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General Physics (PHY 2140) Lecture 2 Electrostatics Electric flux and Gauss s law Electrical energy potential difference and electric potential potential energy of charged conductors http://www.physics.wayne.edu/~alan/ Chapters 15-16 1

Lightning Review Last lecture: 1. Coulomb s law 2. The electric field F = k e q q r 1 2 2 the superposition principle E = F q 0 Review Problem: A free electron and free proton are placed in an identical electric field. Compare the electric force on each particle. Compare their accelerations. 2

Review Solution: Recall F = qe For a proton or an electron the size of the force is the same! Charges are the same in magnitude. Opposite in sign. The direction of the electric forces are opposite. However the accelerations are different! Mass of electron is 9.11x10-31 Kg Mass of a proton is 1.67x10-27 Kg a = F m So, the acceleration of the proton is smaller by m e /m p = 5.5x10-4 3

15.5 Electric Field Lines A convenient way to visualize field patterns is to draw lines in the direction of the electric field. Such lines are called field lines. Remarks: 1. Electric field vector, E, is tangent to the electric field lines at each point in space. 2. The number of lines per unit area through a surface perpendicular to the lines is proportional to the strength of the electric field in a given region. E is large when the field lines are close together and small when far apart. 4

15.5 Electric Field Lines (2) Electric field lines of single positive (a) and (b) negative charges. a) b) q - q 5

15.5 Electric Field Lines (3) Rules for drawing electric field lines for any charge distribution. 1. Lines must begin on positive charges (or at infinity) and must terminate on negative charges or in the case of excess charge at infinity. 2. The number of lines drawn leaving a positive charge or approaching a negative charge is proportional to the magnitude of the charge. 3. No two field lines can cross each other. 6

15.5 Electric Field Lines (4) Electric field lines of a dipole. - 7

15.6 Conductors in Electrostatic Equilibrium Good conductors (e.g. copper, gold) contain charges (electron) that are not bound to a particular atom, and are free to move within the material. When no net motion of these electrons occur the conductor is said to be in electro-static equilibrium. 8

15.6 Conductors in Electrostatic Equilibrium Properties of an isolated conductor (insulated from the ground). 1. Electric field is zero everywhere within the conductor. 2. Any excess charge on an isolated conductor resides entirely on its surface. 3. The electric field just outside a charged conductor is perpendicular to the conductor s s surface. 4. On an irregular shaped conductor, the charge tends to accumulate at locations where the radius of curvature of the surface is smallest at sharp points. 9

Faraday s s ice-pail experiment - - - - - - - - - - - Demonstrates that the charge resides on the surface of a conductor. 10

Mini-quiz Question: Suppose a point charge Q is in empty space. Wearing rubber gloves, we sneak up and surround the charge with a spherical conducting shell. What effect does this have on the field lines of the charge?? q 11

Question: Suppose a point charge Q is in empty space. Wearing rubber gloves, we sneak up and surround the charge with a spherical conducting shell. What effect does this have on the field lines of the charge? Answer: Negative charge will build up on the inside of the shell. Positive charge will build up on the outside of the shell. There will be no field lines inside the conductor but the field lines will remain outside the shell. - - - - - - q - - - - - - 12

15.9 The oscilloscope Changing E field applied on the deflection plate (electrodes) moves the electron beam. d V 2 θ V 1 L 13

Oscilloscope: deflection angle (additional) d V 2 θ V 1 L electron gun between plates y y x y e x v v = a t L = v t v = ev2 L mdv x = a y 2eV m v ev L ev Lm V L tanθ = = = = v m d v m d ev d ev y 2 2 e 2 2 x e x e 2 1 2 e e 1 ee = = m ev2 m d e 14

15.9 Electric Flux and Gauss s s Law A convenient technique was introduced by Karl F. Gauss (1777-1855) 1855) to calculate electric fields. Requires symmetric charge distributions. Technique based on the notion of electrical flux. 15

15.9 Electric Flux To introduce the notion of flux, consider a situation where the electric field is uniform in magnitude and direction. Consider also that the field lines cross a surface of area A which is perpendicular to the field. The number of field lines per unit of area is constant. The flux, Φ,, is defined as the product of the field magnitude by the area crossed by the field lines. Φ=EA Area=A E 16

15.9 Electric Flux Units: Nm 2 /C in SI units. Find the electric flux through the area A = 2 m 2, which is perpendicular to an electric field E=22 N/C Φ=EA Answer: Φ = 44 Nm 2 /C. 17

15.9 Electric Flux If the surface is not perpendicular to the field, the expression of the field becomes: Φ = EAcosθ Where θ is the angle between the field and a normal to the surface. θ N θ 18

15.9 Electric Flux Remark: When an area is constructed such that a closed surface is formed, we shall adopt the convention that the flux lines passing into the interior of the volume are negative and those passing out of the interior of the volume are positive. 19

Example: Question: Calculate the flux of a constant E field (along x) through a cube of side L. y 1 2 E x z 20

Question: Calculate the flux of a constant E field (along x) through a cube e of side L. Reasoning: Dealing with a composite, closed surface. Sum of the fluxes through all surfaces. Flux of field going in is negative Flux of field going out is positive. E is parallel to all surfaces except surfaces labeled 1 and 2. So only those surfaces (1 & 2) contribute to the flux. y 1 2 E x z 21

15.9 Gauss s s Law The net flux passing through a closed surface surrounding a charge Q is proportional to the magnitude of Q: Φ net = EAcosθ Q In free space, the constant of proportionality is 1/ε o where ε o is called the permittivity of of free space. ε o 1 1 = = = 8.85 10 C N m 4π k 4π 8.99 10 Nm / C e ( 9 2 2 ) 12 2 2 23

15.9 Gauss s s Law The net flux passing through any closed surface is equal to the net charge inside the surface divided by ε o. Φ net = EAcosθ = Q ε o Can be used to compute electric fields. Example: point charge 2 Q Q Φ net = EAcosθ = 4πr E E = = k 2 e 2 4πε r r 0 24

16.0 Introduction The Coulomb force is a conservative force A potential energy function can be defined for any conservative force, including Coulomb force The notions of potential and potential energy are important for practical problem solving 25

16.1 Potential difference and electric potential E The electrostatic force is conservative As in mechanics, work is A B W = Fdcosϑ d Work done on the positive charge by moving it from A to B W = Fdcosϑ = qed 26

Potential energy of electrostatic field The work done by a conservative force equals the negative of the change in potential energy, ΔPE Δ PE = W = qed This equation is valid only for the case of a uniform electric field allows us to introduce the concept of the electric potential 27

Electric potential The potential difference between points A and B, V B -V A, is defined as the change in potential energy (final minus initial value) of a charge, q, moved from A to B, divided by the charge Δ V = V V = B A ΔPE q Electric potential is a scalar quantity Electric potential difference is a measure of electric energy per unit charge Potential is often referred to as voltage 28

Electric potential - units Electric potential difference is the work done to move a charge from a point A to a point B divided by the magnitude of the charge. Thus the SI units of electric potential 1V = 1 J C In other words, 1 J of work is required to move a 1 C of charge between two points that are at potential difference of 1 V 29

Electric potential - notes Units of electric field (N/C) can be expressed in terms of the units of potential (as volts per meter) 1NC= 1Vm Because the positive tends to move in the direction of the electric field, work must be done on the charge to move it in the direction, opposite the field. Thus, A positive charge gains electric potential energy when it is moved in a direction opposite the electric field A negative charge looses electrical potential energy when it moves in the direction opposite the electric field 30

Analogy between electric and gravitational fields The same kinetic-potential energy theorem works here A A E q d g m d B B If a positive charge is released from A, it accelerates in the direction of electric field, i.e. gains kinetic energy If a negative charge is released from A, it accelerates in the direction opposite the electric field KE PE = KE PE i i f f 31

Example: motion of an electron What is the speed of an electron accelerated from rest across a potential difference of 100V? What is the speed of a proton accelerated under the same conditions? Given: ΔV=100 V m e = 9.11 10-31 kg m p = 1.67 10-27 kg e = 1.60 10-19 C Find: v e =? v p =? V ab Observations: 1. given potential energy difference, one can find the kinetic energy difference 2. kinetic energy is related to speed KEi PEi = KE f PE f KE KE = KE =Δ PE = qδv f i f 1 2 2qΔV mv f = qδv v f = 2 m ve = 5.9 10 m, vp = 1.3 10 s 6 5 m s 32

16.2 Electric potential and potential energy due to point charges Electric circuits: point of zero potential is defined by grounding some point in the circuit Electric potential due to a point charge at a point in space: point of zero potential is taken at an infinite distance from the charge With this choice, a potential can be found as V = q ke r Note: the potential depends only on charge of an object, q,, and a distance from this object to a point in space, r. 33

Superposition principle for potentials If more than one point charge is present, their electric potential can be found by applying superposition principle 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. Remember that potentials are scalar quantities! 34

Potential energy of a system of point charges Consider a system of two particles If V 1 is the electric potential due to charge q 1 at a point P, then work required to bring the charge q 2 from infinity to P without acceleration is q 2 V 1. If a distance between P and q 1 is r, then by definition q 2 P r q 1 A PE = q V = k 2 1 e qq r 1 2 Potential energy is positive if charges are of the same sign and vice versa. 35

Mini-quiz: potential energy of an ion Three ions, Na, Na, and Cl -, located such, that they form corners of an equilateral triangle of side 2 nm in water. What is the electric potential energy of one of the Na ions?? Cl - Na Na q q q q q PE = k k = k q q r r r [ ] Na Cl Na Na Na e e e Cl Na but : q = q! Cl Na qna PE = ke qna qna = r [ ] 0 36

Recall from last chapter: Electric field lines of a dipole. - 37

16.3 Potentials and charged conductors Recall that work is opposite of the change in potential energy, [ ] W = PE = q V V No work is required to move a charge between two points that are at the same potential. That is, W=0 if V B =V A Recall: 1. all charge of the charged conductor is located on its surface 2. electric field, E, is always perpendicular to its surface, i.e. no work is done if charges are moved along the surface Thus: potential is constant everywhere on the surface of a charged conductor in equilibrium B A but that s not all! 38

Because the electric field in zero inside the conductor, no work is required to move charges between any two points, i.e. [ ] 0 W = q V V = B If work is zero, any two points inside the conductor have the same potential, i.e. potential is constant everywhere inside a conductor Finally, since one of the points can be arbitrarily close to the surface of the conductor, the electric potential is constant everywhere inside a conductor and equal to its value at the surface! A Note that the potential inside a conductor is not necessarily zero, even though the interior electric field is always zero! 39

The electron volt A unit of energy commonly used in atomic, nuclear and particle physics is electron volt (ev( ev) The electron volt is defined as the energy that electron (or proton) gains when accelerating through a potential difference of 1 V Relation to SI: V ab =1 V 1 ev = 1.60 10 10-19 C V V = 1.60 10 10-19 J 40

Problem-solving strategy Remember that potential is a scalar quantity Superposition principle is an algebraic sum of potentials due to a system of charges Signs are important Just in mechanics, only changes in electric potential are significant, hence, the point you choose for zero electric potential is arbitrary. 41

Example : ionization energy of the electron in a hydrogen atom In the Bohr model of a hydrogen atom, the electron, if it is in the ground state, orbits the proton at a distance of r = 5.29 10-11 m. Find the ionization energy of the atom, i.e. the energy required to remove the electron from the atom. Note that the Bohr model, the idea of electrons as tiny balls orbiting the nucleus, is not a very good model of the atom. A better picture is one in which the electron is spread out around the nucleus in a cloud of varying density; however, the Bohr model does give the right answer for the ionization energy 42

In the Bohr model of a hydrogen atom, the electron, if it is in the ground state, orbits the proton at a distance of r = 5.29 x 10-11 m. Find the ionization energy, i.e. the energy required to remove the electron from the atom. Given: r = 5.292 x 10-11 m m e = 9.11 10-31 kg m p = 1.67 10-27 kg e = 1.60 10-19 C Find: E=? The ionization energy equals to the total energy of the electron-proton system, E = PE KE The velocity of e can be found by analyzing the force on the electron. This force is the Coulomb force; because the electron travels in a circular orbit, the acceleration will be the centripetal acceleration: ma c = F c or with v m r Thus, total energy is 2 2 e v PE = ke, KE = m r 2 e = k, e 2 or r 2 2 v 2 = e 2 ke mr, 2 2 2 e m ke e e E = ke = ke = r 2 mr 2r 18 2.18 10 J -13.6 ev 43

16.4 Equipotential surfaces They are defined as a surface in space on which the potential is the same for every point (surfaces of constant voltage) The electric field at every point of an equipotential surface is perpendicular to the surface convenient to represent by drawing equipotential lines 44