EE Science II Laboratory #04 Charge Distributions and Electric Field Visualization in 3D Pre-Laboratory Assignment Your name and group number: _____________ 1. Consider a charge q placed at (x, y) = (0, 0). The magnitude of the charge depends on your group number:  Groups5and6→q=5×10-3 C From Coulomb’s Law, the electric field produced by this charge at any point on the x-y plane is given by the following equation in the cylindrical coordinate system: 𝑞 𝐸 = (4𝜋𝜖)𝑅2 𝑟̂ a. Transform this expression into the Cartesian coordinate system and write down the x and y components of the field below: [1 Point] Ex = ___________________________ Ey = ___________________________ b. Complete Table 1 by calculating the x and y components of the E field for the given (R, φ) values. [1 Point] Table 1: Calculate the electric field in Cartesian coordinates for the given values of (R, φ) ⃗ c. Plot the electric field calculated in Table 1 using Grid #1 attached at the bottom of this document. [1 Point] d. How will the electric field change if the charge was located at (x, y) = (1, 2) instead of (0, 0)? [0.5 Point] 2. A linear charge distribution of length l and charge density λ is located along the x axis as shown in Figure 1, where l and λ depend on your group number – Groups5and6→l=40m, 𝜆=5×10−3 C m Laboratory #4 R 𝝓 (deg) Observation point (x, y) 𝑬𝒙 𝑬𝒚 0.5 0 (,) 1.0 0 (,) 1.5 0 (,) 0.5 45 (,) 0.5 90 (,) 0.5 180 (,) 1.0 270 (,) 1.5 135 (,)  University of South Florida 1 EE204-pre-lab.docx Groups7and8→l=30m. 𝜆=6×10−3 C m Calculate the electric field at any point on the y axis by following the steps listed below – Consider an infinitesimally small charge dQ = λ dx’ as shown in Figure 1. If (x’, y’) is used to denote any point on the charge distribution and (x,y) is used to denote any observation point, then the net electric field at (x,y) due to an infinitesimally small charge dQ placed at (x’, y’) is: 1 𝑑𝑄 1 𝑑𝑄((𝑥−𝑥′)𝑥̂+(𝑦−𝑦′)𝑦̂) 𝑑𝐸=4𝜋𝜀 𝑅2 𝑟̂= 4𝜋𝜀 ((𝑥−𝑥′)2+(𝑦−𝑦′)2)3/2 Show how 𝑟̂ equals ((𝑥−𝑥′)𝑥̂+(𝑦−𝑦′)𝑦̂) in the above step. [1 Point] 𝑅2 ((𝑥−𝑥′)2+(𝑦−𝑦′)2)3/2 d. In this problem, dQ = λ dx’, y’ = 0 (line charge distribution is on the x axis) and x = 0 (observation point is on the y axis). Since the charge distribution is uniform, the x components of the electric field cancel at any point on the y axis. Therefore, the net electric field is due to the y component alone. Calculate the electric field by integrating the y component in the above expression for x’ varying from -l/2 to l/2. Show your work clearly for full credit. [3 Points] Hint: ∫ 𝑑𝑥 = 𝑥 (𝑎2 + 𝑥2)3/2 𝑎2√𝑎2 + 𝑥2 Figure 1: Problem #2 – line charge distribution placed on the x – axis. 3. You will work with MATLAB in this lab. Please review the MATLAB tutorial and answer the following questions: 3.1. Write a MATLAB code line to implement the following equation [1.5 Points] 3.2. Laboratory #4 ⃗ see the file attached more clear.