q}{e}\). In this case, \(N = \frac{-9.84 \times 10^{-9} \, \text{C}}{-1.60 \times 10^{-19} \, \text{C}} = 6.15 \times 10^{10}\). Since the object has a negative charge (\(q\)), it means that \(6.15 \times 10^{10}\) electrons must be added to the object for it to remain motionless above the charged plane.