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Electricity and Magnetism Coulomb’s Law Electric Field Lana Sheridan De Anza College Sept 24, 2015 Last time • charge • charge interactions • charge induction . Warm Up: Worksheet . 3. Do both balloons A and B have a charge? entry by n entering the rds each other ne another rge opposite type (A) yesor r charged (B) no, neither is charged (C) at least 1 is charged. ons being . Warm Up: Worksheet . 3. Do both balloons A and B have a charge? entry by n entering the rds each other ne another rge opposite type (A) yesor r charged (B) no, neither is charged (C) at least 1 is charged. ons being ← charged or Warm Up: Worksheet 5. Does this happen? g straight at _____. re identical other and both (A) yes (B) no charged or Warm Up: Worksheet 5. Does this happen? g straight at _____. re identical other and both (A) yes (B) no ← consider Newton’s 3rd law Overview • Coulomb’s Law • The net force of several charges • Vector review • Charge quantization • Charge conservation • Current • Forces at a fundamental level • Electric field • Conductors and electric fields Electrostatic Forces Charged objects interact via the electrostatic force. The force that one charge exerts on another can be attractive or repulsive, depending on the signs of the charges. • Charges with the same electrical sign repel each other. • Charges with opposite electrical signs attract each other. Charge is written with the symbol q or Q. Electrostatic Forces For a pair of point-particles with charges q1 and q2 , the magnitude of the force on each particle is given by Coulomb’s Law: F1,2 = k |q1 q2 | r2 k is the electrostatic constant and r is the distance between the two charged particles. k= 1 4π0 = 8.99 × 109 N m2 /C2 Electrostatic Forces: Coulomb’s Law F1,2 = k |q1 q2 | r2 Remember however, forces are vectors. The vector version of the law is: F1→2 = k q1 q2 r̂1→2 r2 where F1→2 is the force that particle 1 exerts on particle 2, and r̂1→2 is a unit vector pointing from particle 1 to particle 2. Coulomb’s Law Coulomb’s Law: F1→2 = Does this look a bit familiar? k q1 q2 r̂1→2 r2 Coulomb’s Law Coulomb’s Law: F1→2 = k q1 q2 r̂1→2 r2 Does this look a bit familiar? Similar to this? F1→2 = − G m 1 m2 r̂1→2 r2 Coulomb’s Law F1→2 = Fields When the charges are of the same sign, the force is repulsive. S When the charges are of opposite signs, the force is attractive. " q2 S r ! q1 k q1 q2 r̂1→2 r2 ! q2 F12 S F12 S rˆ12 ! q1 F21 F21 a 1 b Figure from Serway & Jewett, Physics for Scientists and Engineers, 9th ed. Electrostatic Constant The electrostatic constant is: k = 1 4π0 = 8.99 × 109 N m2 C−2 0 is called the permittivity constant or the electrical permittivity of free space. 0 = 8.85 × 10−12 C2 N−1 m−2 F12 ! " (1.15 # 10 Example -8 N)i . (Answer) Sample Prob Finding the net force due to (a) Figure 21-8a shows two positively charged particles fixed in place on an x axis.The charges are q1 ! 1.60 # 10 "19 C and q2 ! 3.20 # 10 "19 C, and the particle separation is R ! 0.0200 m. What are the magnitude andisdirection of the electrostatic force This the first : (a) F12 on particle 1 from particle 2? (b) 3n has par T 1d arrangement. a rged partiKEY IDEAS q2 q1 q1 harges q1 x Because re fixed in both particles are positively charged, particle 1 is re- Th R magnitude given by Eq. 21-4. pelled by particle 2, with a force for n an x axis. : Thus, the direction of force F12(a) on particle 1 is away from partipar free-body cle 2, in the negative direction of the x axis, as indicated in the to free-body diagram of Fig. 21-8b. Be for particle 9 2 −2 −12 2 −1 −2 par This particle kTwo =elec8.99 × 10 N m C Eq. or21-4 0is = the 8.85 × 10 C substituted N m particles: Using with separation R ng the rec F12 of this force as of interest. forcefor onr, we it can write the magnitude gra F12 ! " (1.15 # 10 Example -8 N)i . (Answer) Sample Prob Finding the net force due to (a) Figure 21-8a shows two positively charged particles fixed in place on an x axis.The charges are q1 ! 1.60 # 10 "19 C and q2 ! 3.20 # 10 "19 C, and the particle separation is R ! 0.0200 m. What are the magnitude andisdirection of the electrostatic force This the first : (a) F12 on particle 1 from particle 2? (b) 3n has par T 1d arrangement. a rged partiKEY IDEAS q2 q1 q1 harges q1 x Because re fixed in both particles are positively charged, particle 1 is re- Th R magnitude given by Eq. 21-4. pelled by particle 2, with a force for n an x axis. : Thus, the direction of force F12(a) on particle 1 is away from partipar free-body cle 2, in the negative direction of the x axis, as indicated in the to free-body diagram of Fig. 21-8b. Be for particle −24 par This is the particle Answer: F2→1 = −1.15 10 iN Two particles: Using×Eq. 21-4 with separation R substituted ng the elecrec F12 of this force as of interest. forcefor onr, we it can write the magnitude gra Force from many charges Forces from many charges add up to give a net force This is (very grandly) called the “principle of superposition”. The net force on particle 1 from particles 2, 3, ... n is: Fnet,1 = F2→1 + F3→1 + ... + Fn→1 t is true about S the electric forcesSon the S objects? S S 5S2 F BA (c) 3 F AB 5 2 F BA (d) F AB 5 3 F BA Example F BA Consider three point charges located at the corners of a right triangle as shown, where q1 = q3 = 5.00 µC, q2 = −2.00 µC, and a = 0.100 m. Find the resultant force exerted on q3 . y iangle as shown in 0.100 m. Find the q2 " rge q 3 is near two are exerted in difhown in the figure, egorize this exam- a q1 ! Figure 23.7 S a S F23 F13 ! q3 !2a x (Example 23.2) The S force exerted by q 1 on q 3 is F 13. The 1 S Figure from Serway & Jewett, Physics for Scientists and Engineers, 9th ed. 5S2 F BA (c) 3 F AB 5 2 F BA (d) F AB 5 3 F BA FExample BA Consider three point charges located at the corners of a right triangle as shown, where q1 = q3 = 5.00 µC, q2 = −2.00 µC, and a = 0.100 m. Find the resultant force exerted on q3 . y iangle as shown in 0.100 m. Find the q2 " rge q 3 is near two are exerted in difhown in the figure, a q1 ! S a S F23 F13 ! q3 !2a x egorize this examFigure (ExampleONLY 23.2) The Step 1: What is the force23.7 considering particles 2 and 3? S force exerted by q 1 on q 3 is F 13. The S force q 2 on for q isScientists F 23. Figure from Serway &exerted Jewett, by Physics and Engineers, 9th ed. and q on q are S 3 1 5S2 F BA (c) 3 F AB 5 2 F BA (d) F AB 5 3 F BA FExample BA Consider three point charges located at the corners of a right triangle as shown, where q1 = q3 = 5.00 µC, q2 = −2.00 µC, and a = 0.100 m. Find the resultant force exerted on q3 . y iangle as shown in 0.100 m. Find the rge q 3 is near two are exerted in difhown in the figure, q2 " a q1 ! S a S F23 F13 ! q3 !2a x egorize this examFigure Step 2: What is the force23.7 from(Example particle 23.2) 1Son The particle 3? It has 2 force exerted by q on q is F . The 1 3 components. S13 force exerted by q 2 on q 3 is F 23. 1 q 3 are S for Scientists and Engineers, 9th ed. 1 and q 2 on Figure from Serway & Jewett, Physics Reminder about Vectors scalar A scalar quantity indicates an amount. It is represented by a real number. (Assuming it is a physical quantity.) Reminder about Vectors scalar A scalar quantity indicates an amount. It is represented by a real number. (Assuming it is a physical quantity.) vector A vector quantity indicates both an amount and a direction. It is represented more than one real number. (Assuming it is a physical quantity.) Reminder about Vectors scalar A scalar quantity indicates an amount. It is represented by a real number. (Assuming it is a physical quantity.) vector A vector quantity indicates both an amount and a direction. It is represented more than one real number. (Assuming it is a physical quantity.) There are many ways to represent a vector. • a magnitude and (an) angle(s) • magnitudes in several perpendicular directions Representing Vectors: Angles Bearing angles Example, a plane flies at a bearing of 70◦ N E Reference angles x-axis, CCW A baseball is thrown at 10 ms −1 30◦ above the horizontal. Representing Vectors: Unit Vectors Another useful way to represent vectors is in terms of unit vectors. Unit vectors have a magnitude of one unit. In this course, a unit vector r̂ is a one-unit-long vector parallel to the vector r. Representing Vectors: Unit Vectors Another useful way to represent vectors is in terms of unit vectors. Unit vectors have a magnitude of one unit. In this course, a unit vector r̂ is a one-unit-long vector parallel to the vector r. In two dimensions, a pair of perpendicular unit vectors are usually denoted i and j (or sometimes x̂, ŷ). Representing Vectors: Unit Vectors Another useful way to represent vectors is in terms of unit vectors. Unit vectors have a magnitude of one unit. In this course, a unit vector r̂ is a one-unit-long vector parallel to the vector r. In two dimensions, a pair of perpendicular unit vectors are usually denoted i and j (or sometimes x̂, ŷ). A generic 2 dimensional vector can be written as v = ai + bj, where a and b are numbers. um of two other component vectors A x , which is parallel to the x axis, and A y , whic parallel to the y axis. S From SFigure 3.12b, we see that the three vectors form Components S ight triangle and that A 5 A 1 A . We shall often refer to the “component x y S f a vector A ,” written Ax and Ay (without the boldface notation). The compo S A = Ax i + Ay j, where a and b Consider the 2 dimensional vector ent Ax represents the projection of A along the x axis, and the component A S are numbers. epresents the projection of A along the y axis. These components can be positiv S r negative. The component Ax is positiveSif the component vector A x points i We then say that Ax is the i-component (or x-component) of A he positive x direction and is negative if A x points in the negative x direction. and Ay is the j-component (or y -component) of A. milar statement is made for the component Ay. y y S S S A Ay u O a S Ax A x u O b Notice that Ax = A cos θ and Ay = A sin θ. S Ax S Ay x When two vectors are added, the sum is indepe tion. (This fact may seem trivial, but as you will s Vectors Properties and Operations important when vectors are multiplied. Procedures cussed in Chapters 7 and 11.) This property, which c construction in Figure 3.8, is known as the commut Equality S S S Vectors A = Blawifofand only if the magnitudes and directions areB the A 1 5 B Commutative addition same. (Each component is the same.) Addition A+B S S A " S C S S B S R! S B A" S R ! S B " S C " D S D S B S S A A S Figure 3.6 When vector B is S S added to vector A , the resultant R is the vector that runs from the tail of S S Figure 3.7 Geometric construction for summing four vectors. The S resultant vector R is by definition S 1 A mbers that normal rules C 1 D is the vector that completes the polygon. In other words, R is the vector drawn from the tail of the first vector to the tip of the last vector. This technique for adding vectors is often called the “head to tail method.” When two vectors are added, the sum is independent of the order of the addition. (This fact may seem trivial, but as you will see in Chapter 11, the order is important when vectors are multiplied. Procedures for multiplying vectors are discussed in Chapters 7 and 11.) This property, which can be seen from the geometric construction in Figure 3.8, is known as the commutative law of addition: Vectors Properties and Operations f addition S S S S (3.5) A 1 B 5 B 1 A A Property of Addition S Draw B , S then add A . S A A" S B A + B = B + A (commutative) S S S C A " B S R ! S S B S B S A S vector B is S he resultant R is from the tail of B S S C " B " S S S S B R! S B" D S A! S D Figure 3.7 Geometric construction for summing four vectors. The S resultant vector R is by definition the one that completes the polygon. S A S Draw A , S then add B . Figure 3.8S ThisSconstruction S S shows that A 1 B 5 B 1 A or, in other words, that vector addition is commutative. he bookkeeping nents separately. Vectors Properties and Operations Doing addition: Almost always the right answer is to break each vector into components and sum each component independently. y (3.14) ant vector are (3.15) s, we add all the ctor and use the omponents with Ry S By R Ay S B S A x Bx Ax Rx btained from its Figure 3.16 This geometric S S S S S 5S2 F BA (c) 3 F AB 5 2 F BA (d) F AB 5 3 F BA FExample BA Consider three point charges located at the corners of a right triangle as shown, where q1 = q3 = 5.00 µC, q2 = −2.00 µC, and a = 0.100 m. Find the resultant force exerted on q3 . y iangle as shown in 0.100 m. Find the q2 " rge q 3 is near two are exerted in difhown in the figure, egorize this exam- a q1 ! Figure 23.7 S a S F23 F13 ! q3 !2a x (Example 23.2) The S force exerted by q 1 on q 3 is F 13. The S 1 Figure from Serway &exerted Jewett, by Physics and Engineers, 9th ed. force q 2 on for q 3 isScientists F 23. 5S2 F BA (c) 3 F AB 5 2 F BA (d) F AB 5 3 F BA FExample BA Consider three point charges located at the corners of a right triangle as shown, where q1 = q3 = 5.00 µC, q2 = −2.00 µC, and a = 0.100 m. Find the resultant force exerted on q3 . y iangle as shown in 0.100 m. Find the q2 " rge q 3 is near two are exerted in difhown in the figure, egorize this exam- 1 a q1 ! S a S F23 F13 ! q3 !2a x Figure 23.7 (Example 23.2) The S Answer: Fnet,3 = force (−1.04 i + 7.94 exerted by q 1 j) onN q 3 is F 13. The S force exerted by q 2 on q 3 is F 23. 1 and q 2 on q 3 are S for Scientists and Engineers, 9th ed. Figure from Serway & Jewett, Physics Charge is Quantized quantization A physical quantity is said to be quantized if if can only take discrete values. Originally, charge was thought to be a continuous fluid. Charge is Quantized quantization A physical quantity is said to be quantized if if can only take discrete values. Originally, charge was thought to be a continuous fluid. It is not. Charge is Quantized quantization A physical quantity is said to be quantized if if can only take discrete values. Originally, charge was thought to be a continuous fluid. It is not. Just like water has a smallest unit, the H2 O molecule, charge has a smallest unit, written e, the elementary charge. e = 1.602 × 10−19 C Basic Unit of Charge The elementary charge. e = 1.602 × 10−19 C Any charge must be q = ne , n∈Z The charge of an electron is −e and the proton has a charge +e. Question Initially, sphere A has a charge of −50e and sphere B has a charge of 20e. The spheres are made of conducting material and are identical in size. If the spheres then touch, what is the resulting charge on sphere A? (A) −50e (B) −30e (C) −15e (D) 20e Question Initially, sphere A has a charge of −50e and sphere B has a charge of 20e. The spheres are made of conducting material and are identical in size. If the spheres then touch, what is the resulting charge on sphere A? (A) −50e (B) −30e (C) −15e (D) 20e ← Conservation of Charge Charge can move from one body to another but the net charge of an isolated system never changes. This is called charge conservation. Conservation of Charge Charge can move from one body to another but the net charge of an isolated system never changes. This is called charge conservation. What other quantities are conserved? Conservation of Charge One interesting phenomenon that shows the conservation of charge is pair production. A gamma ray (very high energy photon) converts into an electron and a positron (anti-electron): γ → e− + e+ New mass is created out of light, but charge is still conserved! Current Current is the the rate of flow of charge. Current is written with the symbol I or i. i= (If you like calculus, use i = dq dt .) ∆q ∆t Coulombs and Ampères The unit for current is the Ampère, or more commonly, “Amp”. Using the definition for current, 1 A = 1 C / 1 s. Therefore, we can formally define the unit of charge in terms of the unit of current: 1 C = (1 A)(1 s) Question 574 pg 574, #4 CHAPTER 21 ** View All Solutions Her ELECTRIC CHARGE 4 Figure 21-15 shows two charged –3q –q particles on an axis. The charges are free to move. However, a third Fig. 21-15 Question 4. charged particle can be placed at a certain point such that all three particles are then in equilibrium. (a) Is that point to the left of the first two particles, to their right, or between them? (b) Should the third particle be positively or negatively charged? (c) Is the equilibrium stable or unstable? 5 In Fig. 21-16, a central particle of charge !q is surrounded by two circular rings of charged particles. What are the magnitude and direction of the net electrostatic force on the central particle due to the other parti- +4q –2q +2q –2q +q –7q r –7q Rank elect first. 9 F charg cles o the m an el Are Is the great two f cel? that Question R/2 +8Q particles. d e Question 9. 10 In Fig. 21-20, a central particle of charge !2q is surrounded by a square array of charged particles, separated by either distance d or d/2 along the perimeter of the square. What are the magnitude and direction of the net electrostatic force on the central particle due to the other particles? (Hint: Consideration of symmetry can greatly reduce the amount of work required here.) +2q –7q +3q d –3q +4q +4q –3q –5q e e (4) Fig. 21-19 pg 574, #10 (c) p (3) –2q –5q –7q +2q Of the charge Q initially on a tiny sphere, a portion cles 1 and 2 ha L23 " L12, what be treated as particles. For what value of q/Q will the electrostatic ••8 In Fig. 21 conducting sph the following ch sphere B, #6Q Spheres A and with a centerthat is much lar Two experimen touched to sphe removed. In ex procedure is re (separately) to s the electrostatic that at the end o •1 SSM ILW Question q is to be transferred to a second, nearby sphere. Both spheres can pg 575, #2 force between the two spheres be maximized? •2 Identical isolated conducting spheres 1 and 2 have equal charges and are separated by a distance that is large compared with their diameters (Fig. 21-21a). The electrostatic force acting on : sphere 2 due to sphere 1 is F . Suppose now that a third identical sphere 3, having an insulating handle and initially neutral, is touched first to sphere 1 (Fig. 21-21b), then to sphere 2 (Fig. 21-21c), and finally removed (Fig. 21-21d). The electrostatic force that now acts on sphere 2 has magnitude F!.What is the ratio F!/F? –F 1 2 F 1 3 (a ) 1 (b ) 2 3 (c ) 2 –F' 1 2 (d ) F' ••9 SSM WW place, attract ea their center-toconnected by a the spheres rep Of the initial c with a positive n the negative c and (b) the p other? Forces at a Fundamental Level Often people think about two kinds of forces: contact forces and field forces (ie. forces that act at a distance). In mechanics problems, all forces except gravity are from direct contact. Gravity is a field force. The electric and magnetic forces are also field forces. Forces at a Fundamental Level Often people think about two kinds of forces: contact forces and field forces (ie. forces that act at a distance). In mechanics problems, all forces except gravity are from direct contact. Gravity is a field force. The electric and magnetic forces are also field forces. And actually, at a fundamental level, all forces that we know of are field forces. Forces at a Fundamental Level Contact forces are a result of electrostatic repulsion at very small scales. Forces at a Fundamental Level Contact forces are a result of electrostatic repulsion at very small scales. Fundamental forces: Force Gravitational Electromagnetic Weak Nuclear Strong Nuclear ∼ Rel. strength 10−38 10−2 10−13 1 Range (m) ∞ ∞ < 10−18 < 10−15 Attract/Repel attractive attr. & rep. attr. & rep. attr. & rep. Carrier graviton photon W +, W −, Z 0 gluons Forces at a Fundamental Level Contact forces are a result of electrostatic repulsion at very small scales. Fundamental forces: Force Gravitational Electromagnetic Weak Nuclear Strong Nuclear ∼ Rel. strength 10−38 10−2 10−13 1 Range (m) ∞ ∞ < 10−18 < 10−15 Attract/Repel attractive attr. & rep. attr. & rep. attr. & rep. Carrier graviton photon W +, W −, Z 0 gluons Forces at a Fundamental Level Contact forces are a result of electrostatic repulsion at very small scales. Fundamental forces: Force Gravitational Electromagnetic Weak Nuclear Strong Nuclear ∼ Rel. strength 10−38 10−2 10−13 1 Range (m) ∞ ∞ < 10−18 < 10−15 Attract/Repel attractive attr. & rep. attr. & rep. attr. & rep. Carrier graviton photon W +, W −, Z 0 gluons Gravity is actually quite a weak force, but it is the only one that (typically) matters on large scales - charges cancel out! Fields field A field is any kind of physical quantity that has values specified at every point in space and time. Fields In EM we have vector fields. The electrostatic force is mediated by a vector field. vector field A field is any kind of physical quantity that has values specified as vectors at every point in space and time. Fields Fields were first introduced as a calculation tool. A force-field can be used to identify the force a particular particle will feel at a certain point in space and time based on the other objects in its environment that it will interact with. Imagine a charge q0 . We want to know the force it would feel if we put it at a specific location. Fields Fields were first introduced as a calculation tool. A force-field can be used to identify the force a particular particle will feel at a certain point in space and time based on the other objects in its environment that it will interact with. Imagine a charge q0 . We want to know the force it would feel if we put it at a specific location. The electric field E at that point will tell us that! F = q0 E Fields The source of the field could be another charge. We do not need a description of the sources of the field to describe what their effect is on our particle. All we need to know if the field! Fields The source of the field could be another charge. We do not need a description of the sources of the field to describe what their effect is on our particle. All we need to know if the field! This is also true for gravity. We do not need the mass of the Earth to know something’s weight: FG = m0 g FE = q0 E 22 The rod sets up an electric field, which can create a force on the test charge. Force from a Field F + + + + Test charge q0 + + at point P + + + + + + Charged E 22-1 P W H AT I S The+ physics + + a +particle 1 of c + charge #q + +2. A nagging q + + ence +of particle 2? Tha + object 2 push on particle 1 — ho (b) One purpose of phy (a) (a) A positive test charge F = q0 EFig. 22-1 magnitude and direction The rod sets up an q0 placed at point P near a charged ob: explanation of the w but also: electrostatic force F acts on electric field, which ject. Andeeper : test charge. (b)aThe electric explanati field E at such deeper can create a force point P produced by the charged object F tance. We can answer th E E= on the test charge. q0 fieldTable in the22-1 space surroun space, the particle “know Fields P Halliday, Resnick, Walker.Some Electric 1 Figure from + Electric field on force at point P Field Lines Fields are drawn with lines showing the direction of force that a test particle will feel at that point. The density of the lines at that PA R T 3 point in the diagram indicates the approximate magnitude of the 581 22-3 ELECTRIC FIELD LINES force at that point. ne the electric field of a charged charge. The field at point P in st charge of Fig. 22-1a was put the presence of the test charge arged object, and thus does not – – – – – – – F + Positive test charge e interaction between charged ctric field produced by a given ce that a given field exerts on a Sections 22-4 through 22-7 for ond task in Sections 22-8 and oint charges in an electric field. ic fields. ctric fields in the 19th century, ed with lines of force. Although now usually called electric field erns in electric fields. – (a) – – – – – – – – E Electric field lines (b) Field Lines The electrostatic cally symmetric electric duced bysomething a point charge like: field by an electric dipole system looks An caused electric field produced by an electric dipole q ! c Notice that the lines point outward from a positive charge and lue lines are intersections of these surfaces with the page) and elecinward toward a negative charge. s are perpendicular to the electric field lines at every point. 1 Figure from Serway & Jewett Field Lines C Compare the electrostatic fields for two like charges and two FIELDS opposite charges: ive . ng ge. c- c it t. tric field it represents are said to have rotational symmetry about that axis.The electric field vector at one point is shown; note that it is tangent to the field line through that point. fi n n positive test charge at any point near then of Fig. 22-3a, the net electrostatic force acti s the test charge would be perpendicular t s sheet, because forces acting in all other fi + tions would cancel +one another as a res the symmetry. Moreover, the net force o E test charge would point away from the sh E 2 shown. Thus, the electric field vector at any – + in the space on either side of the sheet iT perpendicular to the sheet and directeda from it (Figs. 22-3b and c). Because the chaF uniformly distributed along the sheet, a Fig.Such 22-5 an Field lines forfield, a positive point field vectors have the same magnitude. electric with the same and a nearby negative point charge nitude and direction at every point, ischarge a uniform electric field. T that are equal in magnitude.The charges atOf course, no real nonconducting sheet (such as a flat expanse of plastic) Gravity figure from http://www.launc.tased.edu.au ; Charge from Halliday, Resnick, Walker 1 Page 582 qual positive ach other. gative -dimenlly rotate axis passing of the page. d the elecave rotahe electric note that it that point. Compare the fields for gravity in an Earth-Sun system and electrostatic repulsion of two charges: positive test charge of Fig. 22-3a, the net the test charge wo sheet, because forc + tions would cancel the symmetry. Mor E test charge would p shown. Thus, the ele + in the space on eit perpendicular to t from it (Figs. 22-3b uniformly distribut field vectors have the same magnitude. Such an elect nitude and direction at every point, is a uniform elect LECTRIC FIELDS 14:16 Field Lines field lines with distance from the sphere tells us that the magnic field decreases with distance from the sphere. of Fig. 22-2 were of uniform positive charge, the electric field ints near the sphere would be directed radially away from he electric field lines would also extend radially away from the ave the following rule: Field Lines es extend away from positive charge (where they originate) and harge (where they terminate). Imagine an infinite sheet of charge. The lines point outward from the positively charged sheet. shows part of an infinitely large, nonconducting sheet (or plane) stribution of positive charge on one side. If we were to place a electrostatic force harge near a very g sheet with uniositive charge on : ctric field vector E test charge, and s in the space eld lines extend vely charged of (b). + + + + + + + + + + + + + + + + F + + + + + (a) 1 Positive test charge + + + + + + + + + + + + (b) Figure from Halliday, Resnick, Walker. E + + + + + + + + + (c) Field from a Point Charge We want an expression for the electric field from a point charge, q. Using Coulomb’s Law the force on the test particle is 0 F→0 = k rqq 2 r̂. F 1 k q q 0 E= = r̂ 2 q0 q r 0 The field at a displacement r from a charge q is: E= kq r̂ r2 charge and a nearby negative point charge that are equal in magnitude.The charges attract each other.The pattern of field lines and the electric field it represents have rotational symmetry about axis passing through both displacement r an from a charge q is: charges in the plane of the page.The electric field vector at one pointkisqshown; the vector is tangent to the E field the point. r̂ =line through Field from a Point Charge The field at a r2 This is a vector field: + The electric field vectors at various points around a positive point charge. Fig. 22-6 : The direction of F is d toward the point charg : The direction of E is directly away from th Because there is gives the field at eve point charge is shown We can quickly fin charge. If we place a p from Eq. 21-7, the net f Therefore, from Eq. 2 : Here Ei is the electri Equation 22-4 shows fields as well as to ele Charges and Conductors Excess charge sits on the outside surface of a conductor. The electric field lines are perpendicular to the surface. 1 Figure from OpenStax College Physics. ormly Conductors and Electric fields charge reachConsider a neutral conductor placed in an electric field: d, proushes, arges, ch cirwhere + + +++ –– – – –– ++ – made –– + 20, all er the – Fig. 24-20 – – E=0 – – – + + + + + ++ An uncharged conduc- Conductors and Electric fields Electric fields exert forces on free charges in conductors. Each charge keeps moving until: 1 the charges reaches the edge of the conductor and can move no further OR 2 the field is cancelled out! Inside a conducting object, the electric field is zero! 12 E (kV/m) om the curve of Fig. 24-18a by differentiating with respect to r, Faraday recall that theCages derivative of any constant is zero). The curve of A conducting can of shield interior from even very be derived from theshell curves Fig. the 24-18b by integrating withstrong electric fields. g Eq. 24-19. 8 4 0 Fig. 24-1 rge spark dy and then ross the t tire (note ving the pered. (Courtesy ric 1 Photo from Halliday, Resnick, Walker inside and cal shell o E(r) for t Faraday Cages 1 Photo found on TheDailySheeple, credits unknown. ) Charges Inside Conductors: The Faraday Ice Pail Gaussian surface _ _ + + _ + + _+ + _ + + _ + _ + _ _ + _+ +_ + _ (b ) _ _ ch ou ge un sk no ut tiv ap th Summary • Coulomb’s law • Quantization of charge • Charge conservation • Current • electric field • field of a point charge Homework worksheets: physicsclassroom.com/getattachment/curriculum/estatics/... • ...static5.pdf • ...static7.pdf Halliday, Resnick, Walker: • Ch 21, onward from page 573. Questions: 1; Sec Qs: 3, 9, 27, 42 & 43