General Physics (PHY 2140)

Transcription

1 General Physics (PHY 2140) Lecture 4 Electrostatics Electric flux and Gauss s law Electrical energy potential difference and electric potential potential energy of charged conductors Chapters

2 Lightning Review Last lecture: 1. Properties of the electric field, field lines 2. Conductors in electrostatic equilibrium Electric field is zero everywhere within the conductor. Any excess charge field on an isolated conductor resides on its surface. The electric field just outside a charged conductor is perpendicular to the conductor s surface. On an irregular shaped conductor, the charge tends to accumulate at locations where the radius of curvature of the surface is smallest. st. Review Problem: Would life be different if the electron were positively charged and the proton were negatively charged? Does the choice of signs have any bearing on physical and chemical interactions? 2

3 15.10 Electric Flux and Gauss s Law A convenient technique was introduced by Karl F. Gauss ( ) 1855) to calculate electric fields. Requires symmetric charge distributions. Technique based on the notion of electrical flux. 3

4 15.10 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 4

5 15.10 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 Nm2/C. 5

6 15.10 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 6

7 15.10 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. 7

8 Example: Question: Calculate the flux of a constant E field (along x) through a cube of side L. y 1 2 E x z 8

9 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 surface contribute to the flux. Solution: Φ = EA cosθ = EL Φ = EA cosθ = EL Φ = + = net EL EL 0 y 1 2 E x z 9

10 15.10 Gauss 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 = = = C N m 4π k 4π Nm / C e ( )

11 15.10 Gauss 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 11

12 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 12

13 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 13

14 Potential energy of electrostatic field The work done by a conservative force equals the negative of the change in potential energy, PE PE = W = qed This equation is valid only for the case of a uniform electric field allows to introduce the concept of electric potential 14

15 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 V = V V = B A PE 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 15

16 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 16

17 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 17

18 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 18

19 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: V=100 V m e = kg m p = kg e = 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 = PE = q V f i f 1 2 2q V mv f = q V v f = 2 m ve = m, vp = s 6 5 m s 19

20 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.. 20

21 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! 21

22 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 q 1 P r 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. 22

23 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 23

24 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! 24

25 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! 25

26 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 = C V V = J 26

27 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. 27

28 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 = 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 28

29 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 m. Find the ionization energy, i.e. the energy required to remove the electron from the atom. Given: r = x m m e = kg m p = kg e = C Find: E=? The ionization energy equals to the total energy of the electron-proton system, E = PE + KE with 2 2 e v PE = ke, KE = m r 2 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 or c Thus, total energy is v m r e = k, e 2 or r 2 2 v 2 = e 2 ke mr, e m ke e e E = ke + = ke = r 2 mr 2r J ev 29

30 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 30

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