Chapter 24 Capacitance A capacitor consists of two spatially separated conductors which can be charged to +Q and -Q. Capacitor is a device that stores electrostatic potential energy. 9/20/2018

Capacitance The capacitance is defined as the ratio of the charge on one conductor of the capacitor to the potential difference between the conductors. Capacitance is measured in Faradays represented by the letter F which is equal to (coulomb/volt) 9/20/2018

Capacitance between Parallel Plates Calculate the capacitance. We assume +, - charge densities on each plate with potential difference V: - - - - - + + + + A d +Q -Q d E x y First determine E field produced by charged conductors: What is s ? A = area of plate Choose integration path,dy, from negative plate to positive plate Dy & E field point in opposite directions The dot porduct of E*dy will be negative V=- integral of – E dy = integral E dy = positive quantity Second, integrate E to find the potential difference V As promised, V is proportional to Q ! C determined by geometry ! 9/20/2018

Example What is the capacitance of a parallel plate capacitor made out of two square plates 10 x10 cm2 separated by 1 mm wide gap? d=1 mm A=0.01 m2 If we connect 1.5V battery to its plates, how much charge could be stored in this capacitor? 9/20/2018

Demos Demo 5A-28: Capacitance versus d 9/20/2018

Question Does the capacitance C of a capacitor increase, decrease or remain the same when the charge q on it is doubled? Decreases Remains the Same Increases 9/20/2018

Question Which line goes with which parallel plate capacitor? q d 1 2 2A 3 2d a b c q V 1:b, 2:a, 3:c 1:b, 2:c, 3:a 1:a, 2:b, 3:c 9/20/2018

A Cylindrical Capacitor: Step 1 9/20/2018

Calculating Capacitance Put + Q , – Q on the plates. Find between the plates i.e., given the charges, using Gauss’ Law Find V across the plates Use the definition 9/20/2018

A Cylindrical Capacitor : Step 2 The electric field in between the plates is radial. Use a cylindrical shaped Gaussian surface of radius R (R1<R<R2) of length l<<L 9/20/2018

A Cylindrical Capacitor : Step 3 -Q In general, the path of integration is taken from the – to the + plate. The vectors E and ds are opposite, so the dot product of E * ds is negative yielding: r R2 +Q ds R1 E For the case of parallel plate capacitors, ds = -dr because the integration path is radially inwards giving: 9/20/2018

A Cylindrical Capacitor : Step 4 9/20/2018

QUIZ (A) A > B > C (B) B > A = C (C) B = C < A An isolated solid metal sphere A of radius R and two spherical capacitors B and C with inner and out radii R and 2R are shown. The inner “spherical” plate of capacitor B is a spherical shell; that of capacitor C is a solid sphere. Rank the objects A, B, and C according to their capacitance, greatest first. (A) A > B > C (B) B > A = C (C) B = C < A (D) C < A < B (E) A < B = C A B C 9/20/2018 13

Energy Stored in a Capacitor If a small amount of additional positive charge dq is transferred from the negative conductor to the positive conductor through a potential increase of V, the potential energy of the charge and thus the capacitor is increased by 9/20/2018

Energy stored in a capacitor Charging capacitor requires work: V must move charges from one plate to another through rising V charging a capacitor involves work: W=q V this work is equal to the energy stored in a capacitor Energy stored in a capacitor: 9/20/2018 15

Energy stored in a capacitor Alternative equations for energy stored in a capacitor Using Using 9/20/2018 16

Demo Demo 5A-29: Energy stored in a capacitor 9/20/2018

QUIZ A parallel plate capacitor is given a charge q. The plates are then pulled a small distance farther apart. Which of the following apply to the situation after the plates have been moved? The energy stored in the capacitor decreases. (A) True (B) False +q -q + - d pull 9/20/2018 18

QUIZ: A parallel plate capacitor is given a charge q. The plates are then pushed a small distance closer together. Which of the following apply to the situation after the plates have been moved? The energy stored in the capacitor decreases. (A) True (B) False +q -q + - d push 9/20/2018 19

QUIZ A capacitor has a capacitance of 25 µF and is initially uncharged. The battery provides a potential difference of 120 Volts. After the switch, S, is closed, how much charge will pass through it? S 3 mC 14 mC 16 mC 30 mC 140 mC 3 milli Coulomb + C - 9/20/2018

Demonstration: Energy Stored in a Capacitor P.S. Power supply (P.S.) supplies voltage V. can produce a big jolt. Application: defibrillator A fibrillating heart is one in which the cardiac muscles go into uncontrolled twitching and quivering. A defibrillator is used to stop this. 9/20/2018

Demonstration: Energy Stored in a Capacitor P.S. Power supply (P.S.) supplies voltage V. Defibrillation is the definitive treatment for the life-threatening cardiac arrhythmias, ventricular fibrillation and pulseless ventricular tachycardia. Defibrillation consists of delivering a therapeutic dose of electrical energy to the affected heart with a device called a defibrillator. This depolarizes a critical mass of the heart muscle, terminates the arrhythmia, and allows normal sinus rhythm to be reestablished by the body's natural pacemaker, in the sinoatrial node of the heart. In 1962 Bernard Lown introduced the external DC defibrillator. This device applied a direct current from a discharging capacitor through the chest wall into the heart to stop heart fibrillation.[5] can produce a big jolt. Application: defibrillator A fibrillating heart is one in which the cardiac muscles go into uncontrolled twitching and quivering. A defibrillator is used to stop this. 9/20/2018

Numbers About 200 J of this is sent through the heart attack victim in 2 milliseconds. short time 9/20/2018

Capacitance Capacitance is defined for any pair of spatially separated conductors. How do we understand this definition ? +Q -Q d Consider two conductors, one with excess charge = +Q and the other with excess charge = -Q V We can integrate the electric field between them to find the potential difference between the conductor These charges create an electric field in the space between them E This potential difference should be proportional to Q ! The ratio of Q to the potential difference is the capacitance and only depends on the geometry of the conductors