L7/L8 FAD1011 (PHYSICS 3) 2. Capacitor and Dielectric 2.1 Capacitors 2.2 Capacitors in series and parallel 2.3 Capacitors with dielectrics Lecturer: Dr. Salmiah Ibrahim Centre for Foundation Studies in Science University of Malaya Session 2020/2021 2.1 Capacitor and Capacitance ▪ A capacitor is an electrical device, sometimes called condenser is a device that can store electric charge. ▪ A capacitor consists of two metal plates separated by an insulator called dielectric. ▪ The insulator can be a vacuum, air or any other materials such as polystyrene, oil or waxed paper. ▪ In a diagram, a capacitor is represented by this symbol. ▪ Capacitors have many applications: – Computer RAM memory and keyboards. – Electronic flashes for cameras. – Electric power surge protectors. – Radios and electronic circuits. Types of Capacitor Parallel-Plate Capacitor • Cylindrical Capacitor A typical capacitor consists of a pair of parallel plates of area, A separated by a small distance, d. Often the two plates are rolled into the form of a cylinder with paper or other insulator separating the plates. ▪ When used in an electric circuit, the plates are connected to the positive and negative terminals of a battery or some other voltage source. ▪ When this connection is made, electrons are pulled off one of the plates, leaving it with a charge of +Q and transferred through the battery to the other plate, leaving it with a charge of –Q. ▪ This charge transfer stops when the potential difference across the plates equals the potential difference of the battery. ▪ Thus, a charged capacitor is a device that stores energy that can be reclaimed when needed for a specific application. Capacitance, C ▪ The measure of the extent to which a capacitor can store charge is called capacitance. ▪ The number of charges stored in capacitor is proportional to the potential difference between the plates, Q = CV where C is capacitance of the capacitor. ▪ The capacitance of a capacitor is defined as the ratio of the magnitude of the charge on either plate to the magnitude of the potential difference, V between them. 𝑄 𝐶= 𝑉 …(1) ▪ S.I unit of the capacitance is farad (F). 1F=1 𝐶 𝑉 ▪ 1 Farad: capacitance of a capacitor where 1 coulomb of charge produces a potential difference of 1 volt. ▪ Practical capacitors usually have the capacitance in the order of F or pF. ▪ For example, if a 3.0 F capacitor is connected to a 12 V battery, the magnitude of the charge on each plate of the capacitor is Q = C V = (3.0 × 10 -6 F) (12 V) = 36 C . A capacitor in a simple electric circuit. Charge stored: Q = CV (The charges stored, Q is directly proportional to the potential difference, V across the conducting plate.) Relationship between Q and V Q Q = CV Q αV V ▪ The capacitance C is a constant (not depend on Q or V. its value depends on the structure and dimensions of the capacitor itself). - for a larger A, more charge can be held on each plate and for a given potential difference, V - the charge on the plates increases with decreasing plate separation, d. Parallel Plate Capacitor ▪ A simple parallel-plate capacitor consists of two conducting plates of area A separated by a distance d. • Charge +Q is placed on one plate and –Q on the other plate. • An electric field E is created between the plates (uniform electric field) Cont.. ▪ The electric field between the plates is 𝐸= 𝜎 𝜀𝑜 = 𝑄 𝜀𝑜 𝐴 𝑄 𝐴 where σ = is surface charge density ▪ Because the field between the plates is uniform, the magnitude of the potential difference between the plates equals Ed, where d is distance between two plates, therefore 𝑉 = 𝐸𝑑 𝑄𝑑 = 𝜀𝑜 𝐴 Cont.. ▪ Substituting into Eq (1) 𝑄 𝑄 𝐶= = 𝑉 𝑄𝑑/𝜀𝑜 𝐴 Therefore, 𝐶= 𝜀𝑜 𝐴 d ▪ The capacitance of a parallel plate capacitor is proportional to the area A of its plate and inversely proportional to the plate separation d. Example 1 : Calculating C for a Parallel-plate Capacitor A parallel plate capacitor has an area A of 2.00 cm2 and a plate separation d of 1.00 mm. Find its capacitance. Solution 1 𝐹𝑟𝑜𝑚 𝐶 = 𝜀𝑜 𝐴 d 𝜀𝑜 𝐴 8.85 × 10−12 C 2 N −1 m−2 2 × 10−4 m2 𝐶= = 𝑑 1 × 10−3 m = 𝟏. 𝟕𝟕 × 𝟏𝟎−𝟏𝟐 𝐅 = 𝟏. 𝟕𝟕 𝐩𝐅 Example 2: 2.2 Combinations of Capacitors (A) Parallel combination (a) (b) (c) (a) A parallel connection of two capacitors (b) The circuit diagram for the parallel combination (c) The potential differences across the capacitors are the same, and the equivalent capacitance is Ceq = C1 + C2 Cont.. ▪ The potential difference across the capacitors is the same and each is equal to the voltage of the battery V = V1 = V2 ▪ The charge in each capacitor is Q1 = C1 V Q2 = C2 V ▪ The total charge is equal to the sum of the charges on the capacitors Qtotal = Q1 + Q2 ▪ It can be written as Q = Ceq V = (C1 + C2 )V Example 3: Four Capacitors connected in Parallel (a) Determine the capacitance of the single capacitor that is equivalent to the parallel combination of capacitors. (b) find the charge on the 12.0 μF capacitor (c) total charge contained in the configuration Solution 3: (a) C e q = C 1 + C 2 + C 3 + C 4 = 4 5 μF (b) The potential difference across the 12.0 μF capacitor is equal to the voltage of the battery and so, Q = C3 ∆V = (1 2 × 10-6 F) (18.0 V) = 2 1 6 × 10-6 C = 2 1 6 μC (c) Total charge QT = Ceq ∆V = (4 5 × 10-6 F) (18.0 V) = 810 × 10-6 C = 810 μ C (B) Series combination (a) A series combination of two capacitors. The charges on the two capacitors are the same. (b) The circuit diagram for the series combination (c) The equivalent capacitance can be calculated from the 1 1 1 relationship = + 𝐶𝑒𝑞 𝐶1 𝐶2 • The potential difference across each capacitor is split between two capacitors ∆ V = ∆ V 1 + ∆ V 2 • ∆V = • 𝑄 𝐶𝑒𝑞 = 𝑄 𝐶1 + 𝑄 𝐶2 Cancelling Q , So 1 𝐶𝑒𝑞 = 1 𝐶1 + 1 𝐶2 • So for three or more capacitors, the equivalent capacitance is 1 𝐶𝑒𝑞 = 1 𝐶1 + 1 𝐶2 + 1 𝐶3 +⋯ Example 3: Four capacitors connected in series Four capacitors are connected in series with a battery, as shown. (a) Find the capacitance of the equivalent capacitor. (b) Find the charge on the 12 μF capacitor. (c) Find the voltage drop across the 12 μF capacitor. Solution 3 (a) 1 𝐶𝑒𝑞 1 𝐶𝑒𝑞 = = 1 1 + 𝐶1 𝐶2 1 3 μF + + 1 𝐶3 1 6 μF + + Note that the equivalent capacitance is less than the capacitance of any of the individual capacitors in the combination. 1 𝐶4 1 12 μF + 1 24 μF Ceq = 1.6 μF (b) Charge on the equivalent capacitor Q = C eq V = (1.6 μF) (18.0 V) = 28.8 μC (c) Voltage drop across the 12 μF capacitor. Q = CV V= 𝑄 𝐶 = 28.8 ×10−6 C 12 ×10−6 𝐹 = 2.4 V Example 4 Find the equivalent capacitance between a and b for the combination of capacitors as shown below. All capacitances are in microfarads.