Chapter 22 Gauss’s Law Chapter 22 opener. Gauss’s law is an elegant relation between electric charge and electric field. It is more general than Coulomb’s.

Chapter 22 Gauss’s Law Chapter 22 opener. Gauss’s law is an elegant relation between electric charge and electric field. It is more general than Coulomb’s law. Gauss’s law involves an integral of the electric field E at each point on a closed surface. The surface is only imaginary, and we choose the shape and placement of the surface so that we can figure out the integral. In this drawing, two different surfaces are shown, both enclosing a point charge Q. Gauss’s law states that the product E·dA, where dA is an infinitesimal area of the surface, integrated over the entire surface, equals the charge enclosed by the surface Qencl divided by ε0. Both surfaces here enclose the same charge Q. Hence ∫E·dA will give the same result for both surfaces. 1

22-1 Electric Flux Flux through a closed surface:

Think in terms of flux lines:

22-2 Gauss’s Law The net number of field lines through the surface is proportional to the charge enclosed, and also to the flux, giving Gauss’s law: This can be used to find the electric field in situations with a high degree of symmetry.

22-2 Gauss’s Law For a point charge, Spherical Therefore,
Solving for E gives the result we expect from Coulomb’s law: Figure A single point charge Q at the center of an imaginary sphere of radius r (our “gaussian surface”—that is, the closed surface we choose to use for applying Gauss’s law in this case). 5

22-2 Gauss’s Law Using Coulomb’s law to evaluate the integral of the field of a point charge over the surface of a sphere surrounding the charge gives: Figure A single point charge surrounded by a spherical surface, A1, and an irregular surface,A2. Looking at the arbitrarily shaped surface A2, we see that the same flux passes through it as passes through A1. Therefore, this result should be valid for any closed surface. 6

22-2 Gauss’s Law Finally, if a gaussian surface encloses several point charges, the superposition principle shows that: Therefore, Gauss’s law is valid for any charge distribution. Note, however, that it only refers to the field due to charges within the gaussian surface – charges outside the surface will also create fields.

22-2 Gauss’s Law Conceptual Example 22-2: Flux from Gauss’s law.
Consider the two gaussian surfaces, A1 and A2, as shown. The only charge present is the charge Q at the center of surface A1. What is the net flux through each surface, A1 and A2? Solution: The net flux through A1 is Q/ε0, as it encloses charge Q. The net flux through surface A2 is zero, even though the field is not zero on the surface. 8

22-3 Applications of Gauss’s Law
Example 22-3: Spherical conductor. A thin spherical shell of radius r0 possesses a total net charge Q that is uniformly distributed on it. Determine the electric field at points (a) outside the shell, and (b) within the shell. (c) What if the conductor were a solid sphere? Figure Cross-sectional drawing of a thin spherical shell of radius r0 carrying a net charge Q uniformly distributed. A1 and A2 represent two gaussian surfaces we use to determine Example 22–3. Solution: a. The gaussian surface A1, outside the shell, encloses the charge Q. We know the field must be radial, so E = Q/(4πε0r2). b. The gaussian surface A2, inside the shell, encloses no charge; therefore the field must be zero. c. All the excess charge on a conductor resides on its surface, so these answers hold for a solid sphere as well. Spherical 9

Example 22-4: Solid sphere of charge. An electric charge Q is distributed uniformly throughout a nonconducting sphere of radius r0. Determine the electric field (a) outside the sphere (r > r0) and (b) inside the sphere (r < r0). Solution: a. Outside the sphere, a gaussian surface encloses the total charge Q. Therefore, E = Q/(4πε0r2). b. Within the sphere, a spherical gaussian surface encloses a fraction of the charge Qr3/r03 (the ratio of the volumes, as the charge density is constant). Integrating and solving for the field gives E = Qr/(4πε0r03). Spherical/radial 10

Example 22-5: Nonuniformly charged solid sphere. Suppose the charge density of a solid sphere is given by ρE = αr2, where α is a constant. (a) Find α in terms of the total charge Q on the sphere and its radius r0. (b) Find the electric field as a function of r inside the sphere. Solution: a. Consider the sphere to be made of a series of spherical shells, each of radius r and thickness dr. The volume of each is dV = 4πr2 dr. To find the total charge: Q = ∫ρE dV = 4παr05/5, giving α = 5Q/4πr05. b. The charge enclosed in a sphere of radius r will be Qr5/r05. Gauss’s law then gives E = Qr3/4πε0r05. 11

Example 22-6: Long uniform line of charge. A very long straight wire possesses a uniform positive charge per unit length, λ. Calculate the electric field at points near (but outside) the wire, far from the ends. Linear, Cylindrical Solution: If the wire is essentially infinite, it has cylindrical symmetry and we expect the field to be perpendicular to the wire everywhere. Therefore, a cylindrical gaussian surface will allow the easiest calculation of the field. The field is parallel to the ends and constant over the curved surface; integrating over the curved surface gives E = λ/2πε0R. 12

Example 22-7: Infinite plane of charge. Charge is distributed uniformly, with a surface charge density σ (σ = charge per unit area = dQ/dA) over a very large but very thin nonconducting flat plane surface. Determine the electric field at points near the plane. Solution: We expect E to be perpendicular to the plane, and choose a cylindrical gaussian surface with its flat sides parallel to the plane. The field is parallel to the curved side; integrating over the flat sides gives E = σ/2ε0. Planar & cylindrical 13

Example 22-8: Electric field near any conducting surface. Show that the electric field just outside the surface of any good conductor of arbitrary shape is given by E = σ/ε0 where σ is the surface charge density on the conductor’s surface at that point. Solution: Again we choose a cylindrical gaussian surface. Now, however, the field inside the conductor is zero, so we only have a nonzero integral over one surface of the cylinder. Integrating gives E = σ/ε0. 14

The difference between the electric field outside a conducting plane of charge and outside a nonconducting plane of charge can be thought of in two ways: The field inside the conductor is zero, so the flux is all through one end of the cylinder. The nonconducting plane has a total charge density σ, whereas the conducting plane has a charge density σ on each side, effectively giving it twice the charge density.

Conceptual Example : Conductor with charge inside a cavity. Suppose a conductor carries a net charge +Q and contains a cavity, inside of which resides a point charge +q. What can you say about the charges on the inner and outer surfaces of the conductor? Solution: The field must be zero within the conductor, so the inner surface of the cavity must have an induced charge totaling –q (so that a gaussian surface just around the cavity encloses no charge). The charge +Q resides on the outer surface of the conductor. 16

Procedure for Gauss’s law problems: Identify the symmetry, and choose a gaussian surface that takes advantage of it (with surfaces along surfaces of constant field). Draw the surface. Use the symmetry to find the direction of E. Evaluate the flux by integrating. Calculate the enclosed charge. Solve for the field.

22-4 Experimental Basis of Gauss’s and Coulomb’s Laws
In the experiment shown, Gauss’s law predicts that the charge on the ball flows onto the surface of the cylinder when they are in contact. This can be tested by measuring the charge on the ball after it is removed – it should be zero. Figure (a) A charged conductor (metal ball) is lowered into an insulated metal can (a good conductor) carrying zero net charge. (b) The charged ball is touched to the can and all of its charge quickly flows to the outer surface of the can. (c) When the ball is then removed, it is found to carry zero net charge. 18

Summary of Chapter 22 Electric flux: Gauss’s law:
Gauss’s law can be used to calculate the field in situations with a high degree of symmetry. Gauss’s law applies in all situations, and therefore is more general than Coulomb’s law.

Chapter 23 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.)

Units of Chapter 23 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

Units of Chapter 23 E Determined from V
Electrostatic Potential Energy; the Electron Volt

23-1 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 Work is done by the electric field in moving the positive charge q from position a to position b.

ConcepTest 23.1b Electric Potential Energy II
A proton and an electron are in a constant electric field created by oppositely charged plates. You release the proton from the positive side and the electron from the negative side. Which has the larger acceleration? 1) proton 2) electron 3) both feel the same acceleration 4) neither – there is no acceleration 5) they feel the same magnitude acceleration but opposite direction Electron electron - + Proton proton

ConcepTest 23.1b Electric Potential Energy II
A proton and an electron are in a constant electric field created by oppositely charged plates. You release the proton from the positive side and the electron from the negative side. Which has the larger acceleration? 1) proton 2) electron 3) both feel the same acceleration 4) neither – there is no acceleration 5) they feel the same magnitude acceleration but opposite direction Since F = ma and the electron is much less massive than the proton, the electron experiences the larger acceleration. Electron electron - + Proton proton

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

Only changes in potential can be measured, allowing free assignment of V = 0:

Analogy between gravitational and electrical potential energy: Figure (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.

Conceptual Example 23-1: 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 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.

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

23-2 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 To find Vba in a nonuniform electric field E, we integrate E·dl from point a to point b.

The simplest case is a uniform field:

Example 23-3: 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 m, calculate the magnitude of the electric field in the space between the plates. Solution: E = V/d = 1000 V/m.

Example 23-4: 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).

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 (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.

23-3 Electric Potential Due to Point Charges
To find the electric potential due to a point charge, we integrate the field along a field line: Figure 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.

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

Example: Work required to bring two positive charges close together. What minimum work must be done by an external force to bring a proton q = 1.60×10-19 C from a great distance away (take r = ∞) to a point 1.00×10-15 m from another proton? 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.

Example: Work required to bring two positive charges close together. What minimum work must be done by an external force to bring a proton q = 1.60×10-19 C from a great distance away (take r = ∞) to a point 1.00×10-15 m from another proton? W = ke2/r = (9×109) (1.6×10-19)2/ (1×10-15) = 2.3×10-13 J = 2.3×10-13 J

ConcepTest 23.3 Electric Potential
1) V > 0 2) V = 0 3) V < 0 What is the electric potential at point B? A B

ConcepTest 23.3 Electric Potential
1) V > 0 2) V = 0 3) V < 0 What is the electric potential at point B? A B Since Q2 and Q1 are equidistant from point B, and since they have equal and opposite charges, the total potential is zero. Follow-up: What is the potential at the origin of the x y axes?

Example 23-7: Potential above two charges. Calculate the electric potential (a) at point A in the figure due to the two charges shown. 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)

23-4 Potential Due to Any Charge Distribution
The potential due to an arbitrary charge distribution can be expressed as a sum or integral (if the distribution is continuous): or

Example 23-8: 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.

Example 23-9: 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.

ConcepTest 23.5 Equipotential Surfaces
Which of these configurations gives V = 0 at all points on the x axis? 1) x +2mC -2mC +1mC -1mC 2) 3) 4) all of the above 5) none of the above

ConcepTest 23.5 Equipotential Surfaces
Which of these configurations gives V = 0 at all points on the x axis? 1) x +2mC -2mC +1mC -1mC 2) 3) 4) all of the above 5) none of the above Only in case (1), where opposite charges lie directly across the x axis from each other, do the potentials from the two charges above the x axis cancel the ones below the x axis.

23-5 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 (E// =0 → ∂V/∂s// = 0) Figure 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 are always perpendicular to field lines; they are always closed surfaces (unlike field lines, which begin and end on charges). Figure Equipotential lines (green, dashed) are always perpendicular to the electric field lines (solid red) shown here for two equal but oppositely charged particles.

A gravitational analogy to equipotential surfaces is the topographical map – the lines connect points of equal gravitational potential (altitude). Figure 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).

23-6 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 Electric dipole. Calculation of potential V at point P.

23-7 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:

23-7 E Determined from V Example 23-11: 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.

Homework Assignment # 3 Chapter 22 – 10, 22, 34 Chapter 23 – 14, 30, 36 Tentative HW # 4: Chapter 24 – 6, 16, 28, 46, 60, 82

Chapter 22 Gauss’s Law Chapter 22 opener. Gauss’s law is an elegant relation between electric charge and electric field. It is more general than Coulomb’s.

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Chapter 22 Gauss’s Law Chapter 22 opener. Gauss’s law is an elegant relation between electric charge and electric field. It is more general than Coulomb’s.

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Presentation on theme: "Chapter 22 Gauss’s Law Chapter 22 opener. Gauss’s law is an elegant relation between electric charge and electric field. It is more general than Coulomb’s."— Presentation transcript:

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