Capacitance and Dielectrics Chapter 26 Capacitance and Dielectrics

Capacitance The capacitance, C, is a measure of the amount of electric charge stored (or separated) for a given electric potential SI unit – Farad (F): 1 F = 1 C / V A 1 Farad capacitance is very large – µF or pF capacitances are more common Michael Faraday 1791 – 1867

Capacitor A capacitor is a device used in a variety of electric circuits to store electric charge Capacitance of a capacitor is the ratio of the magnitude of the charge on either conductor (plate) to the magnitude of the potential difference between the conductors (plates) This capacitance of a device depends on the geometric arrangement of the conductors

Parallel-Plate Capacitor This capacitor consists of two parallel plates (each of area A) separated by a distance d each carrying equal and opposite charges Each plate is connected to a terminal of the battery The battery is a source of potential difference If the capacitor is initially uncharged, the battery establishes an electric field in the connecting wires

Parallel-Plate Capacitor The field applies a force on electrons in the wire just outside of the plates, causing the electrons to move onto the negative plate until equilibrium is achieved The plate, the wire and the terminal are all at the same potential, and there is no field present in the wire and the movement of the electrons ceases At the other plate, electrons are moving away from the plate and leaving it positively charged

Parallel-Plate Capacitor The electric field due to one plate The total electric field between the plates is given by The field outside the plates is zero

Parallel-Plate Capacitor In the final configuration, the potential difference across the capacitor plates is the same as that between the terminals of the battery For a parallel-plate capacitor whose plates are separated by air:

Electric Field in a Parallel-Plate Capacitor The electric field between the plates is uniform near the center and nonuniform near the edges The field may be taken as constant throughout the region between the plates

Spherical Capacitor

Chapter 26 Problem 5 An air-filled capacitor consists of two parallel plates, each with an area of 7.60 cm2, separated by a distance of 1.80 mm. A 20.0-V potential difference is applied to these plates. Calculate (a) the electric field between the plates, (b) the surface charge density, (c) the capacitance, and (d) the charge on each plate.

Electric Circuits A circuit is a collection of objects usually containing a source of electrical energy (such as a battery) connected to elements that convert electrical energy to other forms A circuit diagram – a simplified representation of an actual circuit – is used to show the path of the real circuit Circuit symbols are used to represent various elements (e.g., lines are used to represent wires, battery’s positive terminal is indicated by a longer line)

Capacitors in Parallel When capacitors are first connected in parallel in the circuit, electrons are transferred from the left plates through the battery to the right plates, leaving the left plates positively charged and the right plates negatively charged The flow of charges ceases when the voltage across the capacitors equals that of the battery The capacitors reach their maximum charge when the flow of charge ceases

Capacitors in Parallel The total charge is equal to the sum of the charges on the capacitors: Qtotal = Q1 + Q2 The potential differences across the capacitors is the same and each is equal to the voltage of the battery A circuit diagram for two capacitors in parallel

Capacitors in Parallel The capacitors can be replaced with one capacitor with a equivalent capacitance Ceq – the equivalent capacitor must have exactly the same external effect on the circuit as the original capacitors

Capacitors in Parallel For more than two capacitors in parallel: The equivalent capacitance of a parallel combination of capacitors is greater than any of the individual capacitors

Capacitors in Series When a battery is connected to the circuit, electrons are transferred from the left plate of C1 to the right plate of C2 through the battery As this negative charge accumulates on the right plate of C2, an equivalent amount of negative charge is removed from the left plate of C2, leaving it with an excess positive charge All of the right plates gain charges of –Q and all the left plates have charges of +Q

Capacitors in Series An equivalent capacitor can be found that performs the same function as the series combination The potential differences add up to the battery voltage

Capacitors in Series For more than two capacitors in series: The equivalent capacitance is always less than any individual capacitor in the combination

Problem-Solving Strategy Be careful with the choice of units Combine capacitors: When two or more unequal capacitors are connected in series, they carry the same charge, but the potential differences across them are not the same The capacitances add as reciprocals and the equivalent capacitance is always less than the smallest individual capacitor

Problem-Solving Strategy Be careful with the choice of units Combine capacitors: When two or more capacitors are connected in parallel, the potential differences across them are the same The charge on each capacitor is proportional to its capacitance The capacitors add directly to give the equivalent capacitance

Problem-Solving Strategy Redraw the circuit and continue Repeat the process until there is only one single equivalent capacitor To find the charge on, or the potential difference across, one of the capacitors, start with your final equivalent capacitor and work back through the circuit reductions

Equivalent Capacitance

Chapter 26 Problem 25 Find the equivalent capacitance between points a and b in the combination of capacitors shown in the figure

Energy Stored in a Capacitor Before the switch is closed, the energy is stored as chemical energy in the battery When the switch is closed, the energy is transformed from chemical to electric potential energy The electric potential energy is related to the separation of the positive and negative charges on the plates A capacitor can be described as a device that stores energy as well as charge

Energy Stored in a Capacitor Assume the capacitor is being charged and, at some point, has a charge q on it The work needed to transfer a charge from one plate to the other: The total work required: The work done in charging the capacitor appears as electric potential energy U:

Energy Stored in a Capacitor This applies to a capacitor of any geometry The energy stored increases as the charge increases and as the potential difference increases The energy can be considered to be stored in the electric field For a parallel-plate capacitor, the energy can be expressed in terms of the field as U = ½ (εoAd)E2

Capacitors with Dielectrics A dielectric is an insulating material (e.g., rubber, plastic, etc.) When placed between the plates of a capacitor, it increases the capacitance: C = κ Co = κ εo (A/d) κ - dielectric constant The capacitance is multiplied by the factor κ when the dielectric completely fills the region between the plates

Capacitors with Dielectrics Tubular – metallic foil interlaced with thin sheets of paraffin-impregnated paper rolled into a cylinder Oil filled (for high-V capacitors) – interwoven metallic plates are immersed in silicon oil Electrolytic (to store large amounts of charge at relatively low voltages) – electrolyte is a solution that conducts electricity via motion of ions in the solution

Dielectric Strength For any given plate separation, there is a maximum electric field that can be produced in the dielectric before it breaks down and begins to conduct This maximum electric field is called the dielectric strength

Chapter 26 Problem 37 Determine (a) the capacitance and (b) the maximum voltage that can be applied to a Teflon®-filled parallel-plate capacitor having a plate area of 175 cm2 and an insulation thickness of 0.040 0 mm.

Electric Dipole An electric dipole consists of two charges of equal magnitude and opposite signs separated by 2a The electric dipole moment p is directed along the line joining the charges from –q to +q and has a magnitude of p ≡ 2aq Assume the dipole is placed in a uniform field, external to the dipole (it is not the field produced by the dipole) and makes an angle θ with the field Each charge has a force of F = Eq acting on it

Electric Dipole The net force on the dipole is zero The forces produce a net torque on the dipole: t = 2Fa sin θ = pE sin θ The torque can also be expressed as the cross product of the moment and the field: The potential energy can be expressed as

An Atomic Description of Dielectrics Molecules are said to be polarized when a separation exists between the average position of the negative charges and the average position of the positive charges Polar molecules are those in which this condition is always present Molecules without a permanent polarization are called nonpolar molecules The average positions of the positive and negative charges act as point charges, thus polar molecules can be modeled as electric dipoles

An Atomic Description of Dielectrics A linear symmetric molecule has no permanent polarization (a) Polarization can be induced by placing the molecule in an electric field (b) Induced polarization is the effect that predominates in most materials used as dielectrics in capacitors

An Atomic Description of Dielectrics The molecules that make up the dielectric are modeled as dipoles; in the absence of an electric field they are randomly oriented An external electric field produces a torque on the molecules partially aligning them with the electric field

An Atomic Description of Dielectrics The presence of the positive (negative) charge on the dielectric effectively reduces some of the negative (positive) charge on the metal This allows more charge on the plates for a given applied voltage and the capacitance increases

Answers to Even Numbered Problems Chapter 26: Problem 12 17.0 μF 9.00V 45.0 μC and 108 μC

Answers to Even Numbered Problems Chapter 26: Problem 36 13.3 nC 272 nC

Answers to Even Numbered Problems Chapter 26: Problem 50 (b) 40.0 μF (c) 6.00 V across 50 μF with charge 300 μF; 4.00 V across 30 μF with charge 120 μF; 2.00 V across 20 μF with charge 40 μF; 2.00 V across 40 μF with charge 80 μF