Lab-Capacitance

Capacitance – PHY 202 Introduction: The purpose of this experiment is to explore basic concepts related to capacitance by investigating how the capacitance of a parallel-plate capacitor varies when the plate separation is changed or when the plate area is changed. Other capacitance related parameters, such as stored charge and stored energy, are also examined. Theory: Capacitors store charge. A capacitor can be made with any two conductors kept insulated from each other. If the conductors are connected to a potential difference, V, as for example the opposite terminals of a battery, then the two conductors are charged with equal but opposite amounts of charge Q, which is then referred to as the “charge in the capacitor”. The actual net charge on the capacitor is zero. The capacitance of the device is defined as the amount of charge Q stored in each conductor divided by the potential difference V applied: C = Q/V Rearranging gives: V = Q/C Eq. (1) A simple form of a capacitor consists of two parallel conducting plates, each with area A, separated by a distance d. The charge is uniformly distributed on the surface of the plates. The capacitance of this parallel-plate capacitor is given by: C = κε0A/d where κ is the dielectric constant of the insulating material between the plates (κ = 1 for vacuum; other values are measured experimentally and can be found in tables), and ε0 is the permittivity constant, of universal value ε0 = 8.854 x 10-12 F/m. The SI unit of capacitance is the Farad (F). An insulating material placed between the plates of a capacitor will increase its capacitance by a factor κ called the dielectric constant: C = κC0 Eq. (2) with C0 = ε0A/d being the capacitance when there is vacuum between the plates of the capacitor. Dielectric materials are non-conductive. Any dielectric material can be used to keep the plates in a capacitor insulated from each other (preventing them from touching and discharging). To three significant figures, κ = 1.00 for air. For all other materials, κ > 1. If the charge on a capacitor is kept constant while a dielectric is inserted between the plates (the capacitor is not connected to a voltage source), Equations 1 & 2 yield: Q = CV = C0V0 = (C/κ)V0 so V = V0/κ Where V0 is the voltage before inserting the dielectric and V is the voltage after insertion. Since κ > 1 always, we have: V < V0 Eq. (3) Procedure Open the PhET simulation Capacitor Lab Basics: https://phet.colorado.edu/sims/html/capacitor-lab-basics/latest/capacitor-labbasics_en.html Select Capacitance mode. Click the corresponding squares to turn on the readings for Capacitance, Top Plate Charge, Stored Energy, Plate Charges and Bar Graphs. For the first part of the experiment “The Effect of Plate Separation” on the next page, you should create a table and manually enter your data and at the end provide the corresponding graphs. Notes: • Be careful to use appropriate SI units in your work (the simulation uses different units). • Use 0 = 8.854×10-12 C2 / (N m2) (electric constant: epsilon zero). • p stands for pico or 10-12. 1. The Effect of Plate Separation • • • • • • • • Set the battery voltage to 1.5 V, the plate separation to 2 mm, and the plate area to 200 mm2. Increase the plate separation in increments of 1 mm, from 2 mm to 10 mm. At each step record the capacitance, the top plate charge and the stored energy. Plot the capacitance C (C = κε0A/d) as a function of the plate separation d. Fit the data with a function y = a/x + b. Calculate the dielectric constant κ from the coefficient ‘a’ in the fitting. What value did you obtain for κ? Is this value reasonable? Explain. For the initial configuration: voltage = 1.5 V, plate separation = 2 mm, and plate area = 200 mm2: 1.a) Calculate analytically the charge stored in each capacitor and compare with the simulation value. 1.b) Calculate analytically the energy stored in the capacitor and compare with the simulation value. 1.c) How do capacitance, charge and stored energy change when: • The plate separation is doubled? • The plate area is doubled? • A new dielectric is inserted between the plates with double the dielectric constant? Include a copy of your work along with the table and the graph (with the fitting) in your report as well as the answers to the questions. 2. The Effect of Plate Area • • • • • Set the battery voltage to 1.5 V, the plate separation to 2 mm, and the plate area to 100 mm2. Record the capacitance for this setting. Increase the plate area to 500 mm2 from 100 mm2. Record the capacitance for this setting as well. How did increasing the area of the plates affected the capacitance?

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