Problem 79 - University Physics with Modern Physics

Question

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.

Step-by-Step Solution

Answer

Answer: Approximately \(3.09 \cdot 10^{10}\) electrons need to be added.

Step 1: Identify the relevant forces acting on the object

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.

Step 2: Determine the gravitational force acting on the object

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. \(\textit{F}_g = (0.001\ \text{kg})(9.81\ \text{m/s}^2) = 9.81 \cdot 10^{-3} \mathrm{N}\).

Step 3: Determine the electric field due to the charged sheet

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: \(E = \dfrac{-3.5 \cdot 10^{-5} \mathrm{C} / \mathrm{m}^{2}}{2(8.85 \cdot 10^{-12} \mathrm{C} / \mathrm{N} \cdot \mathrm{m}^{2})} = -1.98 \cdot 10^{6} \mathrm{N} / \mathrm{C}\)

Step 4: Determine the electric force acting on the object

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 \(q\): \(qE = \textit{F}_g \Rightarrow q = \dfrac{\textit{F}_g}{E} = \dfrac{9.81 \cdot 10^{-3} \mathrm{N}}{-1.98 \cdot 10^{6} \mathrm{N} / \mathrm{C}} \approx -4.95 \cdot 10^{-9}\ \mathrm{C}\)

Step 5: Calculate the number of electrons

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.

Key Concepts

Electric field due to charged sheet

The electric field generated by a charged sheet is a fundamental concept in electrostatics. When we deal with an infinitely large, uniformly charged, nonconducting sheet, it creates an electric field that is directed perpendicular to the surface. The strength of this electric field is determined by the surface charge density, \(\sigma\), of the sheet. The formula for calculating this electric field (E) is: \[E = \frac{\sigma}{2 \epsilon_0}\] where \(\epsilon_0\) is the vacuum permittivity, a constant valued at \(8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2\).Notably, the electric field due to the sheet is uniform, meaning its strength does not change with distance from the sheet. This property simplifies calculations as we don't need to consider variations in field strength at different points in space.

Gravitational force

Gravitational force is one of the most familiar forces we experience in our daily lives, pulling objects toward the center of the Earth. It acts on all objects with mass, and its magnitude is calculated using the formula: \[F_g = mg\] where \(m\) is the mass of the object and \(g\) is the acceleration due to gravity, approximately \(9.81 \, \text{m/s}^2\) on Earth's surface.In the exercise we are considering, this gravitational force pulls the object vertically downward. When combined with the electric force, students can learn how different forces come into play to influence an object's motion or keep it steady.

Electric force

The electric force acting on a charged object within an electrical field is determined by the charge of the object and the strength of the field. For our charge \(q\) in the exercise, the resulting electric force can be calculated using:\[F_e = qE\]Here, \(E\) is the electric field strength we calculated earlier. In the context of an object suspended over a charged sheet, the electric force must counteract the gravitational force to keep the object stationary.When \(F_g = F_e\), the object remains motionless, meaning that the forces are in perfect balance.

Charge of an electron

The charge of an electron is one of the most fundamental constants in physics, noted for its small magnitude, \(e = -1.6 \times 10^{-19}\, \text{C}\). Electrons, being negatively charged, are crucial in determining the overall charge on any object when they are added or removed.To find out how many electrons need to be added or removed to achieve a desired charge, the equation \[q = Ne\] is used, where \(N\) is the number of electrons. This relationship outlines how tiny changes in electron count can result in significant changes in an object's net charge owing to the minuscule size of the electron's charge.Understanding this small yet powerful constant helps students to appreciate the collective impact of these tiny particles in electrostatic calculations.