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Physics 2B for Materials and Structural Engineering Lecturer: Daniel Rohrlich Teaching Assistants: Oren Rosenblatt, Irina Segel Ben Gurion University of the Negev Week 2. Gauss’s law – Scalar and vector fields why fields? ring of charge disk of charge lines of electric field electric dipole in a constant field flux Gauss’s law applications Sources: Halliday, Resnick and Krane, 5 th Edition, Chaps ; Purcell (Berkeley course 2), Chap. 1, Sects. 1.8 – 1.13.

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Scalar and vector fields Here are two examples of scalar fields: Time-dependent wave field ·· Instantaneous temperature field ·· In how many spatial dimensions do these scalar fields “live”?

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Wind field ·· Scalar and vector fields Here are two examples of vector fields: Electric dipole field ·· A vector field has a direction and a magnitude at every point in space.

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Why fields? m1m1 m2m2 F 21 F 12

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Why fields? “That gravity should be innate, inherent and essential to matter so that one body may act upon another at a distance through a vacuum without the mediation of anything else, by and through which their action or force may be conveyed from one to another, is to me so great an absurdity that I believe no man who has in philosophical matters any competent faculty of thinking can ever fall into it.” – Sir Isaac Newton m1m1 m2m2 F 21 F 12

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Why fields?

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“Newton himself was better aware of the weaknesses inherent in his intellectual edifice than the generations of learned scientists who followed him. This fact has always aroused my deep admiration.” – Albert Einstein

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Ring of charge What is the electric field of a uniform ring of radius R and linear charge density λ along its axis of symmetry? θ z 0

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Ring of charge What is the electric field of a uniform ring of radius R and linear charge density λ along its axis of symmetry? z Answer: θ 0

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Disk of charge What is the electric field of a uniform disk of radius R, and surface charge density λ, along its axis of symmetry? z 0 θ

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Disk of charge What is the electric field of a uniform disk of radius R, and surface charge density λ, along its axis of symmetry? z Answer: 0 θ

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Disk of charge What is the electric field of a uniform disk of radius R, and surface charge density λ, along its axis of symmetry? z Answer: Substitute ρ = z tan θ dρ = z dθ/cos 2 θ 0 ≤ θ ≤ tan –1 R/z 0 θ

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Disk of charge What is the electric field of a uniform disk of radius R, and surface charge density λ, along its axis of symmetry? z Answer: Substitute ρ = z tan θ dρ = z dθ/cos 2 θ 0 ≤ θ ≤ tan –1 R/z θ 0

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Disk of charge What is the electric field of a uniform disk of radius R, and surface charge density λ, along its axis of symmetry? z θ 0 For R → ∞ we get while for R → 0 we get back the electric field of a point charge q. (We must take the limit with nonvanishing q = λπR 2.)

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Lines of electric field The great experimental physicist Faraday, in his revolutionary papers on electromagnetism, did not write down a single equation! How did he do it? He visualized the electric (and magnetic) fields.

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Particles … elementary particles … I used to study them … calculate them … theorize about them … but now I can see them! based on drawing and text drawingtext

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Lines of electric field The great experimental physicist Faraday, in his revolutionary papers on electromagnetism, did not write down a single equation! How did he do it? He visualized the electric (and magnetic) fields.

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Lines of electric field The great experimental physicist Faraday, in his revolutionary papers on electromagnetism, did not write down a single equation! How did he do it? He visualized the electric (and magnetic) fields.

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Lines of electric field The great experimental physicist Faraday, in his revolutionary papers on electromagnetism, did not write down a single equation! How did he do it? He visualized the electric (and magnetic) fields.

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Lines of electric field Some basic rules and observations about field lines: They never start or stop in empty space – they stop or start on a charge or extend to infinity. They never cross – if they did, a small charge placed at the crossing would show the true direction of the field there. The density of field lines in one direction is proportional to the strength of the field in the perpendicular direction.

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Lines of electric field Some basic rules and observations about field lines: They never start or stop in empty space – they stop or start on a charge or extend to infinity. They never cross – if they did, a small charge placed at the crossing would show the true direction of the field there. The density of field lines in one direction is proportional to the strength of the field in the perpendicular direction.

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Lines of electric field Some basic rules and observations about field lines: They never start or stop in empty space – they stop or start on a charge or extend to infinity. They never cross – if they did, a small charge placed at the crossing would show the true direction of the field there. The density of field lines in one direction is proportional to the strength of the field in the perpendicular direction. E density

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Electric dipole in a constant field A constant electric field is the same E everywhere. So the force on a charge q is everywhere F q = qE. Suppose we have a rigid dipole – two equal and opposite charges of magnitude q separated by a rigid (assumed massless) rod of length d. Since the dipole has a direction, we can represent it by a vector p which points in the direction of the positive charge and has magnitude p = qd. What happens to the dipole in a constant field E? q –q E E θ

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Electric dipole in a constant field A constant electric field is the same E everywhere. So the force on a charge q is everywhere F q = qE. Suppose we have a rigid dipole – two equal and opposite charges of magnitude q separated by a rigid (assumed massless) rod of length d. Since the dipole has a direction, we can represent it by a vector p which points in the direction of the positive charge and has magnitude p = qd. What happens to the dipole in a constant field E? q FqFq θ –q F -q

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Electric dipole in a constant field A constant electric field is the same E everywhere. So the force on a charge q is everywhere F q = qE. Suppose we have a rigid dipole – two equal and opposite charges of magnitude q separated by a rigid (assumed massless) rod of length d. Since the dipole has a direction, we can represent it by a vector p which points in the direction of the positive charge and has magnitude p = qd. What happens to the dipole in a constant field E? 1. Since the charges are equal and opposite, the force is zero. 2. The torque τ on the dipole is τ = p × E. 3. The potential energy of the dipole is U = –p·E.

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Electric dipole in a constant field 3. The potential energy of the dipole is U = –p·E: Let’s start at θ = 90°. The displacement parallel to the force is (d/2) cos θ, so the work is 2F (d/2) cos θ = qEd cos θ = pE cos θ = p·E. The potential energy U is minus the work, so U = –p·E. FqFq θ –q F -q

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Halliday, Resnick and Krane, 5 th Edition, Chap. 26, Prob. 6: R θ E This semi-infinite bar has charge λ per unit length. Prove that θ = 45° regardless of R. x y

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Halliday, Resnick and Krane, 5 th Edition, Chap. 26, Prob. 6: R θ E This semi-infinite bar has charge λ per unit length. Prove that θ = 45° regardless of R. x y Solution:

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Halliday, Resnick and Krane, 5 th Edition, Chap. 26, Prob. 6: R θ E This semi-infinite bar has charge λ per unit length. Prove that θ = 45° regardless of R. x y We substitute x = R tanφ, dx = R dφ/cos 2 φ and integrate. Solution:

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Halliday, Resnick and Krane, 5 th Edition, Chap. 26, Prob. 6: R θ E This semi-infinite bar has charge λ per unit length. Prove that θ = 45° regardless of R. x y Solution:

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Halliday, Resnick and Krane, 5 th Edition, Chap. 26, Prob. 14: Find the frequency of small oscillations of a dipole of strength p and moment of inertia I around its equilibrium position in a constant electric field E. FqFq θ –q F -q

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Halliday, Resnick and Krane, 5 th Edition, Chap. 26, Prob. 14: Find the frequency of small oscillations of a dipole of strength p and moment of inertia I around its equilibrium position in a constant electric field E. Solution: The equilibrium position is θ = 0. The magnitude of the torque τ = p × E is pE sinθ ≈ pEθ so and θ = A sin (ωt+ θ 0 ) where ω =. FqFq θ –q F -q

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Flux Definition of an element of electric flux: E·dA The direction of dA is always perpendicular to the surface. In this example, E · dA is (E)(dA) cos 145 °. Oriented area element dA Electric field E E dAdA

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Flux Definition of an element of electric flux: E·dA The direction of dA is always perpendicular to the surface. In this example, 0 ≤ E · dA ≤ (E)(dA). Oriented area element dA Electric field E E dAdA E E E dAdA dAdA

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Flux The total flux through a surface, denoted Ф E, is the sum of all the flux elements: Ф E =. Flux through back wall: –E(10 cm)(30 cm). Flux through upper wall: E(cos 60º)(30 cm)(10 cm )/(sin 30 º ). Oriented area element dA Electric field E

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Gauss ’ s law The total flux Ф E (the sum of all the fluxes) through a closed surface equals the total charge inside the surface divided by ε 0.

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Gauss ’ s law The total flux Ф E (the sum of all the fluxes) through a closed surface equals the total charge inside the surface divided by ε 0. If there is no charge inside a closed surface, then all the field lines entering must also leave! Then the total flux is zero. ··

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Gauss ’ s law Ф E > 0

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Gauss ’ s law Ф E = 0

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Gauss ’ s law Ф E < 0

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Applications Let’s use Gauss’s law to derive the electric field of a point charge q. Consider a sphere of radius r centered on the point charge. By symmetry, E must be radial and E must be constant and so Now we apply Gauss’s law, and immediately obtain E = q/4πε 0 r 2, which is Coulomb’s law.

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Applications Gauss’s law makes it easy to obtain the electric field when there is symmetry to help us. For example, what is the electric field of an infinite rod with uniform charge density λ? Let the rod be the axis of a cylinder of radius ρ and height h; by symmetry, we have E E

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Applications Similarly, we can use Gauss’s law to calculate the electric field of a plane with uniform charge density σ. If we take a box of any cross-sectional area A and enclose part of the plane in the box, the electric flux will be zero on the sides of the box and 2EA through the top and bottom of the box. Thus E

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Gauss ’ s law Gauss’s law is often more convenient than Coulomb’s law. Gauss’s law implies Coulomb’s law. Coulomb’s law is true only for static charges (just as Newton’s law for the gravitational force is true only for static masses) but Gauss’s law is true also for moving charges.

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Halliday, Resnick and Krane, 5 th Edition, Chap. 27, Prob. 6: This infinite plane has a uniform charge density σ, except that a disk of radius R is missing. The point P is on the symmetry axis at a height z above the plane. What is the electric field E P at P? P

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Halliday, Resnick and Krane, 5 th Edition, Chap. 27, Prob. 6: Answer: Including the disk, we would have The field due to the disk is. P z

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Halliday, Resnick and Krane, 5 th Edition, Chap. 27, Prob. 6: The electric field is a vector field, hence E P + E disk = E plane and the field at the point P is E P = E plane – E disk : P z

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Halliday, Resnick and Krane, 5 th Edition, Chap. 27, Prob. 19: A conducting sphere carrying a charge Q is surrounded by a spherical conducting shell. (a) What is the total charge on the inner surface of the shell? (b) A point charge q is placed outside the shell. Now what is the total charge on the inner surface of the shell? (c) Now the point charge q is between the shell and the sphere. What is the total charge on the inner surface of the shell? (d) Are the answers the same if the sphere and shell are not concentric? Q

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Halliday, Resnick and Krane, 5 th Edition, Chap. 27, Prob. 19: (a) What is the total charge on the inner surface of the shell? Answer: E vanishes everywhere inside a conductor. (Otherwise electrons would rearrange themselves in order to make it vanish.) Hence the flux through the surface shown vanishes. Hence the total charge inside the surface vanishes. Hence the charge on the inner surface of the shell must be –Q. Q

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Halliday, Resnick and Krane, 5 th Edition, Chap. 27, Prob. 19: (b) A point charge q is placed outside the shell. Now what is the total charge on the inner surface of the shell? Answer: E still vanishes everywhere inside a conductor. Hence the flux through the surface shown still vanishes. Hence the total charge inside the surface still vanishes. Hence the charge on the inner surface of the shell must still be –Q. The point charge outside can ’ t change these facts. Q q

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Halliday, Resnick and Krane, 5 th Edition, Chap. 27, Prob. 19: (c) Now the point charge q is between the shell and the sphere. What is the total charge on the inner surface of the shell? Answer: E still vanishes everywhere inside a conductor. Hence the flux through the surface shown still vanishes. Hence the total charge inside the surface still vanishes. But now the charge inside the inner surface of the shell is Q + q. Hence the charge on the inner surface of the shell must be –Q – q. Q q

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Halliday, Resnick and Krane, 5 th Edition, Chap. 27, Prob. 19: (d) Are the answers the same if the sphere and shell are not concentric? Answer: Yes – we never assumed that the sphere and shell were concentric. Q

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