r 2 F ds W = r 1 qe ds = q

Save this PDF as:
 WORD  PNG  TXT  JPG

Size: px
Start display at page:

Download "r 2 F ds W = r 1 qe ds = q"

Transcription

1 Chpter 4 The Electric Potentil 4.1 The Importnt Stuff Electricl Potentil Energy A chrge q moving in constnt electric field E experiences force F = qe from tht field. Also, s we know from our study of work nd energy, the work done on the chrge by the field s it moves from point r 1 to r 2 is W = r2 r 1 F ds where we men tht we re summing up ll the tiny elements of work dw = F ds long the length of the pth. When F is the electrosttic force, the work done is W = r2 r 1 qe ds = q r2 r 1 E ds (4.1) In Fig. 4.1, chrge is shown being moved from r 1 to r 2 long two different pths, with ds nd E shown for bit of ech of the pths. Now it turns out tht from the mthemticl form of the electrosttic force, the work done by the force does not depend on the pth tken to get from r 1 to r 2. As result we sy q r 2 E ds E q ds r 1 Figure 4.1: Chrge is moved from r 1 to r 2 long two seprte pths. Work done by the electric force involves the summing up E ds long the pth. 53

2 54 CHAPTER 4. THE ELECTRIC POTENTIAL tht the electric force is conservtive nd it llows us to clculte n electric potentil energy, which s usul we will denote by U. As before, only the chnges in the potentil hve ny rel mening, nd the chnge in potentil energy is the negtive of the work done by the electric force: r2 U = W = q E ds (4.2) r 1 We usully wnt to discuss the potentil energy of chrge t prticulr point, tht is, we would like function U(r), but for this we need to mke definition for the potentil energy t prticulr point. Usully we will mke the choice tht the potentil energy is zero when the chrge is infinitely fr wy: U = Electric Potentil Recll how we developed the concept of the electric field E: The force on chrge q 0 is lwys proportionl to q 0, so by dividing the chrge out of F we get something which cn conveniently give the force on ny chrge. Likewise, if we divide out the chrge q from Eq. 4.2 we get function which we cn use to get the chnge in potentil energy for ny chrge (simply by multiplying by the chrge). This new function is clled the electric potentil, V : V = U q where U is the chnge in potentil energy of chrge q. Then Eq. 4.2 gives us the difference in electricl potentil between points r 1 nd r 2 : r2 V = E ds (4.3) r 1 The electric potentil is sclr. Reclling tht it ws defined by dividing potentil energy by chrge we see tht its units re J (joules per coulomb). The electric potentil is C of such gret importnce tht we cll this combintion of units volt 1. Thus: 1volt = 1V = 1 J C (4.4) Of course, it is then true tht joule is equl to coulomb-volt=c V. In generl, multiplying chrge times potentil difference gives n energy. It often hppens tht we re multiplying n elementry chrge (e) (or some multiple thereof) nd potentil difference in volts. It is then convenient to use the unit of energy given by the product of e nd volt; this unit is clled the electron-volt: 1eV = (e) (1V) = C (1V) = J (4.5) Eqution 4.3 cn only give us the differences in the vlue of the electric potentil between two points r 1 nd r 2. To rrive t function V (r) defined t ll points we need to specify 1 Nmed in honor of the... uh...french physicist Jim Volt ( ) who did some electricl experiments in... um... Bologn. Tht s it, Bologn.

3 4.1. THE IMPORTANT STUFF 55 point t which the potentil V is zero. Often we will choose this point to be infinity ( ) tht is, s we get very fr wy from the set of chrges which give the electric field, the potentil V becomes very smll in bsolute vlue. However this reference point cn be chosen nywhere nd for ech problem we need to be sure where it is understood tht V = 0 before we cn sensibly tlk bout the function V (r). Then in Eq. 4.3 equl to this reference point nd clculte n potentil function V (r) for ll other points. So we cn write: Equipotentil Surfces r V (r) = E ds r ref (4.6) For given configurtion of chrges, set of points where the electric potentil V (r) hs given vlue is clled n equipotentil surfce. It tkes no work to move chrged prticle from one point on such surfce to nother point on the surfce, for then we hve V = 0. From the reltions between E(r) nd V (r) it follows tht the field lines re perpendiculr to the equipotentil surfces everywhere Finding E from V The definition of V n integrl involving the E field implies tht the electric field comes from V by tking derivtives: E x = V x E y = V y E z = V z (4.7) These reltions cn be written s one eqution using the nottion for the grdient: E = V (4.8) Potentil of Point Chrge nd Groups of Points Chrges Using Eq. 4.3, one cn show tht if we specify tht the electricl potentil is zero t infinity, then the potentil due to point chrge q is V (r) = k q r = 1 q r (4.9) where r is the distnce of the chrge from the point of interest. Furthermore, for set of point chrges q 1, q 2, q 3,... the electricl potentil is V (r) = i k q i r i = i 1 q i r i (4.10) where r i is the distnce of ech chrge from the point of interest.

4 56 CHAPTER 4. THE ELECTRIC POTENTIAL Using Eq. 4.10, one cn show tht the potentil due to n electric dipole with mgnitude p t the origin (pointing upwrd long the z xis) is Here, r nd θ hve the usul mening in sphericl coordintes. V (r) = 1 pcos θ r 2 (4.11) Potentil Due to Continuous Chrge Distribution To get the electricl potentil due to continuous distribution of chrge (with V = 0 t infinity ssumed), dd up the contributions to the potentil; the potentil due to chrge dq t distnce r is dv = 1 dq so tht we must do the integrl r V = 1 dq r = 1 V ρ(r)dτ r (4.12) In the lst expression we re using the chrge density ρ(r) of the distribution to get the element of chrge dq for the volume element dτ Potentil Energy of System of Chrges The potentil energy of pir of point chrges (i.e. the work W needed to bring point chrges q 1 nd q 2 from infinite seprtion to seprtion r) is U = W = 1 q 1 q 2 r For lrger set of chrges the potentil energy is given by the sum U = U 12 + U 23 + U = 1 pirs ij (4.13) q i q j r ij (4.14) Here r ij is the distnce between chrges q i nd q j. Ech pir is only counted once in the sum. 4.2 Worked Exmples Electric Potentil 1. The electric potentil difference between the ground nd cloud in prticulr thunderstorm is V. Wht is the mgnitude of the chnge in energy (in multiples of the electron-volt) of n electron tht moves between the ground nd the cloud?

5 4.2. WORKED EXAMPLES 57 The mgnitude of the chnge in potentil s the electron moves between ground nd cloud (we don t cre which wy) is V = V. Multiplying by the mgnitude of the electron s chrge gives the mgnitude of the chnge in potentil energy. Note tht lumping e nd V together gives the ev (electron-volt), unit of energy: U = q V = e( V) = ev = 1.2GeV 2. An infinite nonconducting sheet hs surfce chrge density σ = 0.10µC/m 2 on one side. How fr prt re equipotentil surfces whose potentils differ by 50V? In Chpter 3, we encountered the formul for the electric field due nonconducting sheet of chrge. From Eq. 3.5, we hd: E z = σ/(2ɛ 0 ), where σ is the chrge density of the sheet, which lies in the xy plne. So the plne of chrge in this problem gives rise to n E field: E z = σ 2ɛ 0 ( C m = 2 2( C 2 ) = N C N m 2 Here the E field is uniform nd lso E x = E y = 0. Now, from Eq. 4.7 we hve V z = E z = N C. nd when the rte of chnge of some quntity (in this cse, with respect to the z coordinte) is constnt we cn write the reltion in terms of finite chnges, tht is, with s: V z = E z = N C nd from this result we cn find the chnge in z corresponding to ny chnge in V. If we re interested in V = 50 V, then z = V E z (50V) = ( N ) = m = 8.8mm C i.e. to get chnge in potentil of +50V we need chnge in z coordinte of 8.8mm. Since the potentil only depends on the distnce from the plne, the equipotentil surfces re plnes. The distnce between plnes whose potentil differs by 50V is 8.8mm. 3. Two lrge, prllel conducting pltes re 12 cm prt nd hve chrges of equl mgnitude nd opposite sign on their fcing surfces. An electrosttic force of N cts on n electron plced nywhere between the two pltes.

6 58 CHAPTER 4. THE ELECTRIC POTENTIAL (Neglect fringing.) () Find the electric field t the position of the electron. (b) Wht is the potentil difference between the pltes? () We re given the mgnitude of the electric force on n electron (whose chrge is e). Then the mgnitude of the E field must be: E = F q = F e = ( N) ( C) = N C = V m (b) The E field in the region between two lrge oppositely chrged pltes is uniform so in tht cse, we cn write E x = V x (where the E field points in the x direction, i.e. perpendiculr to the pltes), nd the potentil difference between the pltes hs mgnitude V = E x x = ( V m )(0.12m) = V 4. The electric field inside nonconducting sphere of rdius R with chrge spred uniformly throughout its volume, is rdilly directed nd hs mgnitude E(r) = qr R 3. Here q (positive or negtive) is the totl chrge within the sphere, nd r is the distnce from the sphere s center. () Tking V = 0 t the center of the sphere, find the electric potentil V (r) inside the sphere. (b) Wht is the difference in electric potentil between point on the surfce nd the sphere s center? (c) If q is positive, which of those two points is t the higher potentil? () We will use Eq. 4.6 to clculte V (r) using r = 0 s the reference point: V (0) = 0. The electric field hs only rdil component E r (r) so tht we will evlute: r r V (r) = E ds = E r (r )dr r ref 0 Using the given expression for E r (r ) (which one cn derive using Guss (s) lw) we get: r V (r) = 0 = q r 2 R 3 2 = qr2 8πɛ 0 R 3 qr R 3dr = q R 3 r 0 r dr

7 4.2. WORKED EXAMPLES 59 x r = r r = 4 Figure 4.2: Pth of integrtion for Exmple 5. Integrtion goes from r = to r = r. (b) Using the result of prt (), the difference between vlues of V (r) on the sphere s surfce nd t its center is V (R) V (0) = qr2 8πɛ 0 R = q 3 8πɛ 0 R (c) For q positive, the nswer to prt (b) is negtive number, so the center of the sphere must be t higher potentil. 5. A chrge q is distributed uniformly throughout sphericl volume of rdius R. () Setting V = 0 t infinity, show tht the potentil t distnce r from the center, where r < R, is given by V = q(3r2 r 2 ) 8πɛ 0 R 3. (b) Why does this result differ from tht of the previous exmple? (c) Wht is the potentil difference between point of the surfce nd the sphere s center? (d) Why doesn t this result differ from tht of the previous exmple? () We find the function V (r) just s we did the lst exmple, but this time the reference point (the plce where V = 0) is t r =. So we will evlute: r r V (r) = E ds = E r (r )dr. (4.15) r ref The integrtion pth is shown in Fig We note tht the integrtion (from r = to r = r with r < R) is over vlues of r both outside nd inside the sphere. Just s before, the E field for points inside the sphere is E r, in (r) = qr R 3, (4.16) but now we will lso need the vlue of the E field outside the sphere. By Guss (s) lw the externl E field is tht sme s tht due to point chrge q t distnce r, so: E r, out (r) = q r 2. (4.17) Becuse E r (r) hs two different forms for the interior nd exterior of the sphere, we will hve to split up the integrl in Eq into two prts. When we go from to R we need

8 60 CHAPTER 4. THE ELECTRIC POTENTIAL to use Eq for E r (r ). When we go from R to r we need to use Eq for E r (r ). So from Eq we now hve R V (r) = = R ( = q r E r, out (r )dr ) q dr r 2 { R dr r r + 2 R R r E r, in (r )dr ( ) qr R r R 3 dr R 3 } dr Now do the individul integrls nd we re done: V (r) = q 1 R r + r 2 r 2R 3 R = q { 1 R + (r2 R 2 } ) 2R 3 ( q 2R 2 = 2R + (R2 r 2 ) ) 3 2R 3 = q(3r2 r 2 ) 8πɛ 0 R 3 (b) The difference between this result nd tht of the previous exmple is due to the different choice of reference point. There is no problem here since it is only the differences in electricl potentil tht hve ny mening in physics. (c) using the result of prt (), we clculte: V (R) V (0) = q(2r2 ) 8πɛ 0 R 3 q(3r2 ) 8πɛ 0 R 3 = qr2 8πɛ 0 R 3 = q 8πɛ 0 R This is the sme s the corresponding result in the previous exmple. (d) Differences in the electricl potentil will not depend on the choice of the reference point, the nswer should be the sme s in the previous exmple... if V (r) is clculted correctly! 6. Wht re () the chrge nd (b) the chrge density on the surfce of conducting sphere of rdius 0.15m whose potentil is 200V (with V = 0 t infinity)? () We re given the rdius R of the conducting sphere; we re sked to find its chrge Q. From our work with Guss (s) lw we know tht the electric field outside the sphere is the sme s tht of point chrge Q t the sphere s center. Then if we were to use Eq. 4.6

9 4.2. WORKED EXAMPLES 61 E r r E r =0 inside V=+400 V Figure 4.3: Conducting chrged sphere, hs potentil of 400 V, from Exmple 7. with the condition V = 0 t infinity (which is outside the sphere!), we would get the sme result for V s we would for point chrge Q t the origin nd V = 0 t infinity, nmely: V (r) = 1 Q r (outside sphere) This eqution holds for r R. Then t the sphere s surfce (r = R) we hve: Solve for Q nd plug in the numbers: Q = V R The chrge on the sphere is C. V = 1 Q R 12 C2 = 4π( )(200V)(0.15m) N m 2 = C (b) The chrge found in () resides on the surfce of the conducting sphere. To get the chrge density, divide the chrge by the surfce re of the sphere: σ = Q 4πR = ( C) 2 4π(0.15 m)2 = C m 2 The chrge density on the sphere s surfce is C m An empty hollow metl sphere hs potentil of +400V with respect to ground (defined to be t V = 0) nd hs chrge of C. Find the electric potentil t the center of the sphere. The problem is digrmmed in Fig From considering sphericl Gussin surfce drwn inside the sphere, we see tht the electric field E r must be zero everywhere in side the sphere becuse such surfce will enclose no chrge. But for sphericl geometries, E r nd V re relted by E r = dv dr

10 62 CHAPTER 4. THE ELECTRIC POTENTIAL Q R V=400 V r = 4 Figure 4.4: Conducting chrged sphere, hs potentil of 400 V (with V = 0 t r = ), from Exmple 8. so tht with E r = 0, V must be constnt throughout the interior of the sphericl conductor. Since the vlue of V on the sphere itself is +400V, V then must lso equl +400V t the center. So V = +400V t the center of the sphere. (There ws no clculting to do on this problem!) 8. Wht is the excess chrge on conducting sphere of rdius R = 0.15m if the potentil of the sphere is 1500 V nd V = 0 t infinity? The problem is digrmmed in Fig If the sphere hs net chrge Q then from Guss lw the rdil component of the electric field for points outside the sphere is E r = k Q r 2 Using Eq. 4.6 with r = s the reference point, the potentil t distnce R from the sphere s center is: r V (R) = = kq r = kq R R r E r dr = = kq R 0 kq r 2 dr (Note tht the integrtion tkes plce over vlues of r outside the sphere so tht the expression for E r is the correct one. E r is zero inside the sphere.) We re given tht V (R) = 400V, so from kq/r = 400V we solve for Q nd get: Q = R(400V) k = (0.15m)(400V) ( N m 2 C 2 ) = C 9. The electric potentil t points in n xy plne is given by V = (2.0 V m 2 )x 2 (3.0 V m 2 )y 2.

11 4.2. WORKED EXAMPLES 63 Wht re the mgnitude nd direction of the electric field t the point (3.0m, 2.0m)? Equtions 4.7 show how to get the components of the E field if we hve the electric potentil V s function of x nd y. Tking prtil derivtives, we find: E x = V x = (4.0 V m 2 )x nd E y = V y = +(6.0 V m 2 )y. Plugging in the given vlues of x = 3.0m nd y = 2.0m we get: E x = 12 V m nd E y = 12 V m So the mgnitude of the E field t the given is nd its direction is given by E = (12.0) 2 + (12.0) 2 V m = 17 V m ( ) θ = tn 1 Ey = tn 1 (1.0) = 135 E x where for θ we hve mde the proper choice so tht it lies in the second qudrnt Potentil Energy of System of Chrges 10. () Wht is the electric potentil energy of two electrons seprted by 2.00 nm? (b) If the seprtion increses, does the potentil energy increse or decrese? Since the chrge of n electron is e, using Eq we find: U = = 1 ( e)( e) r 1 4π( C 2 = J N m 2 ) ( C) 2 ( m) As the chrges re both positive, the potentil energy is positive number nd is inversely proportionl to r. So the potentil energy decreses s r increses. 11. Derive n expression for the work required to set up the four-chrge configurtion of Fig. 4.5, ssuming the chrges re initilly infinitely fr prt. The work required to set up these chrges is the sme s the potentil energy of set of point chrges, given in Eq (Tht is, sum the potentil energies k q iq j r ij over ll pirs of

12 64 CHAPTER 4. THE ELECTRIC POTENTIAL +q + - -q -q - + +q Figure 4.5: Chrge configurtion for Exmple 11. +q + - -q +q + - -q +q + - -q -q - -q - + +q () (b) (c) Figure 4.6: () Second chrge is brought in from nd put in plce. (b) Third chrge is brought in. (c) Lst chrge is brought in. chrges.) We cn rrive t the sme nswer nd understnd tht formul little better if we ssemble the system one chrge t time. Begin with the chrge in the upper left corner of Fig Moving this chrge from infinity to the desired loction requires no work becuse it is never ner ny other chrge. We cn write: W 1 = 0. Now bring up the chrge in the upper right corner ( q). Now we hve the configurtion shown in Fig. 4.6(). While being put into plce it hs experienced force from the first chrge nd the work required of the externl gency is the chnge in potentil energy of this chrge, nmely W 2 = 1 (+q)( q) = q2 Now bring the chrge in the lower left corner ( q), s shown in 4.6(b). When put into plce it is distnce from the first chrge nd 2 from the second chrge. The work required for this step is the potentil energy of the third chrge in this configurtion, nmely: W 3 = 1 (+q)( q) + 1 ( ( q)( q) = q ) 2 2 Finlly, bring in the fourth chrge (+q) to give the configurtion in Fig. 4.6(c). The lst chrge is now distnce from two q chrges nd distnce 2 from the other +q chrge. So the work required for this step is W 4 = 2 1 (+q)( q) + 1 (+q)(+q) 2

13 4.2. WORKED EXAMPLES 65 q 1 - A B + q 2 Figure 4.7: Chrge configurtion for Exmple 12. = q 2 ( ) 2 So now dd up ll the W s to get the totl work done: W Totl = W 1 + W 2 + W 3 + W 4 q 2 ( = ( = ) 2 q ) This is nice nlytic nswer; if we combine ll the numericl fctors (including the 4π) we get: W Totl = ( 0.21)q2 ɛ 0 This is the sme result s we d get by using Eq In the rectngle of Fig. 4.7, the sides hve lengths 5.0cm nd 15cm, q 1 = 5.0µC nd q 2 = +2.0µC. With V = 0 t infinity, wht re the electric potentils () t corner A nd (b) corner B? (c) How much work is required to move third chrge q 3 = +3.0µC from B to A long digonl of the rectngle? (d) Does this work increse or decrese the electric energy of the three chrge system? Is more, less or the sme work required if q 3 is moved long pths tht re (e) inside the rectngle but not on the digonl nd (f) outside the rectngle? () To find the electric potentil due to group of point chrges, use Eq Since point A is 15cm wy from the 5.0µC chrge nd 5.0cm wy from the +2.0µC chrge, we get: V = 1 [ q1 + q ] 2 r 1 r 2 = ( N m2 C 2 ) [ ( C) ( m) + (+2.0 ] 10 6 C) = V ( m)

14 66 CHAPTER 4. THE ELECTRIC POTENTIAL (b) Perform the sme clcultion s in prt (). The chrges q 1 nd q 2 re t different distnces from point B so we get different nswer: [ ( V = ( C) N m2 ) C 2 ( m) + (+2.0 ] 10 6 C) = V ( m) (c) Using the results of prt () nd (b), clculte the chnge in potentil V s we move from point B to point A: V = V A V B = V ( V) = V The chnge in potentil energy for +3.0µC chrge to move from B to A is U = q V = ( C)( V) = 2.5J (d) Since positive mount of work is done by the outside gency in moving the chrge from B to A, the electric energy of the system hs incresed. We cn see tht this must be the cse becuse the +3.0 µc chrge hs been moved closer to nother positive chrge nd frther wy from negtive chrge. (e) The force which point chrge (or set of point chrges) exerts on nother chrge is conservtive force. So the work which it does (or likewise the work required of some outside force) s the chrge moves from one point to nother is independent of the pth tken. Therefore we would require the sme mount of work if the pth tken ws some other pth inside the rectngle. (f) Since the work done is independent of the pth tken, we require the sme mount of work even if the pth from A to B goes outside the rectngle. 13. Two tiny metl spheres A nd B of mss m A = 5.00g nd m B = 10.0g hve equl positive chrges q = 5.00 µc. The spheres re connected by mssless nonconducting string of length d = 1.00 m, which is much greter thn the rdii of the spheres. () Wht is the electric potentil energy of the system? (b) Suppose you cut the string. At tht instnt, wht is the ccelertion of ech sphere? (c) A long time fter you cut the string, wht is the speed of ech sphere? () The initil configurtion of the chrges in shown in Fig. 4.8(). The electrosttic potentil energy of this system (i.e. the work needed to bring the chrges together from fr wy is U = 1 q 1 q 2 r = ( N m2 C 2 ) ( C) 2 (1.00 m) = 0.225J We re justified in using formule for point chrges becuse the problem sttes tht the sizes of the spheres re smll compred to the length of the string (1.00m).

15 4.2. WORKED EXAMPLES 67 q = 5.00 mc q = 5.00 mc m A =5.0 g 1.00 m m B =10.0 g () F F m A =5.0 g m B =10.0 g (b) Figure 4.8: () Chrged spheres ttched to string, in Exmple 13. The electrosttic repulsion is blnced by the string tension. (b) After string is cut there is mutul force of electricl repulsion F. Mgnitude of the force on ech chrge is the sme but their ccelertions re different! (b) From Coulomb s lw, the mgnitude of the mutul force of repulsion of the two chrges is F = 1 q 2 N m2 = ( ) ( C) 2 = 0.225N r2 C 2 (1.00m) 2 but since the msses of the spheres re different their ccelertions hve different mgnitudes. From Newton s 2 nd lw, the ccelertions of the msses re: 1 = F m 1 = 2 = F m 2 = (0.225 N) ( kg) = 45.0 m s 2 (0.225 N) ( kg) = 22.2 m s 2 Of course, the ccelertions re in opposite directions. (c) From the time tht the string breks to the time tht we cn sy tht the spheres re very fr prt, the only force tht ech one experiences is the force of electricl repulsion (rising from the other sphere). This is conservtive force so tht totl mechnicl energy is conserved. It is lso true tht there re no externl forces being exerted on the two sphere system. Then we know tht the totl (vector) momentum of the system is lso conserved. First, let s del with the condition of energy conservtion. The totl energy right fter the string is cut is just the potentil energy found in prt () since the spheres re not yet in motion. So E init = 0.225J. When the spheres re long wys prt, there is no electricl potentil energy, but they re in motion with respective speeds v A nd v B so there is kinetic energy t lrge seprtion. Then energy conservtion tells us: 1 m 2 AvA 2 + 1m 2 BvB 2 = 0.225J (4.18)

16 68 CHAPTER 4. THE ELECTRIC POTENTIAL Momentum conservtion gives us the other eqution tht we need. If mss B hs x velocity v B then mss A hs x velocity v A (it moves in the other direction. The system begins nd ends with totl momentum of zero so then: Substitute this result into 4.18 nd get: m A v A + m B v B = 0 = v B = m A m B v A 1 2 m Av 2 A m B ( ) m 2 A v m 2 A 2 = 0.225J B Fctor out va 2 on the left side nd plug in some numbers: 1 2 ( m A + m2 A m B ) v 2 A = 1 2 So then we get the finl speed of A: ( 5.00g + (5.00g)2 (10.0 g) ) v 2 A = ( kg)v 2 A = 0.225J v 2 A = J kg = 60.0 m2 s 2 = v A = 7.75 m s nd the speed of B: v B = m A m B v A = ( ) 5.00g 7.75 m s 10.0g = 3.87 m s 14. Two electrons re fixed 2.0 cm prt. Another electron is shot from infinity nd stops midwy between the two. Wht is its initil speed? The problem is digrmmed in Fig. 4.9() nd (b). Since the electrosttic force is conservtive force, we know tht energy is conserved between configurtions () nd (b). In picture () there is energy stored in the repulsion of the pir of electrons s well s the kinetic energy of the third electron. (Initilly the third electron is too fr wy to feel the first two electrons.) In picture (b) there is no kinetic energy but the electricl potentil energy hs incresed due to the repulsion between the third electron nd the first two. If we cn clculte the chnge in potentil energy U then by using energy conservtion, U + K = 0 we cn find the initil speed of the electron. The potentil energy of set of point chrges (with V = 0 t ) is given in Eq When the third electron comes from infinity nd stops t the midpoint, the increse in potentil energy the contribution given by the third electron s it sees its new neighbors. With r = 1.0 cm, this increse is U = 1 ( e)( e) + 1 ( e)( e) = e2 r r 2πɛ 0 r

17 4.2. WORKED EXAMPLES 69 -e 2.0 cm -e v 0 -e () 1.0 cm -e -e -e v=0 (b) Figure 4.9: () Electron flies in from with speed v 0. (b) Electron comes to rest midwy between the other two electrons. The chnge in kinetic energy is K = 1 2 mv2 0. Then energy conservtion gives: Solve for v 0 : K = U = 1 2 mv2 0 = e2 2πɛ 0 r v 2 0 = = e 2 πɛ 0 mr ( C) 2 π( C 2 N m 2 )( kg)( m) = m2 s 2 which gives v 0 = m s

18 70 CHAPTER 4. THE ELECTRIC POTENTIAL

Review guide for the final exam in Math 233

Review guide for the final exam in Math 233 Review guide for the finl exm in Mth 33 1 Bsic mteril. This review includes the reminder of the mteril for mth 33. The finl exm will be cumultive exm with mny of the problems coming from the mteril covered

More information

Experiment 6: Friction

Experiment 6: Friction Experiment 6: Friction In previous lbs we studied Newton s lws in n idel setting, tht is, one where friction nd ir resistnce were ignored. However, from our everydy experience with motion, we know tht

More information

PROBLEMS 13 - APPLICATIONS OF DERIVATIVES Page 1

PROBLEMS 13 - APPLICATIONS OF DERIVATIVES Page 1 PROBLEMS - APPLICATIONS OF DERIVATIVES Pge ( ) Wter seeps out of conicl filter t the constnt rte of 5 cc / sec. When the height of wter level in the cone is 5 cm, find the rte t which the height decreses.

More information

The Velocity Factor of an Insulated Two-Wire Transmission Line

The Velocity Factor of an Insulated Two-Wire Transmission Line The Velocity Fctor of n Insulted Two-Wire Trnsmission Line Problem Kirk T. McDonld Joseph Henry Lbortories, Princeton University, Princeton, NJ 08544 Mrch 7, 008 Estimte the velocity fctor F = v/c nd the

More information

5.2. LINE INTEGRALS 265. Let us quickly review the kind of integrals we have studied so far before we introduce a new one.

5.2. LINE INTEGRALS 265. Let us quickly review the kind of integrals we have studied so far before we introduce a new one. 5.2. LINE INTEGRALS 265 5.2 Line Integrls 5.2.1 Introduction Let us quickly review the kind of integrls we hve studied so fr before we introduce new one. 1. Definite integrl. Given continuous rel-vlued

More information

Mathematics. Vectors. hsn.uk.net. Higher. Contents. Vectors 128 HSN23100

Mathematics. Vectors. hsn.uk.net. Higher. Contents. Vectors 128 HSN23100 hsn.uk.net Higher Mthemtics UNIT 3 OUTCOME 1 Vectors Contents Vectors 18 1 Vectors nd Sclrs 18 Components 18 3 Mgnitude 130 4 Equl Vectors 131 5 Addition nd Subtrction of Vectors 13 6 Multipliction by

More information

Week 11 - Inductance

Week 11 - Inductance Week - Inductnce November 6, 202 Exercise.: Discussion Questions ) A trnsformer consists bsiclly of two coils in close proximity but not in electricl contct. A current in one coil mgneticlly induces n

More information

Polynomial Functions. Polynomial functions in one variable can be written in expanded form as ( )

Polynomial Functions. Polynomial functions in one variable can be written in expanded form as ( ) Polynomil Functions Polynomil functions in one vrible cn be written in expnded form s n n 1 n 2 2 f x = x + x + x + + x + x+ n n 1 n 2 2 1 0 Exmples of polynomils in expnded form re nd 3 8 7 4 = 5 4 +

More information

Answer, Key Homework 4 David McIntyre Mar 25,

Answer, Key Homework 4 David McIntyre Mar 25, Answer, Key Homework 4 Dvid McIntyre 45123 Mr 25, 2004 1 his print-out should hve 18 questions. Multiple-choice questions my continue on the next column or pe find ll choices before mkin your selection.

More information

Version 001 CIRCUITS holland (1290) 1

Version 001 CIRCUITS holland (1290) 1 Version CRCUTS hollnd (9) This print-out should hve questions Multiple-choice questions my continue on the next column or pge find ll choices efore nswering AP M 99 MC points The power dissipted in wire

More information

Physics 43 Homework Set 9 Chapter 40 Key

Physics 43 Homework Set 9 Chapter 40 Key Physics 43 Homework Set 9 Chpter 4 Key. The wve function for n electron tht is confined to x nm is. Find the normliztion constnt. b. Wht is the probbility of finding the electron in. nm-wide region t x

More information

Cypress Creek High School IB Physics SL/AP Physics B 2012 2013 MP2 Test 1 Newton s Laws. Name: SOLUTIONS Date: Period:

Cypress Creek High School IB Physics SL/AP Physics B 2012 2013 MP2 Test 1 Newton s Laws. Name: SOLUTIONS Date: Period: Nme: SOLUTIONS Dte: Period: Directions: Solve ny 5 problems. You my ttempt dditionl problems for extr credit. 1. Two blocks re sliding to the right cross horizontl surfce, s the drwing shows. In Cse A

More information

Answer, Key Homework 10 David McIntyre 1

Answer, Key Homework 10 David McIntyre 1 Answer, Key Homework 10 Dvid McIntyre 1 This print-out should hve 22 questions, check tht it is complete. Multiple-choice questions my continue on the next column or pge: find ll choices efore mking your

More information

Operations with Polynomials

Operations with Polynomials 38 Chpter P Prerequisites P.4 Opertions with Polynomils Wht you should lern: Write polynomils in stndrd form nd identify the leding coefficients nd degrees of polynomils Add nd subtrct polynomils Multiply

More information

6.2 Volumes of Revolution: The Disk Method

6.2 Volumes of Revolution: The Disk Method mth ppliction: volumes of revolution, prt ii Volumes of Revolution: The Disk Method One of the simplest pplictions of integrtion (Theorem ) nd the ccumultion process is to determine so-clled volumes of

More information

Double Integrals over General Regions

Double Integrals over General Regions Double Integrls over Generl egions. Let be the region in the plne bounded b the lines, x, nd x. Evlute the double integrl x dx d. Solution. We cn either slice the region verticll or horizontll. ( x x Slicing

More information

MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics. W02D3_0 Group Problem: Pulleys and Ropes Constraint Conditions

MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics. W02D3_0 Group Problem: Pulleys and Ropes Constraint Conditions MSSCHUSES INSIUE OF ECHNOLOGY Deprtment of hysics 8.0 W02D3_0 Group roblem: ulleys nd Ropes Constrint Conditions Consider the rrngement of pulleys nd blocks shown in the figure. he pulleys re ssumed mssless

More information

PHY 222 Lab 8 MOTION OF ELECTRONS IN ELECTRIC AND MAGNETIC FIELDS

PHY 222 Lab 8 MOTION OF ELECTRONS IN ELECTRIC AND MAGNETIC FIELDS PHY 222 Lb 8 MOTION OF ELECTRONS IN ELECTRIC AND MAGNETIC FIELDS Nme: Prtners: INTRODUCTION Before coming to lb, plese red this pcket nd do the prelb on pge 13 of this hndout. From previous experiments,

More information

Example A rectangular box without lid is to be made from a square cardboard of sides 18 cm by cutting equal squares from each corner and then folding

Example A rectangular box without lid is to be made from a square cardboard of sides 18 cm by cutting equal squares from each corner and then folding 1 Exmple A rectngulr box without lid is to be mde from squre crdbord of sides 18 cm by cutting equl squres from ech corner nd then folding up the sides. 1 Exmple A rectngulr box without lid is to be mde

More information

Graphs on Logarithmic and Semilogarithmic Paper

Graphs on Logarithmic and Semilogarithmic Paper 0CH_PHClter_TMSETE_ 3//00 :3 PM Pge Grphs on Logrithmic nd Semilogrithmic Pper OBJECTIVES When ou hve completed this chpter, ou should be ble to: Mke grphs on logrithmic nd semilogrithmic pper. Grph empiricl

More information

1 Numerical Solution to Quadratic Equations

1 Numerical Solution to Quadratic Equations cs42: introduction to numericl nlysis 09/4/0 Lecture 2: Introduction Prt II nd Solving Equtions Instructor: Professor Amos Ron Scribes: Yunpeng Li, Mrk Cowlishw Numericl Solution to Qudrtic Equtions Recll

More information

Volumes of solids of revolution

Volumes of solids of revolution Volumes of solids of revolution We sometimes need to clculte the volume of solid which cn be obtined by rotting curve bout the x-xis. There is strightforwrd technique which enbles this to be done, using

More information

AREA OF A SURFACE OF REVOLUTION

AREA OF A SURFACE OF REVOLUTION AREA OF A SURFACE OF REVOLUTION h cut r πr h A surfce of revolution is formed when curve is rotted bout line. Such surfce is the lterl boundr of solid of revolution of the tpe discussed in Sections 7.

More information

COMPONENTS: COMBINED LOADING

COMPONENTS: COMBINED LOADING LECTURE COMPONENTS: COMBINED LOADING Third Edition A. J. Clrk School of Engineering Deprtment of Civil nd Environmentl Engineering 24 Chpter 8.4 by Dr. Ibrhim A. Asskkf SPRING 2003 ENES 220 Mechnics of

More information

Integration by Substitution

Integration by Substitution Integrtion by Substitution Dr. Philippe B. Lvl Kennesw Stte University August, 8 Abstrct This hndout contins mteril on very importnt integrtion method clled integrtion by substitution. Substitution is

More information

Math 135 Circles and Completing the Square Examples

Math 135 Circles and Completing the Square Examples Mth 135 Circles nd Completing the Squre Exmples A perfect squre is number such tht = b 2 for some rel number b. Some exmples of perfect squres re 4 = 2 2, 16 = 4 2, 169 = 13 2. We wish to hve method for

More information

Vectors 2. 1. Recap of vectors

Vectors 2. 1. Recap of vectors Vectors 2. Recp of vectors Vectors re directed line segments - they cn be represented in component form or by direction nd mgnitude. We cn use trigonometry nd Pythgors theorem to switch between the forms

More information

Integration. 148 Chapter 7 Integration

Integration. 148 Chapter 7 Integration 48 Chpter 7 Integrtion 7 Integrtion t ech, by supposing tht during ech tenth of second the object is going t constnt speed Since the object initilly hs speed, we gin suppose it mintins this speed, but

More information

Use Geometry Expressions to create a more complex locus of points. Find evidence for equivalence using Geometry Expressions.

Use Geometry Expressions to create a more complex locus of points. Find evidence for equivalence using Geometry Expressions. Lerning Objectives Loci nd Conics Lesson 3: The Ellipse Level: Preclculus Time required: 120 minutes In this lesson, students will generlize their knowledge of the circle to the ellipse. The prmetric nd

More information

Example 27.1 Draw a Venn diagram to show the relationship between counting numbers, whole numbers, integers, and rational numbers.

Example 27.1 Draw a Venn diagram to show the relationship between counting numbers, whole numbers, integers, and rational numbers. 2 Rtionl Numbers Integers such s 5 were importnt when solving the eqution x+5 = 0. In similr wy, frctions re importnt for solving equtions like 2x = 1. Wht bout equtions like 2x + 1 = 0? Equtions of this

More information

Physics 2102 Lecture 2. Physics 2102

Physics 2102 Lecture 2. Physics 2102 Physics 10 Jonthn Dowling Physics 10 Lecture Electric Fields Chrles-Augustin de Coulomb (1736-1806) Jnury 17, 07 Version: 1/17/07 Wht re we going to lern? A rod mp Electric chrge Electric force on other

More information

1. In the Bohr model, compare the magnitudes of the electron s kinetic and potential energies in orbit. What does this imply?

1. In the Bohr model, compare the magnitudes of the electron s kinetic and potential energies in orbit. What does this imply? Assignment 3: Bohr s model nd lser fundmentls 1. In the Bohr model, compre the mgnitudes of the electron s kinetic nd potentil energies in orit. Wht does this imply? When n electron moves in n orit, the

More information

EQUATIONS OF LINES AND PLANES

EQUATIONS OF LINES AND PLANES EQUATIONS OF LINES AND PLANES MATH 195, SECTION 59 (VIPUL NAIK) Corresponding mteril in the ook: Section 12.5. Wht students should definitely get: Prmetric eqution of line given in point-direction nd twopoint

More information

The Chain Rule. rf dx. t t lim " (x) dt " (0) dx. df dt = df. dt dt. f (r) = rf v (1) df dx

The Chain Rule. rf dx. t t lim  (x) dt  (0) dx. df dt = df. dt dt. f (r) = rf v (1) df dx The Chin Rule The Chin Rule In this section, we generlize the chin rule to functions of more thn one vrible. In prticulr, we will show tht the product in the single-vrible chin rule extends to n inner

More information

Mechanics Cycle 1 Chapter 5. Chapter 5

Mechanics Cycle 1 Chapter 5. Chapter 5 Chpter 5 Contct orces: ree Body Digrms nd Idel Ropes Pushes nd Pulls in 1D, nd Newton s Second Lw Neglecting riction ree Body Digrms Tension Along Idel Ropes (i.e., Mssless Ropes) Newton s Third Lw Bodies

More information

Scalar and Vector Quantities. A scalar is a quantity having only magnitude (and possibly phase). LECTURE 2a: VECTOR ANALYSIS Vector Algebra

Scalar and Vector Quantities. A scalar is a quantity having only magnitude (and possibly phase). LECTURE 2a: VECTOR ANALYSIS Vector Algebra Sclr nd Vector Quntities : VECTO NLYSIS Vector lgebr sclr is quntit hving onl mgnitude (nd possibl phse). Emples: voltge, current, chrge, energ, temperture vector is quntit hving direction in ddition to

More information

Basic Analysis of Autarky and Free Trade Models

Basic Analysis of Autarky and Free Trade Models Bsic Anlysis of Autrky nd Free Trde Models AUTARKY Autrky condition in prticulr commodity mrket refers to sitution in which country does not engge in ny trde in tht commodity with other countries. Consequently

More information

Curve Sketching. 96 Chapter 5 Curve Sketching

Curve Sketching. 96 Chapter 5 Curve Sketching 96 Chpter 5 Curve Sketching 5 Curve Sketching A B A B A Figure 51 Some locl mximum points (A) nd minimum points (B) If (x, f(x)) is point where f(x) reches locl mximum or minimum, nd if the derivtive of

More information

PROF. BOYAN KOSTADINOV NEW YORK CITY COLLEGE OF TECHNOLOGY, CUNY

PROF. BOYAN KOSTADINOV NEW YORK CITY COLLEGE OF TECHNOLOGY, CUNY MAT 0630 INTERNET RESOURCES, REVIEW OF CONCEPTS AND COMMON MISTAKES PROF. BOYAN KOSTADINOV NEW YORK CITY COLLEGE OF TECHNOLOGY, CUNY Contents 1. ACT Compss Prctice Tests 1 2. Common Mistkes 2 3. Distributive

More information

SOLUTIONS TO CONCEPTS CHAPTER 5

SOLUTIONS TO CONCEPTS CHAPTER 5 1. m k S 10m Let, ccelertion, Initil velocity u 0. S ut + 1/ t 10 ½ ( ) 10 5 m/s orce: m 5 10N (ns) 40000. u 40 km/hr 11.11 m/s. 3600 m 000 k ; v 0 ; s 4m v u ccelertion s SOLUIONS O CONCEPS CHPE 5 0 11.11

More information

Lecture 3 Gaussian Probability Distribution

Lecture 3 Gaussian Probability Distribution Lecture 3 Gussin Probbility Distribution Introduction l Gussin probbility distribution is perhps the most used distribution in ll of science. u lso clled bell shped curve or norml distribution l Unlike

More information

Factoring Polynomials

Factoring Polynomials Fctoring Polynomils Some definitions (not necessrily ll for secondry school mthemtics): A polynomil is the sum of one or more terms, in which ech term consists of product of constnt nd one or more vribles

More information

6 Energy Methods And The Energy of Waves MATH 22C

6 Energy Methods And The Energy of Waves MATH 22C 6 Energy Methods And The Energy of Wves MATH 22C. Conservtion of Energy We discuss the principle of conservtion of energy for ODE s, derive the energy ssocited with the hrmonic oscilltor, nd then use this

More information

Or more simply put, when adding or subtracting quantities, their uncertainties add.

Or more simply put, when adding or subtracting quantities, their uncertainties add. Propgtion of Uncertint through Mthemticl Opertions Since the untit of interest in n eperiment is rrel otined mesuring tht untit directl, we must understnd how error propgtes when mthemticl opertions re

More information

9.3. The Scalar Product. Introduction. Prerequisites. Learning Outcomes

9.3. The Scalar Product. Introduction. Prerequisites. Learning Outcomes The Sclr Product 9.3 Introduction There re two kinds of multipliction involving vectors. The first is known s the sclr product or dot product. This is so-clled becuse when the sclr product of two vectors

More information

Mathematics Higher Level

Mathematics Higher Level Mthemtics Higher Level Higher Mthemtics Exmintion Section : The Exmintion Mthemtics Higher Level. Structure of the exmintion pper The Higher Mthemtics Exmintion is divided into two ppers s detiled below:

More information

Version 001 Summer Review #03 tubman (IBII20142015) 1

Version 001 Summer Review #03 tubman (IBII20142015) 1 Version 001 Summer Reiew #03 tubmn (IBII20142015) 1 This print-out should he 35 questions. Multiple-choice questions my continue on the next column or pge find ll choices before nswering. Concept 20 P03

More information

Applications to Physics and Engineering

Applications to Physics and Engineering Section 7.5 Applictions to Physics nd Engineering Applictions to Physics nd Engineering Work The term work is used in everydy lnguge to men the totl mount of effort required to perform tsk. In physics

More information

9 CONTINUOUS DISTRIBUTIONS

9 CONTINUOUS DISTRIBUTIONS 9 CONTINUOUS DISTIBUTIONS A rndom vrible whose vlue my fll nywhere in rnge of vlues is continuous rndom vrible nd will be ssocited with some continuous distribution. Continuous distributions re to discrete

More information

Pure C4. Revision Notes

Pure C4. Revision Notes Pure C4 Revision Notes Mrch 0 Contents Core 4 Alger Prtil frctions Coordinte Geometry 5 Prmetric equtions 5 Conversion from prmetric to Crtesin form 6 Are under curve given prmetriclly 7 Sequences nd

More information

Algebra Review. How well do you remember your algebra?

Algebra Review. How well do you remember your algebra? Algebr Review How well do you remember your lgebr? 1 The Order of Opertions Wht do we men when we write + 4? If we multiply we get 6 nd dding 4 gives 10. But, if we dd + 4 = 7 first, then multiply by then

More information

v T R x m Version PREVIEW Practice 7 carroll (11108) 1

v T R x m Version PREVIEW Practice 7 carroll (11108) 1 Version PEVIEW Prctice 7 crroll (08) his print-out should he 5 questions. Multiple-choice questions y continue on the next colun or pge find ll choices before nswering. Atwood Mchine 05 00 0.0 points A

More information

www.mathsbox.org.uk e.g. f(x) = x domain x 0 (cannot find the square root of negative values)

www.mathsbox.org.uk e.g. f(x) = x domain x 0 (cannot find the square root of negative values) www.mthsbo.org.uk CORE SUMMARY NOTES Functions A function is rule which genertes ectl ONE OUTPUT for EVERY INPUT. To be defined full the function hs RULE tells ou how to clculte the output from the input

More information

addition, there are double entries for the symbols used to signify different parameters. These parameters are explained in this appendix.

addition, there are double entries for the symbols used to signify different parameters. These parameters are explained in this appendix. APPENDIX A: The ellipse August 15, 1997 Becuse of its importnce in both pproximting the erth s shpe nd describing stellite orbits, n informl discussion of the ellipse is presented in this ppendix. The

More information

LINEAR TRANSFORMATIONS AND THEIR REPRESENTING MATRICES

LINEAR TRANSFORMATIONS AND THEIR REPRESENTING MATRICES LINEAR TRANSFORMATIONS AND THEIR REPRESENTING MATRICES DAVID WEBB CONTENTS Liner trnsformtions 2 The representing mtrix of liner trnsformtion 3 3 An ppliction: reflections in the plne 6 4 The lgebr of

More information

A.7.1 Trigonometric interpretation of dot product... 324. A.7.2 Geometric interpretation of dot product... 324

A.7.1 Trigonometric interpretation of dot product... 324. A.7.2 Geometric interpretation of dot product... 324 A P P E N D I X A Vectors CONTENTS A.1 Scling vector................................................ 321 A.2 Unit or Direction vectors...................................... 321 A.3 Vector ddition.................................................

More information

Rotating DC Motors Part II

Rotating DC Motors Part II Rotting Motors rt II II.1 Motor Equivlent Circuit The next step in our consiertion of motors is to evelop n equivlent circuit which cn be use to better unerstn motor opertion. The rmtures in rel motors

More information

Radius of the Earth - Radii Used in Geodesy James R. Clynch February 2006

Radius of the Earth - Radii Used in Geodesy James R. Clynch February 2006 dius of the Erth - dii Used in Geodesy Jmes. Clynch Februry 006 I. Erth dii Uses There is only one rdius of sphere. The erth is pproximtely sphere nd therefore, for some cses, this pproximtion is dequte.

More information

MATH 150 HOMEWORK 4 SOLUTIONS

MATH 150 HOMEWORK 4 SOLUTIONS MATH 150 HOMEWORK 4 SOLUTIONS Section 1.8 Show tht the product of two of the numbers 65 1000 8 2001 + 3 177, 79 1212 9 2399 + 2 2001, nd 24 4493 5 8192 + 7 1777 is nonnegtive. Is your proof constructive

More information

, and the number of electrons is -19. e e 1.60 10 C. The negatively charged electrons move in the direction opposite to the conventional current flow.

, and the number of electrons is -19. e e 1.60 10 C. The negatively charged electrons move in the direction opposite to the conventional current flow. Prolem 1. f current of 80.0 ma exists in metl wire, how mny electrons flow pst given cross section of the wire in 10.0 min? Sketch the directions of the current nd the electrons motion. Solution: The chrge

More information

2005-06 Second Term MAT2060B 1. Supplementary Notes 3 Interchange of Differentiation and Integration

2005-06 Second Term MAT2060B 1. Supplementary Notes 3 Interchange of Differentiation and Integration Source: http://www.mth.cuhk.edu.hk/~mt26/mt26b/notes/notes3.pdf 25-6 Second Term MAT26B 1 Supplementry Notes 3 Interchnge of Differentition nd Integrtion The theme of this course is bout vrious limiting

More information

Module 2. Analysis of Statically Indeterminate Structures by the Matrix Force Method. Version 2 CE IIT, Kharagpur

Module 2. Analysis of Statically Indeterminate Structures by the Matrix Force Method. Version 2 CE IIT, Kharagpur Module Anlysis of Stticlly Indeterminte Structures by the Mtrix Force Method Version CE IIT, Khrgpur esson 9 The Force Method of Anlysis: Bems (Continued) Version CE IIT, Khrgpur Instructionl Objectives

More information

SPECIAL PRODUCTS AND FACTORIZATION

SPECIAL PRODUCTS AND FACTORIZATION MODULE - Specil Products nd Fctoriztion 4 SPECIAL PRODUCTS AND FACTORIZATION In n erlier lesson you hve lernt multipliction of lgebric epressions, prticulrly polynomils. In the study of lgebr, we come

More information

1. Find the zeros Find roots. Set function = 0, factor or use quadratic equation if quadratic, graph to find zeros on calculator

1. Find the zeros Find roots. Set function = 0, factor or use quadratic equation if quadratic, graph to find zeros on calculator AP Clculus Finl Review Sheet When you see the words. This is wht you think of doing. Find the zeros Find roots. Set function =, fctor or use qudrtic eqution if qudrtic, grph to find zeros on clcultor.

More information

P.3 Polynomials and Factoring. P.3 an 1. Polynomial STUDY TIP. Example 1 Writing Polynomials in Standard Form. What you should learn

P.3 Polynomials and Factoring. P.3 an 1. Polynomial STUDY TIP. Example 1 Writing Polynomials in Standard Form. What you should learn 33337_0P03.qp 2/27/06 24 9:3 AM Chpter P Pge 24 Prerequisites P.3 Polynomils nd Fctoring Wht you should lern Polynomils An lgeric epression is collection of vriles nd rel numers. The most common type of

More information

Physics 6010, Fall 2010 Symmetries and Conservation Laws: Energy, Momentum and Angular Momentum Relevant Sections in Text: 2.6, 2.

Physics 6010, Fall 2010 Symmetries and Conservation Laws: Energy, Momentum and Angular Momentum Relevant Sections in Text: 2.6, 2. Physics 6010, Fll 2010 Symmetries nd Conservtion Lws: Energy, Momentum nd Angulr Momentum Relevnt Sections in Text: 2.6, 2.7 Symmetries nd Conservtion Lws By conservtion lw we men quntity constructed from

More information

and thus, they are similar. If k = 3 then the Jordan form of both matrices is

and thus, they are similar. If k = 3 then the Jordan form of both matrices is Homework ssignment 11 Section 7. pp. 249-25 Exercise 1. Let N 1 nd N 2 be nilpotent mtrices over the field F. Prove tht N 1 nd N 2 re similr if nd only if they hve the sme miniml polynomil. Solution: If

More information

Binary Representation of Numbers Autar Kaw

Binary Representation of Numbers Autar Kaw Binry Representtion of Numbers Autr Kw After reding this chpter, you should be ble to: 1. convert bse- rel number to its binry representtion,. convert binry number to n equivlent bse- number. In everydy

More information

4 Geometry: Shapes. 4.1 Circumference and area of a circle. FM Functional Maths AU (AO2) Assessing Understanding PS (AO3) Problem Solving HOMEWORK 4A

4 Geometry: Shapes. 4.1 Circumference and area of a circle. FM Functional Maths AU (AO2) Assessing Understanding PS (AO3) Problem Solving HOMEWORK 4A Geometry: Shpes. Circumference nd re of circle HOMEWORK D C 3 5 6 7 8 9 0 3 U Find the circumference of ech of the following circles, round off your nswers to dp. Dimeter 3 cm Rdius c Rdius 8 m d Dimeter

More information

LECTURE #05. Learning Objective. To describe the geometry in and around a unit cell in terms of directions and planes.

LECTURE #05. Learning Objective. To describe the geometry in and around a unit cell in terms of directions and planes. LECTURE #05 Chpter 3: Lttice Positions, Directions nd Plnes Lerning Objective To describe the geometr in nd round unit cell in terms of directions nd plnes. 1 Relevnt Reding for this Lecture... Pges 64-83.

More information

RIGHT TRIANGLES AND THE PYTHAGOREAN TRIPLETS

RIGHT TRIANGLES AND THE PYTHAGOREAN TRIPLETS RIGHT TRIANGLES AND THE PYTHAGOREAN TRIPLETS Known for over 500 yers is the fct tht the sum of the squres of the legs of right tringle equls the squre of the hypotenuse. Tht is +b c. A simple proof is

More information

Unit 6: Exponents and Radicals

Unit 6: Exponents and Radicals Eponents nd Rdicls -: The Rel Numer Sstem Unit : Eponents nd Rdicls Pure Mth 0 Notes Nturl Numers (N): - counting numers. {,,,,, } Whole Numers (W): - counting numers with 0. {0,,,,,, } Integers (I): -

More information

Reasoning to Solve Equations and Inequalities

Reasoning to Solve Equations and Inequalities Lesson4 Resoning to Solve Equtions nd Inequlities In erlier work in this unit, you modeled situtions with severl vriles nd equtions. For exmple, suppose you were given usiness plns for concert showing

More information

Lecture 15 - Curve Fitting Techniques

Lecture 15 - Curve Fitting Techniques Lecture 15 - Curve Fitting Techniques Topics curve fitting motivtion liner regression Curve fitting - motivtion For root finding, we used given function to identify where it crossed zero where does fx

More information

Brillouin Zones. Physics 3P41 Chris Wiebe

Brillouin Zones. Physics 3P41 Chris Wiebe Brillouin Zones Physics 3P41 Chris Wiebe Direct spce to reciprocl spce * = 2 i j πδ ij Rel (direct) spce Reciprocl spce Note: The rel spce nd reciprocl spce vectors re not necessrily in the sme direction

More information

Econ 4721 Money and Banking Problem Set 2 Answer Key

Econ 4721 Money and Banking Problem Set 2 Answer Key Econ 472 Money nd Bnking Problem Set 2 Answer Key Problem (35 points) Consider n overlpping genertions model in which consumers live for two periods. The number of people born in ech genertion grows in

More information

Regular Sets and Expressions

Regular Sets and Expressions Regulr Sets nd Expressions Finite utomt re importnt in science, mthemtics, nd engineering. Engineers like them ecuse they re super models for circuits (And, since the dvent of VLSI systems sometimes finite

More information

CHAPTER 11 Numerical Differentiation and Integration

CHAPTER 11 Numerical Differentiation and Integration CHAPTER 11 Numericl Differentition nd Integrtion Differentition nd integrtion re bsic mthemticl opertions with wide rnge of pplictions in mny res of science. It is therefore importnt to hve good methods

More information

Review Problems for the Final of Math 121, Fall 2014

Review Problems for the Final of Math 121, Fall 2014 Review Problems for the Finl of Mth, Fll The following is collection of vrious types of smple problems covering sections.,.5, nd.7 6.6 of the text which constitute only prt of the common Mth Finl. Since

More information

15.6. The mean value and the root-mean-square value of a function. Introduction. Prerequisites. Learning Outcomes. Learning Style

15.6. The mean value and the root-mean-square value of a function. Introduction. Prerequisites. Learning Outcomes. Learning Style The men vlue nd the root-men-squre vlue of function 5.6 Introduction Currents nd voltges often vry with time nd engineers my wish to know the verge vlue of such current or voltge over some prticulr time

More information

AAPT UNITED STATES PHYSICS TEAM AIP 2010

AAPT UNITED STATES PHYSICS TEAM AIP 2010 2010 F = m Exm 1 AAPT UNITED STATES PHYSICS TEAM AIP 2010 Enti non multiplicnd sunt preter necessittem 2010 F = m Contest 25 QUESTIONS - 75 MINUTES INSTRUCTIONS DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD

More information

Bayesian Updating with Continuous Priors Class 13, 18.05, Spring 2014 Jeremy Orloff and Jonathan Bloom

Bayesian Updating with Continuous Priors Class 13, 18.05, Spring 2014 Jeremy Orloff and Jonathan Bloom Byesin Updting with Continuous Priors Clss 3, 8.05, Spring 04 Jeremy Orloff nd Jonthn Bloom Lerning Gols. Understnd prmeterized fmily of distriutions s representing continuous rnge of hypotheses for the

More information

4.11 Inner Product Spaces

4.11 Inner Product Spaces 314 CHAPTER 4 Vector Spces 9. A mtrix of the form 0 0 b c 0 d 0 0 e 0 f g 0 h 0 cnnot be invertible. 10. A mtrix of the form bc d e f ghi such tht e bd = 0 cnnot be invertible. 4.11 Inner Product Spces

More information

Homework 3 Solutions

Homework 3 Solutions CS 341: Foundtions of Computer Science II Prof. Mrvin Nkym Homework 3 Solutions 1. Give NFAs with the specified numer of sttes recognizing ech of the following lnguges. In ll cses, the lphet is Σ = {,1}.

More information

Module Summary Sheets. C3, Methods for Advanced Mathematics (Version B reference to new book) Topic 2: Natural Logarithms and Exponentials

Module Summary Sheets. C3, Methods for Advanced Mathematics (Version B reference to new book) Topic 2: Natural Logarithms and Exponentials MEI Mthemtics in Ection nd Instry Topic : Proof MEI Structured Mthemtics Mole Summry Sheets C, Methods for Anced Mthemtics (Version B reference to new book) Topic : Nturl Logrithms nd Eponentils Topic

More information

Lectures 8 and 9 1 Rectangular waveguides

Lectures 8 and 9 1 Rectangular waveguides 1 Lectures 8 nd 9 1 Rectngulr wveguides y b x z Consider rectngulr wveguide with 0 < x b. There re two types of wves in hollow wveguide with only one conductor; Trnsverse electric wves

More information

Slow roll inflation. 1 What is inflation? 2 Equations of motions for a homogeneous scalar field in an FRW metric

Slow roll inflation. 1 What is inflation? 2 Equations of motions for a homogeneous scalar field in an FRW metric Slow roll infltion Pscl udrevnge pscl@vudrevnge.com October 6, 00 Wht is infltion? Infltion is period of ccelerted expnsion of the universe. Historiclly, it ws invented to solve severl problems: Homogeneity:

More information

Newton s Three Laws. d dt F = If the mass is constant, this relationship becomes the familiar form of Newton s Second Law: dv dt

Newton s Three Laws. d dt F = If the mass is constant, this relationship becomes the familiar form of Newton s Second Law: dv dt Newton s Three Lws For couple centuries before Einstein, Newton s Lws were the bsic principles of Physics. These lws re still vlid nd they re the bsis for much engineering nlysis tody. Forml sttements

More information

Harvard College. Math 21a: Multivariable Calculus Formula and Theorem Review

Harvard College. Math 21a: Multivariable Calculus Formula and Theorem Review Hrvrd College Mth 21: Multivrible Clculus Formul nd Theorem Review Tommy McWillim, 13 tmcwillim@college.hrvrd.edu December 15, 2009 1 Contents Tble of Contents 4 9 Vectors nd the Geometry of Spce 5 9.1

More information

69. The Shortest Distance Between Skew Lines

69. The Shortest Distance Between Skew Lines 69. The Shortest Distnce Between Skew Lines Find the ngle nd distnce between two given skew lines. (Skew lines re non-prllel non-intersecting lines.) This importnt problem is usully encountered in one

More information

Section 7-4 Translation of Axes

Section 7-4 Translation of Axes 62 7 ADDITIONAL TOPICS IN ANALYTIC GEOMETRY Section 7-4 Trnsltion of Aes Trnsltion of Aes Stndrd Equtions of Trnslted Conics Grphing Equtions of the Form A 2 C 2 D E F 0 Finding Equtions of Conics In the

More information

Math Review 1. , where α (alpha) is a constant between 0 and 1, is one specific functional form for the general production function.

Math Review 1. , where α (alpha) is a constant between 0 and 1, is one specific functional form for the general production function. Mth Review Vribles, Constnts nd Functions A vrible is mthemticl bbrevition for concept For emple in economics, the vrible Y usully represents the level of output of firm or the GDP of n economy, while

More information

CHAPTER 6 MAGNETIC EFFECT OF AN ELECTRIC CURRENT

CHAPTER 6 MAGNETIC EFFECT OF AN ELECTRIC CURRENT CHAPTER 6 MAGNETIC EFFECT OF AN ELECTRIC CURRENT 6. Introduction Most of us re fmilir with the more obvious properties of mgnets nd compss needles. A mgnet, often in the form of short iron br, will ttrct

More information

Lecture 5. Inner Product

Lecture 5. Inner Product Lecture 5 Inner Product Let us strt with the following problem. Given point P R nd line L R, how cn we find the point on the line closest to P? Answer: Drw line segment from P meeting the line in right

More information

Finite Automata. Informatics 2A: Lecture 3. John Longley. 25 September School of Informatics University of Edinburgh

Finite Automata. Informatics 2A: Lecture 3. John Longley. 25 September School of Informatics University of Edinburgh Lnguges nd Automt Finite Automt Informtics 2A: Lecture 3 John Longley School of Informtics University of Edinburgh jrl@inf.ed.c.uk 25 September 2015 1 / 30 Lnguges nd Automt 1 Lnguges nd Automt Wht is

More information

Understanding 22. 23. The frictional force acting to the left is missing. It is equal in magnitude to the applied force acting to the right.

Understanding 22. 23. The frictional force acting to the left is missing. It is equal in magnitude to the applied force acting to the right. Chpter 3 Review, pges 154 159 Knowledge 1. (c) 2. () 3. (d) 4. (d) 5. (d) 6. (c) 7. (b) 8. (c) 9. Flse. One newton is equl to 1 kg /s 2. 10. Flse. A norl force is perpendiculr force cting on n object tht

More information

Strong acids and bases

Strong acids and bases Monoprotic Acid-Bse Equiliri (CH ) ϒ Chpter monoprotic cids A monoprotic cid cn donte one proton. This chpter includes uffers; wy to fi the ph. ϒ Chpter 11 polyprotic cids A polyprotic cid cn donte multiple

More information

Helicopter Theme and Variations

Helicopter Theme and Variations Helicopter Theme nd Vritions Or, Some Experimentl Designs Employing Pper Helicopters Some possible explntory vribles re: Who drops the helicopter The length of the rotor bldes The height from which the

More information

Homework #4: Answers. 1. Draw the array of world outputs that free trade allows by making use of each country s transformation schedule.

Homework #4: Answers. 1. Draw the array of world outputs that free trade allows by making use of each country s transformation schedule. Text questions, Chpter 5, problems 1-5: Homework #4: Answers 1. Drw the rry of world outputs tht free trde llows by mking use of ech country s trnsformtion schedule.. Drw it. This digrm is constructed

More information

MODULE 3. 0, y = 0 for all y

MODULE 3. 0, y = 0 for all y Topics: Inner products MOULE 3 The inner product of two vectors: The inner product of two vectors x, y V, denoted by x, y is (in generl) complex vlued function which hs the following four properties: i)

More information