In this section we shall look at the motion of a projectile MOTION IN FIELDS 9.1 PROJECTILE MOTION PROJECTILE MOTION

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1 MOTION IN FIELDS MOTION IN FIELDS 9 9. Pojectile motion 9. Gavitational field, potential and enegy 9.3 Electic field, potential and enegy 9. PROJECTILE MOTION 9.. State the independence of the vetical and the hoizontal components of velocity fo a pojectile in a unifom field. 9.. Descibe and sketch the tajectoy of pojectile motion as paabolic in the absence of ai esistance. cliff h v h gms d 9..3 Descibe qualitatively the effect of ai esistance on the tajectoy of a pojectile Solve poblems on pojectile motion. IBO PROJECTILE MOTION Pojectile launched hoizontally In this section we shall look at the motion of a pojectile that is launched hoizontally fom a point above the suface of the Eath. Figue 90 The path of a hoizontal pojectile Since thee is no foce acting in the hoizontal diection the hoizontal velocity will emain unchanged thoughout the light of the paticle. Howeve, the vetical acceleation of the pojectile will be equal to g. We can ind the time of light t by inding the time it takes the paticle to fall a height h. To stat with, we conside only the vetical motion of the object: u0 In the Figue 90a pojectile is ied hoizontally fom a clif of height h with an initial hoizontal velocity v h h ag sh Ou poblem is efectively to ind whee it will land, with what velocity and the time of light. We shall assume that we can ignoe ai esistance and that the acceleation due to gavity g is constant. Figue 90 v v v t? The vetical motion of the object 49

2 CHAPTER 9 Time of flight his is calculated fom the deinition of acceleation i.e., using v u + at, we have that v v 0 + g t t v v --- g whee v v is the vetical velocity with which the object stikes the gound. To ind v v we use the equation v u + as, so that as the initial vetical velocity (u) is zeo, the acceleation a g and s h. hen v v 0 + g h v v g h Fom which we have that and so, the time of light is given by t g h g t h ---- g h ---- g Figue 903 Using multi-flash photogaphy his is iespective of the speed with which the paticle is ied hoizontally. he geate the hoizontal speed, the futhe this pojectile will tavel fom the base of the clif. It is also possible to show that the path of the paticle is paabolic. Since the hoizontal velocity is constant, the hoizontal distance d that the paticle tavels befoe stiking the gound is v h t. (i.e., using s ut + ½ at ut, whee in the hoizontal diection we have that a 0 and u v h constant) his gives To ind the velocity with which the paticle stikes the gound we must emembe that velocity is a vecto quantity. So, using Pythagoas theoem at the point of impact (to take into account both the vetical component of velocity and the hoizontal component of velocity) we have that the velocity has a magnitude of h d v v --- g his is the geneal solution to the poblem and it is not epected that you should emembe the fomula fo this geneal esult. You should always wok fom ist pinciples with such poblems. v h θ v v V v v + v h V An inteesting point to note is that, since thee is no hoizontal acceleation, then if you wee to dop a pojectile fom the top of the clif vetically down, at the moment that the othe pojectile is ied hoizontally, then both would each the gound at the same time. his is illustated by the copy of a multilash photogaph, as shown in Figue 903. and the diection will be given by inding θ v v ac tan --- whee the angle is quoted elative to the hoizontal. If the angle is to be given elative to the vetical then we evaluate ( 90 θ) v h 50

3 MOTION IN FIELDS o θ v h ac tan --- Notice that at impact the velocity vecto is tangential to the path of motion. As a matte of fact, the velocity vecto is always tangential to the path of motion and is made up of the hoizontal and vetical components of the velocities of the object. Pojectiles launched at an angle to the hoizontal Conside the poblem of a pojectile that is launched fom the suface of the Eath and at an angle to the suface of the Eath. Ignoe ai esistance and assume that g is constant. In Figue 905 the paticle is launched with velocity v at angle θ to the suface. v v y θ v v h v v 0, a g maimum height (H) Range v v g +ve v v T g time to impact so that y s ut + --at v v t -- ( g )t + y ( vsinθ)t --g t If we now substitute fo into this equation we get t vcos θ y ( v sinθ) g v cosθ vcos θ sinθ cosθ -g v cosθ tanθ - g v - sec θ his is the geneal equation of the motion of the pojectile that elates the vetical and hoizontal distances. his equation is plotted below fo a pojectile that is launched with an initial speed of 0 m s at 60 to the hoizontal. he path followed by the pojectile is a paabola. y 0 0 v 0 v 0 0 0sin Figue 905 Pojectile launched at an angle 60 0 cos he vetical component of the velocity, v v, is v v vsinθ he hoizontal component of the velocity, v h, is Figue 906 The paabolic path v h vcos θ As in the case of the pojectile launched hoizontally, thee is no acceleation in the hoizontal diection and the acceleation in the vetical diection is g. If we efe the motion of the pojectile to a Catesian coodinate system, then ate a time t, the hoizontal distance tavelled will be given by v h t ( vcosθ)t and the vetical distance can be found by using the equation he maimum height H that the pojectile eaches can be found fom the equation v u + as whee u is the initial vetical component of the velocity and v the inal (vetical component) of the velocity at the highest point, whee at this point, the vetical component is zeo. So that, 0 ( vsinθ) + g H H v sin θ g If we use the igues in the eample above (v 0 m s, θ 60 ) ( 0 sin60 ) with g 0 m s then we see that H

4 CHAPTER 9 i.e., the object eaches a maimum height of 5 m. he time T to each the maimum height is found using v u + a t, such v 0, u v sin θ and a g, to give 0 v sin θ g T g T v sin θ Hence, T v sinθ g Fo the eample above the value of T is.73 s. his means (using symmety) that the pojectile will stike the gound 3.46 s ate the launch. he hoizontal ange R is given by R ( v cos θ) T which fo the eample gives R 34.6 m. Epeiment shows that both the hoizontal and vetical dag foces depend on the speed of the pojectile. he efect of the hoizontal dag will be to foeshoten the ange of the pojectile and the efect of the vetical dag will be to educe the maimum height eached by the pojectile. Howeve, the pesence o ai esistance also means that the mass of the pojectile will now afect the path followed by the pojectile. In the absence of ai esistance thee is no acceleation in the hoizontal diection and the acceleation in the vetical diection is g, the acceleation of fee fall. With ai esistance pesent, to ind the hoizontal (a H )and vetical (a V ) acceleations we have to apply Newton s second law to both the diections. If we let the hoizontal dag equal kv H and the vetical dag equal Kv V whee k and K ae constants and v H and v V ae the hoizontal and vetical speeds espectively at any instant, then we can wite (We could also ind the time fo the pojectile to stike the gound by putting y 0 in the equation y ( vsinθ)t --g t Although we have established a geneal solution, essentially solving pojectile poblems, emembe that the hoizontal velocity does not change and that when using the equations of unifom motion you must use the component values of the espective velocities. Do not ty to emembe the fomulae. kv H mah and mg KvV mav Fom this we can now see why the mass afects the path since both a H and a V depend on height. (Fo those of you doing HL maths, you will ealise that the above equations can be witten as difeential equations but inding thei solution is no easy matte!) We have hee, anothe eample of the Newton method fo solving the geneal mechanics poblem- know the foces acting at a paticula instant and you can in pinciple pedict the futue behaviou of the system. The effect of ai esistance on pojectile motion We have seen that in the absence of ai esistance, the path followed by a pojectile is a paabola and that the path depends only on the initial speed and angle of pojection. Of couse, in the eal wold all pojectiles ae subject to ai esistance. Fig 907 shows the fee body foce diagam fo a pojectile subject to ai esistance SOLVE PROJECTILE PROBLEMS Notice that at impact the velocity vecto is tangential to the path of motion. As a matte of fact, the velocity vecto is always tangential to the path of motion and is made up of the hoizontal and vetical components of the velocities of the object. Eample vetical dag A paticle is ied hoizontally with a speed of 5 ms - fom the top of a vetical clif of height 80 m. Detemine hoizontal dag (a) the time of light weight (b) he distance fom the base of the clif whee it stikes the gound Figue 907 The effect of ai esistance (c) the velocity with which it stikes the gound 5

5 MOTION IN FIELDS Solution he magnitude of this velocity is m s 80 m Vetical: u 0 a g cliff Hoizontal: u 5 m s - a 0 +ve and it makes an angle to the hoizontal of θ ac tan (o to the vetical of 3 ). (a) he vetical velocity with which it stikes the gound can be found using the equation v u + as, with u 0, a g and s 80 ( h). his then gives v v g h hat is, the vetical velocity at impact is 40 m s. Consevation of enegy and pojectile poblems In some situations the use of consevation of enegy can be a much simple method than using the kinematics equations. Solving pojectile motion poblems makes use of the fact that E k + E p constant at evey point in the object s light (assuming no loss of enegy due to fiction). he time to stike the gound can be found using v u + at, with u 0, a g and v v v. So that, In Figue 909, using the consevation of enegy pinciple we have that the t v v g 0 Total enegy at A Total enegy at B Total enegy at C (b) hat is, 4 seconds. he distance tavelled fom the base of the clif using s ut + --at, with u 5, i.e., --mv A --mv B +mgh -mv C +mgh Notice that at A, the potential enegy is set at zeo (h 0). (c) a 0 and t 4 is given by s hat is, the ange is 00 m. he velocity with which it stikes the gound is given by the esultant of the vetical and hoizontal velocities as shown. v v -mv A + 0 A θ v A v h Figue 909 -mv B + mg H B H v B Enegy poblem C h --mv C + mg h v C Gound level θ v h 5 v v 40 V 53

6 CHAPTER 9 Eample A ball is pojected at 50 ms - at an angle of 40 o above the hoizontal. he ball is eleased.00 m above gound level. Taking g 0 m s -, detemine (a) (b) the maimum height eached by the ball the speed of the ball as it hits the gound Solution B H H 53.6 hat is, the maimum height eached is 53.6 m. b. At C, the total enegy is given by E k + E p -mv C + mg 0 Using the total enegy at A, E k Equating, we have that -mv C + E p 70 m 70 m --mv C v C 540 A m H R ange C v C hat is, the ball hits the gound with a speed of 50.4 m s. Eecise a. he total enegy at A is given by E k + E p -m( 50.0 ) + mg m + 0m 70 m Net, to ind the total enegy at B we need to ist detemine the speed at B, which is given by the hoizontal component of the speed at A. Hoizontal component: 50.0 cos m s. heefoe, we have that E k + E p Equating, we have -m( 38.3 ) + mg H y 5t 5t m + 0 mh. A pojectile is ied fom the edge of a vetical clif with a speed of 30 m s at an angle of 30 to the hoizontal. he height of the clif above the suface of the sea is 00 m. (a) If g 0 m s and ai esistance is ignoed show that at any time t ate the launch the vetical displacement y of the pojectile as measued fom the top of the clif is given by: y 5t - 5t Hence show that the pojectile will hit the suface of the sea about 6 s ate it is launched. (b) (c) Suggest the signiicance of the negative value of t that can be obtained in solving the equation? Detemine the maimum height eached by the pojectile and the hoizontal distance to whee it stikes the sea as measued fom the base of the clif. 70 m m + 0 mh 54

7 MOTION IN FIELDS 9. GRAVITATIONAL FIELD, POTENTIAL AND ENERGY 9.. Define gavitational potential and gavitational potential enegy. 9.. State and apply the epession fo gavitational potential due to a point mass State and apply the fomula elating gavitational field stength to gavitational potential gadient. IBO GRAVITATIONAL POTENTIAL We have seen that if we lit an object of mass m to a height h above the suface of the Eath then its gain in gavitational potential enegy is mgh. Howeve, this is by no means the full stoy. Fo a stat we now know that g vaies with h and also the epession eally gives a difeence in potential enegy between the value that the object has at the Eath s suface and the value that it has at height h. So what we eally need is a zeo point. Can we ind a point whee the potential enegy is zeo and use this point fom which to measue changes in potential enegy? Well, the point that is chosen is in fact ininity. At ininity the gavitational ield stength of any object will in fact be zeo. So let us see if we can deduce an epession fo the gain in potential enegy of an object when it is lited fom the suface of the Eath to ininity. his in efect means inding the wok necessay to pefom this task. M e Figue 9 δ m g A + δ B Gavitational foces If the paticle is moved to B, then since δ is vey small, we can assume that the ield emains constant ove the distance AB. he wok δw done against the gavitational ield of the Eath in moving the distance AB is δw GM e m δ (emembe that wok done against a foce is negative) To ind the total wok done, W, in going fom the suface of the Eath to ininity we have to add all these little bits of wok. his is done mathematically by using integal calculus. G M e m W d G M e m ---- d G M e m -- R R R GM e m 0 -- R G M e m R Hence we have, whee R is the adius of the Eath, that the wok done by the gavitational ield in moving an object of mass m fom R (suface of the Eath) to ininity, is given by W GM e m R We can genealise the esult by calculating the wok necessay pe unit mass to take a small mass fom the suface of the Eath to ininity. his we call the gavitational potential, V, i.e., We would get eactly the same esult if we calculated the wok done to bing the point mass fom ininity to the suface of Eath. In this espect the fomal deinition of gavitational potential at a point in a gavitational ield is theefoe deined as the wok done pe unit mass in binging a point mass fom ininity to that point. In the diagam we conside the wok necessay to move the paticle of mass m a distance δ in the gavitational ield of the Eath. he foce on the paticle at A is F G M e m Clealy then, the gavitational potential at any point in the Eath s ield distance fom the cente of the Eath (poviding > R) is GM V e

8 CHAPTER 9 he potential is theefoe a measue of the amount of wok that has to be done to move paticles between points in a gavitational ield and its unit is the J kg. We also note that the potential is negative so that the potential enegy as we move away fom the Eath s suface inceases until it eaches the value of zeo at ininity. If the gavitational ield is due to a point mass of mass m, then we have the same epession as above ecept that M e is eplaced by m and must also eclude the value of the potential at the point mass itself i.e. at 0. We can epess the gavitational potential due to the Eath (o due to any spheical mass) in tems of the gavitational ield stength at its suface GRAVITATIONAL POTENTIAL GRADIENT Let us conside now a egion in space whee the gavitational ield is constant. In Figue 9 the two points A and B ae sepaated by the distance. A Figue 9 B diection of unifom gavitational field of stength I The gavitational potential gadient At the suface of the Eath we have GM R e g 0 R e So that, g 0 R e GM he gavitational ield is of stength I and is in the diection shown. he gavitational potential at A is V and at B is V + V. he wok done is taking a point mass m fom A to B is F mi. Howeve, by deinition this wok is also equal to -m V. Hence at a distance fom the cente of the Eath the gavitational potential V can be witten as GM e g 0 R e V he potential at the suface of the Eath ( R e ) is theefoe -g 0 R e It is inteesting to see how the epession fo the gavitational potential ties in with the epession mgh. he potential at the suface of the Eath is -g 0 R e (see the eample above) and at a height h will be g 0 ( R e + h) if we assume that g 0 does not change ove the distance h. he difeence in potential between the suface and the height h is theefoe g 0 h. So the wok needed to aise an object of mass m to a height h is mgh, i.e., m difeence in gavitational potential his we have efeed to as the gain in gavitational potential enegy (see.3.5). Howeve, this epession can be etended to any two points in any gavitational ield such that if an object of mass m moves between two points whose potentials ae V and V espectively, then the change in gavitational potential enegy of the object is m(v V ). heefoe mi -m V o V I Efectively this says that the magnitude of the gavitational ield stength is equal to the negative gadient of the potential. If I constant then V is a linea function of and I is equal to the negative gadient of the staight line gaph fomed by plotted V against. If I is not constant (as usually the case), then the magnitude of I at any point in the ield can be found by ind the gadient of the V- gaph at that point. An eample of such a calculation can be found in Section Fo those of you who do HL maths the elationship between ield and potential is seen to follow fom the epession fo the potential of a point mass viz: m V G dv m G I d + 56

9 MOTION IN FIELDS 9..4 Detemine the potential due to one o moe point masses. Figue 93 shows the ield lines and equipotentials fo two point masses m Descibe and sketch the patten of equipotential sufaces due to one and two point masses State the elation between equipotential sufaces and gavitational field lines. m m 9..7 Eplain the concept of escape speed fom a planet Deive an epession fo the escape speed of an object fom the suface of a planet. IBO POTENTIAL DUE TO ONE OR Figue 93 Equipotentials fo two point masses It is woth noting that we would get eactly the same patten if we wee to eplace the point masses with two equal point chages. (See 9.3.5) MORE POINT MASSES Gavitational potential is a scala quantity so calculating the potential due to moe than one point mass is a matte of simple addition. So fo eample, the potential V due to the Moon and Eath and a distance fom the cente of Eath is given by the epession M M + E M V G ESCAPE SPEED he potential at the suface of Eath is which means that the enegy equied to take a paticle of mass m fom the suface to ininity is equal to But what does it actually mean to take something to ininity? When the paticle is on the suface of the Eath we can think of it as sitting at the bottom of a potential well as in igue 94. whee M E mass of Eath, M M mass of Moon and distance between cente of Eath and Moon. infinity EQUIPOTENTIALS AND FIELD LINES If the gavitational potential has the same value at all points on a suface, the suface is said to be an equipotential suface. So fo eample, if we imagine a spheical shell about Eath whose cente coincides with the cente of Eath, this shell will be an equipotential suface. Clealy, if we epesent the gavitational ield stength by ield lines, since the lines adiate out fom the cente of Eath, then these lines will be at ight angles to the suface If the ield lines wee not nomal to the equipotential suface then thee would be a component of the ield paallel to the suface. his would mean that points on the suface would be at difeent potentials and so it would no longe be an equipotential suface! his of couse holds tue fo any equipotential suface. GM R FIgue 94 satellite A potential well sufaceof Eath he depth of the well is and if the satellite gains an amount of kinetic enegy equal to whee m is its mass then it will have just enough enegy to lit it out of the well. In eality it doesn t actually go to ininity it just means that the ocket is efectively fee of the gavitational attaction of the Eath. We say that it has escaped the Eath s gavitational pull. We meet this idea in connection with molecula foces. Two molecules in a solid will sit at thei 57

10 CHAPTER 9 equilibium position, the sepaation whee the epulsive foce is equal to the attactive foce. If we supply just enough enegy to incease the sepaation of the molecules such that they ae an ininite distance apat then the molecules ae no longe afected by intemolecula foces and the solid will have become a liquid. hee is no incease in the kinetic enegy of the molecules and so the solid melts at constant tempeatue. We can calculate the escape speed of a satellite vey easily by equating the kinetic enegy to the potential enegy such that --mv GM e m esc ape R e v escape GM e g R 0 R e e Substituting fo g 0 and R e gives a value fo v escape of about km s fo the Eath. Detemine also the gavitational ield stength at a distance of m above the suface of Mas. Solution 3 M V G N kg - R But V -g 0 R heefoe g 7 V m s - R To detemine the ield stength g h at m above the M suface, we use the fact that g 0 G such that GM g R 0 R heefoe g GM g R 3.8 (3.4) 0.4 m s - 0 h Rh Rh (0.) (the distance fom the cente is m) You will note that the escape speed does not depend on the mass of the satellite since both kinetic enegy and potential enegy ae popotional to the mass. In theoy if you want to get a ocket to the moon it can be done without eaching the escape speed. Howeve this would necessitate an enomous amount of fuel and it is likely that the ocket plus fuel would be so heavy that it would neve get of the gound. It is much moe pactical to acceleate the ocket to the escape speed and then in theoy just point it at the Moon to whee it will now coast at constant speed SOLVE PROBLEMS INVOLVING Eample GRAVITATIONAL POTENTIAL ENERGY AND GRAVITATIONAL POTENTIAL Use the following data to detemine the potential at the suface of Mas and the magnitude of the acceleation of fee fall mass of Mas adius of Mas kg m Eecise. he gaph below shows how the gavitational potential outside of the Eath vaies with distance fom the cente. V /Jkg /m 0 6 (a) Use the gaph to detemine the gain in gavitational potential enegy of a satellite of mass 00 kg as it moves fom the suface of the Eath to a height of m above the Eath s suface. Answe: 0 0 J (b) Calculate the enegy equied to take it to ininity? Answe; J 58

11 MOTION IN FIELDS (c) Detemine the slope of the gaph at the suface of the Eath, m, above the suface of the Eath? Comment on you answes. Answe: 0, and equals the gavitational ield stength at the suface of Eath (acceleation of fee fall) 9.3 ELECTRIC FIELD, POTENTIAL AND ENERGY Let us ist look at a case of two positive point chages each of μc that ae initially bound togethe by a thead in a vacuum in space with a distance between them of 0 cm as shown in Figue 96. When the thead is cut, the point chages, initially at est would move in opposite diections, moving with velocities v and v along the diection of the electostatic foce of epulsion. BEFORE Figue 96 v v AFTER Inteaction of two positive paticles 9.3. Define electic potential and electic potential enegy State and apply the epession fo electic potential due to a point chage State and apply the fomula elating electic field stength to electical potential gadient Detemine the potential due to one o moe point chages Descibe and sketch the patten of equipotential sufaces due to one and two point chages State the elation between equipotential sufaces and electic field lines Solve poblems involving electic potential enegy and electic potential. IBO DEFINING ELECTRIC POTENTIAL & ELECTRIC POTENTIAL ENERGY he electic potential enegy between two point chages can be found by simply adding up the enegy associated with each pai of point chages. Fo a pai of inteacting chages, the electic potential enegy is given by: U Ep +E k W F kqq / kqq Because no etenal foce is acting on the system, the enegy and momentum must be conseved. Initially, E k 0 and E p k qq / / 0. m 0.09 J. When they ae a geat distance fom each othe, E p will be negligible. he inal enegy will be equal to ½ mv + ½ mv 0.09 J. Momentum is also conseved and the velocities would be the same magnitude but in opposite diections. Electic potential enegy is moe oten deined in tems of a point chage moving in an electic ield as: the electic potential enegy between any two points in an electic ield is deined as negative of the wok done by an electic ield in moving a point electic chage between two locations in the electic ield. U Ep -W -Fd qe whee is the distance moved along (o opposite to) the diection of the electic ield. Electic potential enegy he concept of electic potential enegy was developed with gavitational potential enegy in mind. Just as an object nea the suface of the Eath has potential enegy because of its gavitational inteaction with the Eath, so too thee is electical potential enegy associated with inteacting chages. Electic potential enegy is measued in joule (J). Just as wok is a scala quantity, so too electical potential enegy is a scala quantity. he negative of the wok done by an electic ield in moving a unit electic chage between two points is independent of the path taken. In physics, we say the electic ield is a consevative ield. Suppose an etenal foce such as you hand moves a small positive point test chage in the diection of a unifom electic ield. As it is moving it must be gaining kinetic 59

12 CHAPTER 9 enegy. If this occus, then the electic potential enegy of the unit chage is changing. In Figue 97 a point chage +q is moved between points A and B though a distance in a unifom electic ield. +q B A Figue 97 Movement of a positive point chage in a unifom field In ode to move a positive point chage fom point A to point B, an etenal foce must be applied to the chage equal to qe (F qe). Since the foce is applied though a distance, then negative wok has to be done to move the chage because enegy is gained, meaning thee is an incease electic potential enegy between the two points. Remembe that the wok done is equivalent to the enegy gained o lost in moving the chage though the electic ield. he concept of electic potential enegy is only meaningful if the electic ield which geneates the foce in question is consevative. Figue 98 W F E q cosθ θ Chage moved at an angle to the field If a chage moves at an angle θ to an electic ield, the component of the displacement paallel to the electic ield is used as shown in Figue 98 W F Eq cos θ he electic potential enegy is stoed in the electic ield, and the electic ield will etun the enegy to the point chage when equied so as not to violate the Law of consevation of enegy. Electic potential he electic potential at a point in an electic ield is deined as being the wok done pe unit chage in binging a small positive point chage fom ininity to that point. ΔV V V f -W /q If we designate the potential enegy to be zeo at ininity then it follows that electic potential must also be zeo at ininity and the electic potential at any point in an elctic ield will be: ΔV -W / q Now suppose we apply an etenal foce to a small positive test chage as it is moved towads an isolated positive chage. he etenal foce must do wok on the positive test chage to move it towads the isolated positive chage and the wok must be positive while the wok done by the electic ield must theefoe be negative. So the electic potential at that point must be positive accoding to the above equation. If a negative isolated chage is used, the electic potential at a point on the positive test chage would be negative. Positive point chages of thei own accod, move fom a place of high electic potential to a place of low electic potential. Negative point chages move the othe way, fom low potential to high potential. In moving fom point A to point B in the diagam, the positive chage +q is moving fom a low electic potential to a high electic potential. he electic potential is theefoe difeent at both points. In the deinition given, the tem wok pe unit chage has signiicance. If the test chage is C whee the chage has a potential enegy of J, then the potential enegy would be J / C 00 JC -. Now if the chage was doubled, the potential enegy would become J. Howeve, the potential enegy pe unit chage would be the same. Electic potential is a scala quantity and it has units JC - o volts whee volt equals one joule pe coloumb. he volt allows us to adopt a unit fo the electic ield in tems of the volt. Peviously, the unit fo the electic ield was NC -. W qv and F qe, 60 so W / V F / E E FV / W NV / Nm Vm -.

13 MOTION IN FIELDS he wok done pe unit chage in moving a point chage between two points in an electic ield is again independant of the path taken ELECTRIC POTENTIAL DUE TO A POINT CHARGE Let us take a point metes fom a chaged object. he potential at this point can be calculated using the following: W -F -qv and F q q 4 π ε 0 heefoe, ELECTRIC FIELD STRENGTH AND ELECTRIC POTENTIAL GRADIENT Let us look back at Figue 97. Suppose again that the chage +q is moved a small distance by a foce F fom A to B so that the foce can be consideed constant. he wok done is given by: W F he foce F and the electic ield E ae oppositely diected, and we know that: F -qe and ΔW q ΔV heefoe, the wok done can be given as: W hat is q q q πε 0 q q q 4πε q 4πε 0 V O, simply Eample V V q πε 0 kq ---- heefoe q ΔV -q E Δ E ΔV Δ he ate of change of potential ΔV at a point with espect to distance Δ in the diection in which the change is maimum is called the potential gadient. We say that the electic ield - the potential gadient and the units ae Vm -. Fom the equation we can see that in a gaph of electic potential vesus distance, the gadient of the staight line equals the electic ield stength. Detemine how much wok is done by the electic ield of point chage 5.0 μc when a chage of.00 μc is moved fom ininity to a point m fom the point chage. (Assume no acceleation of the chages). Solution The wok done by the electic ield is W -qv -/4πε 0 q (Q / - Q / ) W ( C NmC C) m J An etenal foce would have to do J of wok. In eality, if a chaged paticle entes a unifom electic ield, it will be acceleated unifomly by the ield and its kinetic enegy will incease. his is why we had to assume no acceleation in the last woked eample. E k Eample -mv q E q V -- q V Detemine how fa apat two paallel plates must be situated so that a potential difeence of.50 0 V poduces an electic ield stength of NC -. 6

14 CHAPTER 9 Solution Using E V V E he plates ae m apat..5 0 V N C he electic ield and the electic potential at a point due to an evenly distibuted chage +q on a sphee can be epesented gaphically as in Figue 99. Some futhe obsevations of the gaphs in Figue 95 ae: Outside the sphee, the gaphs obey the elationships given as E α / and V α / At the suface, 0. heefoe, the electic ield and potential have the minimum value fo at this point and this infes a maimum ield and potential. Inside the sphee, the electic ield is zeo. Inside the sphee, no wok is done to move a chage fom a point inside to the suface. heefoe, thee is no potential difeence and the potential is the same as it is when 0. 0 Chage of +Q evenly distibuted ove suface Simila gaphs can be dawn fo the electic ield intensity and the electic potential as a function of distance fom conducting paallel plates and sufaces, and these ae given in Figue 90. E V 0 E V Q 4πε 0, > Q 4πε ---, 0 > 0 E field: Potential plot E field plot 0 Figue 99 Electic field and potential due to a chaged sphee When the sphee becomes chaged, we know that the chage distibutes itself evenly ove the suface. heefoe evey pat of the mateial of the conducto is at the same potential. As the electic potential at a point is deined as being numeically equal to the wok done in binging a unit positive chage fom ininity to that point, it has a constant value in evey pat of the mateial of the conducto Since the potential is the same at all points on the conducting suface, then V / is zeo. But E V /. heefoe, the electic ield inside the conducto is zeo. hee is no electic ield inside the conducto. Figue 90 Electic field and electic potential at a distance fom a chaged suface 6

15 MOTION IN FIELDS POTENTIAL DUE TO ONE OR MORE POINT CHARGES he potential due to one point chage can be detemined by using the equation fomula V kq /. Eample Detemine the electic potential at a point m fom the cente of an isolated conducting sphee with a point chage of 4.0 pc in ai. Solution he electic potential of the + μc chage due to the 6 μc chage is: V (9 0 9 Nm C C) ( ) m V he electic potential of the + μc chage due to the +3 μc chage is: V (9 0 9 Nm C C) 3m V he net absolute potential is the sum of the potentials V V Solution Using the fomula V he absolute potential at the point is V. V kq /, we have V ( ) ( ) ( V ) the potential at the point is V. he potential due to a numbe of point chages can be detemined by adding up the potentials due to individual point chages because the electic potential at any point outside a conducting sphee will be the same as if all the chage was concentated at its cente. Eample hee point chages of ae placed at the vetices of a ightangled tiangle as shown in the diagam below. Detemine the absolute potential of the +.0 μc chage. m - 6µC 3µC 3m +µc EQUIPOTENTIAL SURFACES Regions in space whee the electic potential of a chage distibution has a constant value ae called equipotentials. he places whee the potential is constant in thee dimensions ae called equipotential sufaces, and whee they ae constant in two dimensions they ae called equipotential lines. hey ae in some ways analogous to the contou lines on topogaphic maps. In this case, the gavitational potential enegy is constant as a mass moves aound the contou lines because the mass emains at the same elevation above the Eath s suface. he gavitational ield stength acts in a diection pependicula to a contou line. Similaly, because the electic potential on an equipotential line has the same value, no wok can be done by an electic foce when a test chage moves on an equipotential. heefoe, the electic ield cannot have a component along an equipotential, and thus it must be eveywhee pependicula to the equipotential suface o equipotential line. his fact makes it easy to plot equipotentials if the lines of foce o lines of electic lu of an electic ield ae known. Fo eample, thee ae a seies of equipotential lines between two paallel plate conductos that ae pependicula to the electic ield. hee will be a seies of concentic cicles (each cicle futhe apat than the pevious one) that map out the equipotentials aound an 63

16 CHAPTER 9 isolated positive sphee. he lines of foce and some equipotential lines fo an isolated positive sphee ae shown in Figue V 40V Lines of equipotential 30V 0V 0V Figue 95 Fquipotential lines between chaged paallel plates gavitational fields and electic fields Figue 9 Equipotentials aound an isolated positive sphee In summay, we can conclude that houghout this chapte the similaities and difeences between gavitational ields and electic ields have been discussed. he elationships that eists between gavitational and electic quantities and the efects of point masses and chages is summaised in Table 96 No wok is done to move a chage along an equipotential. Equipotentials ae always pependicula to the electic lines of foce. Figue 93 and 94 show some equipotential lines fo two oppositely chaged and indentically positive sphees sepaated by a distance. equipotential lines Quantities Gavitational quantity V g ---- E Electical quantity V ve ve Figue 93 Equipotential lines between two opposite chages Point masses and chages V G m --- V g G m --- E F G m m F Figue 96 Fomulas (table) q 4πε q -- 4πε q q πε 0 equipotential lines SOLVING PROBLEMS hee ae a numbe of woked eamples that have been given in section 9.3. Hee ae two moe eamples. Figue 94 Equipotential lines between two chages which ae the same 64

17 MOTION IN FIELDS Eample Deduce the electic potential on the suface of a gold nucleus that has a adius of 6. fm. Solution Using the fomula Total enegy ½ kqq / + - kqq / -½ kqq / Nm C - ( C) m J J ev. he ionisation enegy is 3.6 ev. Eecise 9.3 V kq /, and knowing the atomic numbe of gold is 79. We will assume the nucleus is spheical and it behaves as if it wee a point chage at its cente (elative to outside points). V Nm C C m V he potential at the point is 8 MV. Eample. A point chage P is placed midway between two identical negative chages. Which one of the following is coect with egads to electic ield and electic potential at point P? Electic ield Electic potential A non-zeo zeo B zeo non-zeo C non-zeo non-zeo D zeo zeo Deduce the ionisation enegy in electon-volts of the electon in the hydogen atom if the electon is in its gound state and it is in a cicula obit at a distance of m fom the poton. Solution his poblem is an enegy, coulombic, cicula motion question based on Boh s model of the atom (not the accepted quantum mechanics model). he ionisation enegy is the enegy equied to emove the electon fom the gound state to ininity. he electon tavels in a cicula obit and theefoe has a centipetal acceleation. he ionisation enegy will counteact the coulombic foce and the movement of the electon will be in the opposite diection to the centipetal foce Total enegy E k electon + E p due to the poton-electon inteaction ΣF kqq / mv / and as such mv kqq /. heefoe, E k electon ½ kqq /. E p due to the poton-electon inteaction - kqq /.. Two positive chaged sphees ae tied togethe in a vacuum somewhee in space whee thee ae no etenal foces. A has a mass of 5 g and a chage of.0 μc and B has a mass of 5 g and a chage of 3.0 μc. he distance between them is 4.0 cm. hey ae then eleased as shown in the diagam. A BEFORE (a) (b) B v v AFTER Detemine thei initial electic potential enegy in the befoe situation. Detemine the speed of sphee B ate elease. 3. he diagam below epesents two equipotential lines in sepaated by a distance of 5 cm in a unifom electic ield. 40V 0V cm Detemine the stength of the electic ield. 65

18 CHAPTER 9 4. his question is about the electic ield due to a chaged sphee and the motion of electons in that ield. he diagam below shows an isolated, metal sphee in a vacuum that caies a negative electic chage of 6.0 μc. (a) (b) (c) On the diagam daw the conventional way to epesent the electic ield patten due to the chaged sphee and lines to epesent thee equipotential sufaces in the egion outside the sphee. Eplain how the lines epesenting the equipotential sufaces that you have sketched indicate that the stength of the electic ield is deceasing with distance fom the cente of the sphee. he electic ield stength at the suface of the sphee and at points outside the sphee can be detemined by assuming that the sphee acts as a point chage of magnitude 6.0 μc at its cente. he adius of the sphee is.5 0 m. Deduce that the magnitude of the ield stength at the suface of the sphee is Vm. 8. he gap between two paallel plates is m, and thee is a potential difeence of V between the plates. Calculate i. the wok done by an electon in moving fom one plate to the othe ii. the speed with which the electon eaches the second plate if eleased fom est. iii. the electic ield intensity between the plates. 9. An electon gun in a pictue tube is acceleated by a potential V. Detemine the kinetic enegy gained by the electon in electon-volts. 0. Detemine the electic potential m fom a chage of C.. Detemine the electic potential at a point midway between a chage of 0 pc and anothe of + 5 pc on the line joining thei centes if the chages ae 0 cm apat.. Duing a thundestom the electic potential difeence between a cloud and the gound is V. Detemine the magnitude of the change in electic potential enegy of an electon that moves between these points in electon-volts. An electon is initially at est on the suface of the sphee. (d) (i) Descibe the path followed by the electon as it leaves the suface of the sphee. (ii) Calculate the initial acceleation of the electon. 5. Detemine the amount of wok that is done in moving a chage of 0.0 nc though a potential difeence of.50 0 V. 6. hee identical.0 μc conducting sphees ae placed at the cones of an equilateal tiangle of sides 5 cm. he tiangle has one ape C pointing up the page and base angles A and B. Detemine the absolute potential at B. 7. Detemine how fa apat two paallel plates must be situated so that a potential difeence of.50 0 V poduces an electic ield stength of NC A chage of.5 μc is placed in a unifom electic ield of two oppositely chaged paallel plates with a magnitude of NC -. (a) (b) (c) Detemine the wok that must be done against the ield to move the point chage a distance of 5.5 cm. Calculate the potential difeence between the inal and initial positions of the chage. Detemine the potential difeence between the plates if thei sepaation distance is 5 cm. 4. Duing a lash of lightning, the potential difeence between a cloud and the gound was. 0 9 V and the amount of tansfeed chage was 3 C. (a) (b) (c) Detemine the change in enegy of the tansfeed chage. If the enegy eleased was all used to acceleate a tonne ca, deduce its inal speed. If the enegy eleased could be used to melt ice at 0 C, deduce the amount of ice that could be melted. 66

19 MOTION IN FIELDS 5. Suppose that when an electon moved fom A to B in the diagam along an electic ield line that the electic ield does J of wok on it. Detemine the difeences in electic potential: (a) (b) (c) A V B V A V C V A V C V B 9.4 ORBITAL MOTION Although obital motion may be cicula, elliptical o paabolic, this sub-topic only deals with cicula obits. his sub-topic is not fundamentally new physics, but an application that synthesizes ideas fom gavitation, cicula motion, dynamics and enegy State that gavitation povides the centipetal foce fo cicula obital motion. B 6. Detemine the potential at point P that is located at the cente of the squae as shown in the diagam below. -6µC C 5µC 9.4. Deive Keple s thid law Deive epessions fo the kinetic enegy, potential enegy and total enegy of an obiting satellite Sketch gaphs showing the vaiation with obital adius of the kinetic enegy, gavitational potential enegy and total enegy of a satellite. m P Discuss the concept of weightlessness in obital motion, in fee fall and in deep space Solve poblems involving obital motion. IBO µC m +µc 9.4. SATELLITES he Moon obits the Eath and in this sense it is oten efeed to as a satellite of the Eath. Befoe 957 it was the only Eath satellite! Howeve, in 957 the Russians launched the ist man made satellite, Sputnik. Since this date many moe satellites have been launched and thee ae now liteally thousands of them obiting the Eath. Some ae used to monito the weathe, some used to enable people to ind accuately thei position on the suface of the Eath, many ae used in communications, and no doubt some ae used to spy on othe counties. Figue 93 shows how, in pinciple, a satellite can be put into obit. he peson (whose size is geatly eaggeated with espect to Eath) standing on the suface on the Eath thows some stones. he geate the speed with which a stone is thown the futhe it will land fom he. he paths followed by the thown stones ae paabolas. By a stetch of the imagination we can visualise a situation in which a stone is thown with such a speed that, because of the cuvatue of the Eath, it will not land on the suface of the Eath but go into obit. (Path 4 on igue 93). 67

20 CHAPTER 9 Eath 3 4 Satellite obit Eath Tangential component of velocity Satellite caied by ocket to hee Figue 933 Getting a satellite into obit Figue 93 Thowing a stone into obit he foce that causes the stones to follow a paabolic path and to fall to Eath is gavity and similaly the foce that keeps the stone in obit is gavity. Fo cicula motion to occu we have seen that a foce must act at ight angles to the velocity of an object, that is thee must be a centipetal foce. Hence in the situation we descibe hee the centipetal foce fo cicula obital motion about the Eath is povided by gavitational attaction of the Eath. We can calculate the speed with which a stone must be thown in ode to put it into obit just above the suface of the Eath. If the stone has mass m and speed v then we have fom Newton s nd law mv G M E m R E R E whee R E is the adius of the Eath and M E is the mass of the Eath. Beaing in mind that g 0 G M E R, then E v g R E hat is, the stone must be thown at m s. Clealy we ae not going to get a satellite into obit so close to the suface of the Eath. Moving at this speed the fiction due to ai esistance would melt the satellite befoe it had tavelled a couple of kilometes! In eality theefoe a satellite is put into obit about the Eath by sending it, attached to a ocket, way beyond the Eath s atmosphee and then giving it a component of velocity pependicula to a adial vecto fom the Eath. See Figue KEPLER S THIRD LAW (his wok of Keple and Newton s synthesis of the wok is an ecellent eample of the scientiic method and makes fo a good TOK discussion) In 67 Johannes Keple (57-630) published his laws of planetay motion. he laws ae empiical in natue and wee deduced fom the obsevations of the Danish astonome Tycho de Bahe (546-60). he thid law gives a elationship between the adius of obit R of a planet and its peiod T of evolution about the Sun. he law is epessed mathematically as T constant 3 R We shall now use Newton s Law of Gavitation to show how it is that the planets move in accodance with Keple s thid law. In essence Newton was able to use his law of gavity to pedict the motion of the planets since all he had to do was facto the F given by this law into his second law, F ma, to ind thei acceleations and hence thei futue positions. In Figue 934 the Eath is shown obiting the Sun and the distance between thei centes is R. Eath F es Figue 934 R Fse Sun Planets move accoding to Keple s thid law F es is the foce that the Eath eets on the Sun and F se is the foce that the Sun eets on the Eath. he foces ae equal and opposite and the Sun and the Eath will actually obit about a common cente. Howeve since the Sun is so vey much moe massive than the Eath this common cente will be close to the cente of the Sun and so we can egad the Eath as obiting about the cente of the Sun. he othe 68

21 MOTION IN FIELDS thing that we shall assume is that we can ignoe the foces that the othe planets eet on the Eath. (his would not be a wise thing to do if you wee planning to send a space ship to the Moon fo eample!). We shall also assume that we have followed Newton s eample and indeed poved that a sphee will act as a point mass situated at the cente of the sphee. Keple had postulated that the obits of the planets ae elliptical but since the eccenticity of the Eath s obit is small we shall assume a cicula obit. he acceleation of the Eath towads the Sun is a Rω whee ω Hence, π ---- T a R π --- T 4π R But the acceleation is given by Newton s Second Law, F ma, whee F is now given by the Law of Gavitation. So in this situation GM s M e F ma R, but, we also have that 4π R a T and m M e so that T G M s M e 4π R GM R M e s T π But the quantity R T is a constant that has the same value fo each of the planets so we have fo all the planets, not just Eath, that R T k whee k is a constant. Which is of couse Keple s thid law. his is indeed an amazing beakthough. It is diicult to efute the idea that all paticles attact each othe in accodance with the Law of Gavitation when. the law is able to account fo the obseved motion of the planets about the Sun. he gavitational efects of the planets upon each othe should poduce petubations in thei obits. Such is the pedictive powe of the Univesal Gavitational Law that it enabled physicists to compute these petubations. he telescope had been invented in 608 and by the middle of the 8th Centuy had eached a degee of pefection in design that enabled astonomes to actually measue the obital petubations of the planets. hei measuements wee always in ageement with the pedictions made by Newton s law. Howeve, in 78 a new planet, Uanus was discoveed and the obit of this planet did not it with the obit pedicted by Univesal Gavitation. Such was the physicist s faith in the Newtonian method that they suspected that the discepancy was due to the pesence of a yet undetected planet. Using the Law of Gavitation the Fench astonome J.Leveie and the English astonome. J. C. Adams wee able to calculate just how massive this new planet must be and also whee it should be. In 846 the planet Neptune was discoveed just whee they had pedicted. In a simila way, discepancies in the obit of Neptune led to the pediction and subsequent discovey in 930 of the planet Pluto. Newton s Law of Gavitation had passed the ultimate test of any theoy; it is not only able to eplain eisting data but also to make pedictions SATELLITE ENERGY When a satellite is in obit about a planet it will have both kinetic enegy and gavitational potential enegy. Suppose we conside a satellite of mass m that is in an obit of adius about a planet of mass M. he gavitational potential due to the planet at distance fom its cente is. he gavitational potential enegy of the satellite V sat GM e m is theefoe GM hat is, V e m sat he gavitational ield stength at the suface of the planet is given by Hence we can wite g 0 GM e R e g V 0 R e sat he kinetic enegy of the satellite K sat is equal to ½mv, whee v is its obital speed. 69