Electric Potential Chapter 23 opener. We are used to voltage in our lives—a 12-volt car battery, 110 V or 220 V at home, 1.5 volt flashlight batteries, and so on. Here we see a Van de Graaff generator, whose voltage may reach 50,000 V or more. Voltage is the same as electric potential difference between two points. Electric potential is defined as the potential energy per unit charge. The children here, whose hair stands on end because each hair has received the same sign of charge, are not harmed by the voltage because the Van de Graaff cannot provide much current before the voltage drops. (It is current through the body that is harmful, as we will see later.)

Electric Potential Energy and Potential Difference Relation between Electric Potential and Electric Field Electric Potential Due to Point Charges Potential Due to Any Charge Distribution Equipotential Surfaces Electric Dipole Potential

E Determined from V Electrostatic Potential Energy; the Electron Volt Cathode Ray Tube: TV and Computer Monitors, Oscilloscope

Electrostatic Potential Energy and Potential Difference The electrostatic force is conservative – potential energy can be defined. Change in electric potential energy is negative of work done by electric force: Figure 23-1. Work is done by the electric field in moving the positive charge q from position a to position b.

Electrostatic Potential Energy and Potential Difference Electric potential is defined as potential energy per unit charge: Unit of electric potential: the volt (V): 1 V = 1 J/C.

Electrostatic Potential Energy and Potential Difference Only changes in potential can be measured, allowing free assignment of V = 0:

Electrostatic Potential Energy and Potential Difference A negative charge. Suppose a negative charge, such as an electron, is placed near the negative plate at point b, as shown here. If the electron is free to move, will its electric potential energy increase or decrease? How will the electric potential change? Figure 23-2. Central part of Fig. 23–1, showing a negative point charge near the negative plate, where its potential energy (PE) is high. Example 23–1. Solution: The electron will move towards the positive plate if released, thereby increasing its kinetic energy. Its potential energy must therefore decrease. However, it is moving to a region of higher potential V; the potential is determined only by the existing charge distribution and not by the point charge. U and V have different signs due to the negative charge.

Electrostatic Potential Energy and Potential Difference Analogy between gravitational and electrical potential energy: Figure 23-3. (a) Two rocks are at the same height. The larger rock has more potential energy. (b) Two charges have the same electric potential. The 2Q charge has more potential energy.

Electrostatic Potential Energy and Potential Difference Electrical sources such as batteries and generators supply a constant potential difference. Here are some typical potential differences, both natural and manufactured:

Electrostatic Potential Energy and Potential Difference Electron in CRT. Suppose an electron in a cathode ray tube is accelerated from rest through a potential difference Vb – Va = Vba = +5000 V. (a) What is the change in electric potential energy of the electron? (b) What is the speed of the electron (m = 9.1 × 10-31 kg) as a result of this acceleration? Solution: a. The change in potential energy is qV = -8.0 x 10-16 J. b. The change in potential energy is equal to the change in kinetic energy; solving for the final speed gives v = 4.2 x 107 m/s.

Relation between Electric Potential and Electric Field The general relationship between a conservative force and potential energy: Substituting the potential difference and the electric field: Figure 23-5. To find Vba in a nonuniform electric field E, we integrate E·dl from point a to point b.

Relation between Electric Potential and Electric Field The simplest case is a uniform field:

Relation between Electric Potential and Electric Field Electric field obtained from voltage. Two parallel plates are charged to produce a potential difference of 50 V. If the separation between the plates is 0.050 m, calculate the magnitude of the electric field in the space between the plates. Solution: E = V/d = 1000 V/m.

Relation between Electric Potential and Electric Field Charged conducting sphere. Determine the potential at a distance r from the center of a uniformly charged conducting sphere of radius r0 for (a) r > r0, (b) r = r0, (c) r < r0. The total charge on the sphere is Q. Solution: The electric field outside a conducting sphere is Q/(4πε0r2). Integrating to find the potential, and choosing V = 0 at r = ∞: a. V = Q/4πε0r. b. V = Q/4πε0r0. c. V = Q/4πε0r0 (the potential is constant, as there is no field inside the sphere).

Relation between Electric Potential and Electric Field The previous example gives the electric potential as a function of distance from the surface of a charged conducting sphere, which is plotted here, and compared with the electric field: Figure 23-8. (a) E versus r, and (b) V versus r, for a uniformly charged solid conducting sphere of radius r0 (the charge distributes itself on the surface); r is the distance from the center of the sphere.

Relation between Electric Potential and Electric Field Breakdown voltage. In many kinds of equipment, very high voltages are used. A problem with high voltage is that the air can become ionized due to the high electric fields: free electrons in the air (produced by cosmic rays, for example) can be accelerated by such high fields to speeds sufficient to ionize O2 and N2 molecules by collision, knocking out one or more of their electrons. The air then becomes conducting and the high voltage cannot be maintained as charge flows. The breakdown of air occurs for electric fields of about 3.0 × 106 V/m. (a) Show that the breakdown voltage for a spherical conductor in air is proportional to the radius of the sphere, and (b) estimate the breakdown voltage in air for a sphere of diameter 1.0 cm. Solution: a. Combining the equations for the field and the potential gives V = r0E. b. Substituting gives V = 15,000 V.

Point Charges To find the electric potential due to a point charge, we integrate the field along a field line: Figure 23-9. We integrate Eq. 23–4a along the straight line (shown in black) from point a to point b. The line ab is parallel to a field line.

Point Charges Setting the potential to zero at r = ∞ gives the general form of the potential due to a point charge: Figure 23-10. Potential V as a function of distance r from a single point charge Q when the charge is positive. Figure 23-11. Potential V as a function of distance r from a single point charge Q when the charge is negative.

Point Charges Work required to bring two positive charges close together. What minimum work must be done by an external force to bring a charge q = 3.00 μC from a great distance away (take r = ∞) to a point 0.500 m from a charge Q = 20.0 µC? Solution: The work is equal to the change in potential energy; W = 1.08 J. Note that the field, and therefore the force, is not constant.

Point Charges Potential above two charges. Calculate the electric potential (a) at point A in the figure due to the two charges shown, and (b) at point B. Solution: The total potential is the sum of the potential due to each charge; potential is a scalar, so there is no direction involved, but we do have to keep track of the signs. a. V = 7.5 x 105 V b. V = 0 (true at any point along the perpendicular bisector)

Charge Distribution The potential due to an arbitrary charge distribution can be expressed as a sum or integral (if the distribution is continuous): or

Charge Distribution Potential due to a ring of charge. A thin circular ring of radius R has a uniformly distributed charge Q. Determine the electric potential at a point P on the axis of the ring a distance x from its center. Solution: Each point on the ring is the same distance from point P, so the potential is just that of a charge Q a distance (R2 + x2)1/2 from point P.

Charge Distribution Potential due to a charged disk. A thin flat disk, of radius R0, has a uniformly distributed charge Q. Determine the potential at a point P on the axis of the disk, a distance x from its center. Solution: Consider the disk to be made up of infinitely thin rings, each at a radius R with a thickness dR. Each ring then carries a charge dq = 2QR dR/R02. Integrating to find V then gives the solution in the text.

Equipotential Surfaces An equipotential is a line or surface over which the potential is constant. Electric field lines are perpendicular to equipotentials. The surface of a conductor is an equipotential. Figure 23-16. Equipotential lines (the green dashed lines) between two oppositely charged parallel plates. Note that they are perpendicular to the electric field lines (solid red lines).

Equipotential Surfaces Point charge equipotential surfaces. For a single point charge with Q = 4.0 × 10-9 C, sketch the equipotential surfaces (or lines in a plane containing the charge) corresponding to V1 = 10 V, V2 = 20 V, and V3 = 30 V. Solution: Equipotential surfaces are spheres surrounding the charge; radii are shown in the figure (in meters).

Equipotential Surfaces Equipotential surfaces are always perpendicular to field lines; they are always closed surfaces (unlike field lines, which begin and end on charges). Figure 23-18. Equipotential lines (green, dashed) are always perpendicular to the electric field lines (solid red) shown here for two equal but oppositely charged particles.

Equipotential Surfaces A gravitational analogy to equipotential surfaces is the topographical map – the lines connect points of equal gravitational potential (altitude). Figure 23-19. A topographic map (here, a portion of the Sierra Nevada in California) shows continuous contour lines, each of which is at a fixed height above sea level. Here they are at 80 ft (25 m) intervals. If you walk along one contour line, you neither climb nor descend. If you cross lines, and especially if you climb perpendicular to the lines, you will be changing your gravitational potential (rapidly, if the lines are close together).

Electric Dipole Potential The potential due to an electric dipole is just the sum of the potentials due to each charge, and can be calculated exactly. For distances large compared to the charge separation: Figure 23-20. Electric dipole. Calculation of potential V at point P.

E E Determined from V If we know the field, we can determine the potential by integrating. Inverting this process, if we know the potential, we can find the field by differentiating: This is a vector differential equation; here it is in component form:

E Determined from V E for ring and disk. Use electric potential to determine the electric field at point P on the axis of (a) a circular ring of charge and (b) a uniformly charged disk. Solution: a. Just do the derivatives of the result of Example 23-8; the only nonzero component is in the x direction. b. Same as (a); use the result of Example 23-9.

Electrostatic Potential Energy; the Electron Volt The potential energy of a charge in an electric potential is U = qV. To find the electric potential energy of two charges, imagine bringing each in from infinitely far away. The first one takes no work, as there is no field. To bring in the second one, we must do work due to the field of the first one; this means the potential energy of the pair is:

Electrostatic Potential Energy; the Electron Volt One electron volt (eV) is the energy gained by an electron moving through a potential difference of one volt: 1 eV = 1.6 × 10-19 J. The electron volt is often a much more convenient unit than the joule for measuring the energy of individual particles.

Electrostatic Potential Energy; the Electron Volt Disassembling a hydrogen atom. Calculate the work needed to “disassemble” a hydrogen atom. Assume that the proton and electron are initially separated by a distance equal to the “average” radius of the hydrogen atom in its ground state, 0.529 × 10-10 m, and that they end up an infinite distance apart from each other. Solution: After separation, the total energy – potential plus kinetic – is zero. The energy (work) needed to separate the proton and electron is the negative of the potential energy plus the kinetic energy of the atom before disassembly. The potential energy is -27.2 eV (convert from joules), and the kinetic energy of the electron in its orbit is 13.6 eV (calculate its speed from F = ma; assume it is moving in a circle). Therefore the energy needed to separate the proton and electron is -(-27.2 eV + 13.6 eV) = 13.6 eV. (Although it is classical, this calculation actually gives the correct ionization energy).

Summary Electric potential is potential energy per unit charge: Potential difference between two points: Potential of a point charge:

Summary Equipotential: line or surface along which potential is the same. Electric dipole potential is proportional to 1/r2. To find the field from the potential:

Capacitance and Dielectrics Chapter 24 opener. Capacitors come in a wide range of sizes and shapes, only a few of which are shown here. A capacitor is basically two conductors that do not touch, and which therefore can store charge of opposite sign on its two conductors. Capacitors are used in a wide variety of circuits, as we shall see in this and later Chapters.

Capacitors Capacitance Capacitors in Series and Parallel Electric Energy Storage Dielectrics Molecular Description of Dielectrics*

Capacitors A capacitor is a device that stores charge and electrical energy. A capacitor consists of two conductors separated by an insulator. Figure 24-1. Capacitors: diagrams of (a) parallel plate, (b) cylindrical (rolled up parallel plate).

Capacitors Parallel-plate capacitor connected to battery. (b) is a circuit diagram. Figure 24-2. (a) Parallel-plate capacitor connected to a battery. (b) Same circuit shown using symbols.

Capacitance When a capacitor is connected to a battery, the charge on its plates is proportional to the voltage: The quantity C is called the capacitance. Unit of capacitance: the farad (F): 1 F = 1 C/V.

Capacitance For a parallel-plate capacitor as shown, the field between the plates is E = Q/ε0A. In a uniform field, V=Ed: Vba = Qd/ε0A. This gives the capacitance: C=Q/V= ε0A/d Figure 24-4. Parallel-plate capacitor, each of whose plates has area A. Fringing of the field is ignored.

Capacitance Capacitor calculations. (a) Calculate the capacitance of a parallel-plate capacitor whose plates are 20 cm × 3.0 cm and are separated by a 1.0-mm air gap. (b) What is the charge on each plate if a 12-V battery is connected across the two plates? (c) What is the electric field between the plates? (d) Estimate the area of the plates needed to achieve a capacitance of 1 F, given the same air gap d. Solution: a. C = 53 pF. b. Q = CV = 6.4 x 10-10 C. c. E = V/d = 1.2 x 104 V/m. d. A = Cd/ε0 = 108 m2.

Capacitance Cylindrical capacitor. A cylindrical capacitor consists of a cylinder (or wire) of radius Rb surrounded by a coaxial cylindrical shell of inner radius Ra. Both cylinders have length l which we assume is much greater than the separation of the cylinders, so we can neglect end effects. The capacitor is charged (by connecting it to a battery) so that one cylinder has a charge +Q (say, the inner one) and the other one a charge –Q. Determine a formula for the capacitance. Figure 24-6. (a) Cylindrical capacitor consists of two coaxial cylindrical conductors. (b) The electric field lines are shown in cross-sectional view. Solution: We need to find the potential difference between the cylinders; we can do this by integrating the field (which was calculated for a long wire already). The field is proportional to 1/R, so the potential is proportional to ln Ra/Rb. Then C = Q/V.

Capacitance Spherical capacitor. A spherical capacitor consists of two thin concentric spherical conducting shells of radius ra and rb as shown. The inner shell carries a uniformly distributed charge Q on its surface, and the outer shell an equal but opposite charge –Q. Determine the capacitance of the two shells. Figure 24-7. Cross section through the center of a spherical capacitor. The thin inner shell has radius rb and the thin outer shell has radius ra. Solution: As in the previous example, we find the potential difference by integrating the field (that of a point charge) from rb to ra. Then C = Q/V.

Determination of Capacitance Capacitance of two long parallel wires. Estimate the capacitance per unit length of two very long straight parallel wires, each of radius R, carrying uniform charges +Q and –Q, and separated by a distance d which is large compared to R (d >> R). Solution: The field between the wires is the superposition of the fields due to each wire. Integrating gives the potential; then C = Q/V. See text for details. Since the charge is a charge density (charge per unit length), the capacitance is also quoted as C/L – capacitance per unit length.

Capacitors in Parallel Capacitors in parallel have the same voltage across each one: Figure 24-9. Capacitors in parallel: Ceq = C1 + C2 + C3 .

Capacitors in Series Capacitors in series have the same charge: Figure 24-10. Capacitors in series: 1/Ceq = 1/C1 + 1/C2 + 1/C3.

Capacitors in Series and Parallel Determine the capacitance of a single capacitor that will have the same effect as the combination shown. Solution: First, find the equivalent capacitance of the two capacitors in parallel (2C); then the equivalent of that capacitor in series with the third (2/3 C).

Solution: In parallel, In series,

Capacitors in Series and Parallel Determine the charge on each capacitor and the voltage across each, assuming C = 3.0 μF and the battery voltage is V = 4.0 V. Solution: First, find the charge on C1 from Q = CVeq. This charge is split equally between C2 and C3. Now we can calculate the voltages across each capacitor, using V = Q/C.

Capacitors in Series and Parallel Capacitors reconnected. Two capacitors, C1 = 2.2 μF and C2 = 1.2 μF, are connected in parallel to a 24-V source as shown. After they are charged they are disconnected from the source and from each other and then reconnected directly to each other, with plates of opposite sign connected together. Find the charge on each capacitor and the potential across each after equilibrium is established. Solution: When the capacitors are connected to the voltage source, the charges on them are given by Q = CV: Q1 = 52.8 μC and Q2 = 28.8 μC. After disconnection and reconnection, we know three things: 1. The voltage across each capacitor is the same. 2. The charge on each capacitor is given by Q = CV (using the appropriate values of C and V). 3. The sum of the charges on the two capacitors equals the total charge they started with. This gives three equations (two of the form Q = CV and the charge conservation one), and can be solved for the three unknowns, the two charges and the potential. V = 7.1 V, Q’1 = 16 µC, Q’2 = 8.5 µC.

Electric Energy Storage A charged capacitor stores electric energy; the energy stored is equal to the work done to charge the capacitor. The work needed to transfer charge dq is dW=Vdq=(q/C)dq. The total work done is This work is stored as electrical potential energy:

Electric Energy Storage A camera flash unit stores energy in a 150-μF capacitor at 200 V. (a) How much electric energy can be stored? (b) What is the power output if nearly all this energy is released in 1.0 ms? Solution: a. U = ½ CV2 = 3.0 J. b. P = U/t = 3000 W.

Electric Energy Storage A parallel-plate capacitor carries charge Q and is then disconnected from a battery. The two plates are initially separated by a distance d. Suppose the plates are pulled apart until the separation is 2d. How has the energy stored in this capacitor changed? Solution: Increasing the plate separation decreases the capacitance, but the charge remains the same. Therefore the energy, U = ½ Q2/C, doubles.

Electric Energy Storage The plates of a parallel-plate capacitor have area A, separation x, and are connected to a battery with voltage V. While connected to the battery, the plates are pulled apart until they are separated by 3x. (a) What are the initial and final energies stored in the capacitor? (b) How much work is required to pull the plates apart (assume constant speed)? (c) How much energy is exchanged with the battery? Solution: a. U = ½ CV2, and increasing the plate separation decreases the capacitance, so the initial and final energies can be calculated from this. b. The work is equal to the negative of change in energy. c. The energy of the battery is increased by the work done minus the change in potential energy of the capacitor (which is negative).

Electric Energy Storage The capacitance of a parallel-plate capacitor is C=ε0A/d, and the potential difference between its plates is V=Ed. The energy stored can be written as Since the volume where the electric fieled exists is Ad, the energy density is defined as:

Dielectrics A dielectric is an insulator, introduced between the plates of a capacitor, and will increase the capacitance. It is characterized by a dielectric constant κ. Capacitance of a parallel-plate capacitor filled with dielectric: Using the dielectric constant, we define the permittivity:

Dielectrics Dielectric strength is the maximum field a dielectric can experience without breaking down.

Dielectrics The capacitor connected to a battery The voltage remaining constant A dielectric inserted Figure 24-15a. Dielectric inserted into a capacitor, voltage held constant.

Dielectrics The capacitor connected to a battery The capacitor then disconnected from the battery The charge remaining constant A dielectric inserted Figure 24-15b. Dielectric inserted into a capacitor, charge held constant.

Dielectrics A parallel-plate capacitor, filled with a dielectric with K = 3.4, is connected to a 100-V battery. After the capacitor is fully charged, the battery is disconnected. The plates have area A = 4.0 m2 and are separated by d = 4.0 mm. (a) Find the capacitance, the charge on the capacitor, the electric field strength, and the energy stored in the capacitor. (b) The dielectric is carefully removed, without changing the plate separation nor does any charge leave the capacitor. Find the new values of capacitance, electric field strength, voltage between the plates, and the energy stored in the capacitor. Solution: a. C = Kε0A/d = 3.0 x 10-8 F. Q = CV = 3.0 x 10-6 C. E = V/d = 25 kV/m. U = ½ CV2 = 1.5 x 10-4 J. B. Now C = 8.8 x 10-9 F, Q = 3.0 x 10-6 C (no change), V = 340 V, E = 85 kV/m, U = 5.1 x 10-4 J. The increase in energy comes from the work it takes to remove the dielectric.

+Q d E -Q d t +Q -Q E Ed

Molecular Description of Dielectrics The molecules in a dielectric, when in an external electric field, tend to become oriented in a way that reduces the external field. Figure 24-17. Molecular view of the effects of a dielectric.

Molecular Description of Dielectrics This means that the electric field within the dielectric is less than it would be in air, allowing more charge to be stored for the same potential. This reorientation of the molecules results in an induced charge – there is no net charge on the dielectric, but the charge is asymmetrically distributed. The magnitude of the induced charge depends on the dielectric constant:

Summary Capacitor: nontouching (separated) conductors carrying equal and opposite charge. Capacitance: Capacitance of a parallel-plate capacitor:

Summary Capacitors in parallel: Capacitors in series:

Summary Energy density in electric field: A dielectric is an insulator. Dielectric constant gives ratio of total field to external field. For a parallel-plate capacitor: