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

Chpter 4 The Electric Potentil 4.1 The Importnt Stuff 4.1.1 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

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 = 0. 4.1.2 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) = 1.60 10 19 C (1V) = 1.60 10 19 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 (1813 1743) who did some electricl experiments in... um... Bologn. Tht s it, Bologn.

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: 4.1.3 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. 4.1.4 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) 4.1.5 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.

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) 4.1.6 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. 4.1.7 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 13 +... = 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 4.2.1 Electric Potentil 1. The electric potentil difference between the ground nd cloud in prticulr thunderstorm is 1.2 10 9 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?

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 = 1.2 10 9 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(1.2 10 9 V) = 1.2 10 9 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 (0.10 10 6 C m = 2 2(8.85 10 12 C 2 ) = 5.64 103 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 = 5.64 10 3 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 = 5.64 10 3 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) = (5.64 10 3 N ) = 8.8 10 3 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 3.9 10 15 N cts on n electron plced nywhere between the two pltes.

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 = (3.9 10 15 N) (1.60 10 19 C) = 2.4 104 N C = 2.4 104 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 = (2.4 10 4 V m )(0.12m) = 2.9 103 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

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. 4.2. 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. 4.15 into two prts. When we go from to R we need

60 CHAPTER 4. THE ELECTRIC POTENTIAL to use Eq. 4.17 for E r (r ). When we go from R to r we need to use Eq. 4.16 for E r (r ). So from Eq. 4.15 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

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 3.3 10 9 C. V = 1 Q R 12 C2 = 4π(8.85 10 )(200V)(0.15m) N m 2 = 3.3 10 9 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 = (3.3 10 9 C) 2 4π(0.15 m)2 = 1.2 10 8 C m 2 The chrge density on the sphere s surfce is 1.2 10 8 C m 2. 7. An empty hollow metl sphere hs potentil of +400V with respect to ground (defined to be t V = 0) nd hs chrge of 5.0 10 9 C. Find the electric potentil t the center of the sphere. The problem is digrmmed in Fig. 4.3. 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

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. 4.4. 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) (8.99 10 9 N m 2 C 2 ) = 2.5 10 8 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.

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. 4.2.2 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. 4.13 we find: U = = 1 ( e)( e) r 1 4π(8.85 10 12 C 2 = 1.15 10 19 J N m 2 ) (1.60 10 19 C) 2 (2.00 10 9 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. 4.14. (Tht is, sum the potentil energies k q iq j r ij over ll pirs of

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. 4.5. 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) = q2 1 + 1 ) 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

4.2. WORKED EXAMPLES 65 q 1 - A B + q 2 Figure 4.7: Chrge configurtion for Exmple 12. = q 2 ( 2 + 1 ) 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 ( = ( = 4 + 2 ) 2 q 2 1 1 + 1 2 2 + 1 2 ) 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. 4.14. 12. 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. 4.10. 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 = (8.99 10 9 N m2 C 2 ) [ ( 5.0 10 6 C) (15 10 2 m) + (+2.0 ] 10 6 C) = 6.0 10 4 V (5.0 10 2 m)

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: [ ( 5.0 10 V = (8.99 10 9 6 C) N m2 ) C 2 (5.0 10 2 m) + (+2.0 ] 10 6 C) = 7.8 10 5 V (15 10 2 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 = 6.0 10 4 V ( 7.8 10 5 V) = 8.4 10 5 V The chnge in potentil energy for +3.0µC chrge to move from B to A is U = q V = (3.0 10 6 C)(8.4 10 5 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 = (8.99 10 9 N m2 C 2 ) (5.00 10 6 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).

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 = (8.99 109 ) (5.00 10 6 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) (5.00 10 3 kg) = 45.0 m s 2 (0.225 N) (10.0 10 3 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)

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 + 1 2 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 = (3.75 10 3 kg)v 2 A = 0.225J v 2 A = 0.225 J 3.75 10 3 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. 4.14. 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

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 (1.60 10 19 C) 2 π(8.85 10 12 C 2 N m 2 )(9.11 10 31 kg)(1.0 10 2 m) = 1.01 105 m2 s 2 which gives v 0 = 3.18 10 2 m s

70 CHAPTER 4. THE ELECTRIC POTENTIAL