Chapter 16 Electrical Energy and Capacitance Electrical Potential

Chapter 16 Electrical Energy and Capacitance

Electrical Potential Energy • The electrostatic force is a conservative force • It is possible to define an electrical potential energy function associated with this force • Work done by a conservative force is equal to the negative of the change in potential energy W = – Δ PE

Work and Potential Energy • There is a uniform field between the two plates • As the charge moves from A to B, work is done on it W = F d = q Ex (xf – xi) Δ PE = – W = – q Ex (xf – xi)

Potential Difference • The potential difference between points A and B – the change in the potential energy of a charge q moved from A to B divided by the charge ΔV = VB – VA = ΔPE / q ΔPE = q ΔV • Alessandro Giuseppe Antonio Anastasio Both electrical potential energy and potential Volta 1745 -1827 difference are scalar quantities • SI unit of potential difference is Volt (V): 1 V = 1 J/C • For a uniform electric field: ΔV = VB – VA= – Ex Δx

Energy and the Direction of Electric Field • When the electric field is directed downward, point B is at a lower potential than point A • When a positive test charge moves from A to B, the charge-field system loses potential energy • The system loses electric potential energy when the positive charge moves in the direction of the field (an electric field does work on the charge) and the charge gains kinetic energy equal to the potential energy lost by the charge-field system

Energy and the Direction of Electric Field • If qo is negative, then ΔPE is positive • A system consisting of a negative charge and an electric field gains potential energy when the charge moves in the direction of the field • In order for a negative charge to move in the direction of the field, an external agent must do positive work on the charge

Energy and Charge Movements • A positive (negative) charge gains (loses) electrical potential energy when it is moves in the direction opposite the electric field • If a charge is released in the electric field, it experiences a force and accelerates, gaining kinetic energy and losing an equal amount of electrical potential energy • When the electric field is directed downward, point B is at a lower potential than point A: a positive test charge moving from A to B loses electrical potential energy

Chapter 16 Problem 6 A point charge q = +40. 0 µC moves from A to B separated by a distance d = 0. 180 m in the presence of an external electric field E of magnitude 275 N/C directed toward the right as in the figure. Find (a) the electric force exerted on the charge, (b) the work done by the electric force, (c) the change in the electric potential energy of the charge, and (d) the potential difference between A and B.

The Electron Volt • The electron volt (e. V) is defined as the energy that an electron gains when accelerated through a potential difference of 1 V: 1 e. V = 1. 6 x 10 -19 J

Electric Potential of a Point Charge • The point of zero electric potential is taken to be at an infinite distance from the charge • The potential created by a point charge q at a distance r from the charge is • A potential exists at some point in space whether or not there is a test charge at that point

Electric Potential of a Point Charge • The electric potential is proportional to 1/r while the electric field is proportional to 1/r 2

Potential of Multiple Point Charges and Potential Energy • Superposition principle applies • The total electric potential at some point P due to several point charges is the algebraic sum of the electric potentials due to the individual charges (potentials are scalar quantities) • V 1: the electric potential due to q 1 at P • The work required to bring q 2 from infinity to P without acceleration is q 2 V 1 and it is equal to the potential energy of the two particle system

Potential of Multiple Point Charges and Potential Energy • If the charges have the same sign, PE is positive (positive work must be done to force the two charges near one another), so the charges would repel • If the charges have opposite signs, PE is negative (work must be done to hold back the unlike charges from accelerating as they are brought close together), so the force would be attractive

Solving Problems with Electric Potential (Point Charges) • Note the point of interest and draw a diagram of all charges • Calculate the distance from each charge to the point of interest • Use the basic equation V = keq/r and include the sign – the potential is positive (negative) if the charge is positive (negative) • Use the superposition principle when you have multiple charges and take the algebraic sum (potential is a scalar quantity and there are no components to worry about)

Chapter 16 Problem 17 The three charges in the figure at the vertices of an isosceles triangle. Let q = 7. 00 n. C, and calculate the electric potential at the midpoint of the base.

Potentials and Charged Conductors • Since W = – q (VB – VA), no work is required to move a charge between two points that are at the same electric potential (W = 0 when VA = VB) • For a charged conductor in electrostatic equilibrium the electric field just outside the conductor is perpendicular to the surface • Path AB is perpendicular to the electric field lines at every point – the work will be zero along AB, so all points on the surface of are at the same potential

Potentials and Charged Conductors • Since all of the charge resides at the surface, E = 0 inside the conductor • Therefore work will be zero along any path inside the conductor, so the potential everywhere inside the conductor is constant and equal to its value at the surface

Equipotentials • Point B is at a lower potential than point A • Points B and C are at the same potential, since all points in a plane perpendicular to a uniform electric field are at the same electric potential • Equipotential surface is a continuous distribution of points having the same electric potential

Equipotential Surfaces • An equipotential surface is a surface on which all points are at the same potential • No work is required to move a charge at a constant speed on an equipotential surface • The electric field at every point on an equipotential surface is perpendicular to the surface • For a uniform electric field the equipotential surfaces are everywhere perpendicular to the field lines

Equipotential Surfaces • An equipotential surface is a surface on which all points are at the same potential • No work is required to move a charge at a constant speed on an equipotential surface • The electric field at every point on an equipotential surface is perpendicular to the surface • For a point charge the equipotential surfaces are a family of spheres centered on the point charge

Equipotential Surfaces • An equipotential surface is a surface on which all points are at the same potential • No work is required to move a charge at a constant speed on an equipotential surface • The electric field at every point on an equipotential surface is perpendicular to the surface • For a dipole the equipotential surfaces are shown in blue

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 Michael Faraday 1791 – 1867 • A 1 Farad capacitance is very large – µF or p. F capacitances are more common

Capacitor • A capacitor is a device used in a variety of electric circuits • 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 • When connected to the battery, charge is pulled off one plate and transferred to the other plate (the transfer stops when ΔVcap = Δ Vbattery) • For a parallel-plate capacitor whose plates are separated by air:

Chapter 16 Problem 27 An air-filled parallel-plate capacitor has plates of area 2. 30 cm 2 separated by 1. 50 mm. The capacitor is connected to a 12. 0 -V battery. (a) Find the value of its capacitance. (b) What is the charge on the capacitor? (c) What is the magnitude of the uniform electric field between the plates?

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

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 = Q 1 + Q 2 • 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 C 1 to the right plate of C 2 through the battery • As this negative charge accumulates on the right plate of C 2, an equivalent amount of negative charge is removed from the left plate of C 2, 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 16 Problem 37 For the system of capacitors shown in the figure, find (a) the equivalent capacitance of the system, (b) the charge on each capacitor, and (c) the potential difference across each capacitor.

Energy Stored in a Capacitor • Energy stored = ½ Q ΔV • From the definition of capacitance, this can be rewritten in different forms

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 paraffinimpregnated 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

An Atomic Description of Dielectrics • Polarization occurs when there is a separation between the average positions of its negative charge and its positive charge • In a capacitor, the dielectric becomes polarized because it is in an electric field that exists between the plates

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

Chapter 16 Problem 51 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 cm 2 and an insulation thickness of 0. 0400 mm.

Answers to Even Numbered Problems Chapter 16: Problem 4 6. 67 × 1011 electrons

Answers to Even Numbered Problems Chapter 16: Problem 16 (b) 3 keq/a

Answers to Even Numbered Problems Chapter 16: Problem 26 (a) 3. 00 μF (b) 36. 0 μC

Answers to Even Numbered Problems Chapter 16: Problem 32 1. 23 k. V

Answers to Even Numbered Problems Chapter 16: Problem 42 (a) 6. 04 μF (b) 83. 6 μC

Answers to Even Numbered Problems Chapter 16: Problem 46 (a) Ceq= 12. 0 μF, Estored, total= 8. 64 × 10− 4 J (b) Estored, 1= 5. 76 × 10− 4 J, Estored, 2= 2. 88 × 10− 4 J, Estored, total= Estored, 1 + Estored, 2 (c) 5. 66 V; C 2, with the largest capacitance stores the most energy