University Physics (PHYS 2326)

Transcription

1 Chapters University Physics (PHYS 2326) Lecture 4 Electrostatics Electric flux and Gauss s law Electrical energy potential difference and electric potential potential energy of charged conductors 3/26/2015 1

2 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 3/26/2015 2

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4 E The electrostatic force is conservative As in mechanics, work is A B W Fd cos d Work done on the positive charge by moving it from A to B W Fd cos qed 3/26/2015 4

5 The work done by a conservative force equals the negative of the change in potential energy, DPE DPE W qed This equation is valid only for the case of a uniform electric field allows to introduce the concept of electric potential 3/26/2015 5

6 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 DV V V B A DPE 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 3/26/2015 6

7 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 3/26/2015 7

8 Units of electric field (N/C) can be expressed in terms of the units of potential (as volts per meter) 1N C 1V m 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 3/26/2015 8

9 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 KEi PEi KE f PE f 3/26/2015 9

10 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: DV=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 DPE qdv f i f 1 2 2qDV mv f qdv v f 2 m ve m, vp s 6 5 m s 3/26/

11 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. 3/26/

12 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! 3/26/

13 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. 3/26/

14 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 3/26/

15 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! 3/26/

16 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! Note that the potential inside a conductor is not necessarily zero, even though the interior electric field is always zero! A 3/26/

17 A unit of energy commonly used in atomic, nuclear and particle physics is electron volt (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 = J 3/26/

18 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. 3/26/

19 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 3/26/

20 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 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 k e e r 2, or v 2 e 2 ke mr, e m ke e e E ke ke r 2 mr 2r J ev 3/26/

21 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 3/26/

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23 dv ( V ˆ V ˆ E i j V kˆ ) dr x y z 3/26/

24 Three point charges, which initially are infinitely far apart, are placed at the corners of an equilateral triangle with sides d.two of the point charges are identical and have charge q. If zero net work is required to place the three charges at the corners of the triangle, what must the value of the third charge be? A thin spherical shell with radius R 1 = 3.00cm is concentric with a larger thin spherical shell with radius R 2 = 5.00cm. Both shells are made of insulating material. The smaller shell has charge q 1 = +6.00nC distributed uniformly over its surface, and the larger shell has charge q 2 = -9.00nC distributed uniformly over its surface. Take the electric potential to be zero at an infinite distance from both shells. (a) What is the electric potential due to the two shells at the following distance from their common center: (i) r = 0; (ii) r = 4.00 cm; (iii) r = 6.00cm? (b) What is the magnitude of the potential difference between the surfaces of the two shells? Which shell is at higher potential: the inner shell or the outer shell? 3/26/

25 1. a) E = 0 V/m in throughout some region of space, can you conclude that the potential V = 0 in this region? b) V = 0 V throughout some region of space. Can you conclude that the electric field E = 0 V/m in this region? --Find the electric potential everywhere for a sphere (radius R) with charge (Q) uniformly distributed. Take V=0 at infinity. --Sketch V vs r and E r vs r. Given ì ï K Q E = R r r ˆ í 3 ï K Q r î r ˆ 2 r < R R < r 3/26/

26 Find the electric potential everywhere for a sphere (radius R) with charge (Q) uniformly distributed. V = ì ï í ï î ï 3KQ 2R - KQ 2R 3 r2 K Q r (r < R) (R < r) E r = ì K Q ï R r í 3 ï K Q î r 2 (r < R) (R < r) 3/26/

27 Find the x,y and z components of the electric field, given that the electric potential of a disk is given by V disk = Q 2pR 2 e 0 ( z 2 + R 2 - z) 3/26/

28 Find the z component of the electric field, given that the electric potential of a disk is given by V disk = Q 2pR 2 e 0 ( z 2 + R 2 - z) ( E ) disk z = - dv dz = h æ 1-2e ç 0 è z z 2 + R 2 ö ø 3/26/

29 Geometry of potential/field is perp to equipotential surfaces points downhill (decreasing V) --strength proportional to spacing equipotentials 3/26/

30 Conductor in equilibrium: field and potential --field is zero inside conductor --field is perpendicular at surface --conductor is at equipotential (no work to move) 3/26/

31 Conductor in equilibrium: equipotentials --equipotentials are parallel to nearby conductor 3/26/

32 Problem: Finding Potential --Find the electric potential everywhere for a point charge (q) at the center of a hollow metal sphere (inner radius a, outer radius b) with charge Q. (Take V = 0 at infinity.) --Sketch V vs r and E r vs r. b a ì ï ï E r (r) = í ï ï î Kq r, 2 r < a 0, a < r < b K(Q + q) r 2, b < r 3/26/

33 Problem: Finding Potential (ans) Finding V r For all r For b < r = é = - - ë ê K(Q + q) r K(Q + q) r - Defined V = 0 V r - V = V r = K(Q + q) r K(Q + q) r (R < r) ù û r ú K(Q + q) For r < a é = - - Kq ù ë ê r û ì ï ï E r (r) = í ï ï î Kq r, 2 r < a 0, a < r < b K(Q + q) r 2, b < r 3/26/ ú a = Kq r - Kq a K(Q + q) From before V a = V b = b V r - V r = K(Q + q) b r = Kq r - Kq a K(Q + q) æ + Kq b r - Kq ö è a ø (r < a) V r V V a V r For a < r < b E dr from b < r; V b = V r = K(Q + q) b V V r K(Q + q) b (a < r < b)

34 Problem: finding Potential (Answer) V = ì ï K q + Q r ï í K q + Q ï b ï K q + Q æ + Kq 1 b r - 1 ö î ï è aø b < r a < r < b r < a ì ï ï E r (r) = í ï ï î Kq r, 2 r < a 0, a < r < b K(Q + q) r 2, b < r not origin 3/26/

35 Sources of potential: Capacitor -charge separation -not sustained 3/26/

36 Finding V r For all r For R < r definition For r < R V r ì ï K Q E = R r r ˆ í 3 ï K Q r î r ˆ 2 V V R V r V r E dr V r < R R < r V r é = - - KQ ë ê r r ù û ú = KQ r - KQ Defined V = 0 V r - V = KQ r V r = KQ r (R < r) é = - KQ ë ê 2R 3 r2 û ú R 3/26/ ù æ = - KQ 2R 3 r2 - KQ è r 2R 3 R2 From R < r : V R = KQ R = KQ R 3 R2 V r - KQ R = - æ KQ 2R 3 r2 - KQ è 2R 3 R2 ö ø ö ø V r = 3KQ 2R 3 R2 - KQ 2R 3 r2 (r < R)