Electric Potential. Concepts and Principles. An Alternative Approach. A Gravitational Analogy

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1 . Electric Potentil Concepts nd Principles An Alterntive Approch The electric field surrounding electric chrges nd the mgnetic field surrounding moving electric chrges cn both be conceptulized s informtion embedded in spce. In both cses, the informtion is embedded s vectors, detiling both the mgnitude nd direction of ech field. Moreover, when this informtion is red by other moving electric chrges, the result is force cting on the chrge. These forces cn be clculted to llow us to determine the subsequent motion of the chrge. Just s in mechnics, there is n lterntive to this force-bsed pproch to nlyzing the behvior of electric chrges. In this chpter I will define new field, the electric potentil, which surrounds every electric chrge. This field differs from the electric field becuse it is sclr field, mening tht this field hs only mgnitude t every point in spce nd no ssocited direction. Moreover, the informtion in this field, when red by other electric chrges, does not result in force on the chrge but rther determines the electric potentil energy the chrge possesses t tht point in spce. Understnding how to clculte this new field, how this field reltes to energy, nd how this field is relted to the electric field will be the focus of this chpter. A Grvittionl Anlogy Rther thn thinking in terms of the grvittionl force nd Newton s Second Lw, n lterntive wy to exmine mechnics scenrios is by using the concept of grvittionl potentil energy nd the conservtion of energy. In the force pproch, we envision vector field surrounding the erth, regrdless of whether second mss is nerby to interct with this field. If mss is present, the mss intercts with this field nd feels grvittionl force. This ide is cptured in the eqution: mg F G 1

2 In the energy pproch, we cn envision sclr field, the grvittionl potentil, which is present regrdless of whether second mss is nerby to interct with this field. If mss is present, the mss intercts with this field nd hs grvittionl potentil energy. Ner the surfce of the erth, the fmilir expression for grvittionl potentil energy: U G mgh cn be thought of s the product of the mss of the object nd this pre-existing grvittionl potentil, G : if we define the grvittionl potentil by: U G m G G gh Although we didn t use the concept of grvittionl potentil while studying mechnics, it will prove to be very useful concept in our study of electricl phenomenon. The generl expression for grvittionl potentil, vlid regrdless of distnce from mssive object, is: G Gm r In summry, just s mss will interct with the vector grvittionl field s force, mss will interct with the sclr grvittionl potentil field s potentil energy. The sitution is very similr for electricl phenomenon. We cn envision sclr field, the electric potentil, which is present regrdless of whether second chrge is nerby to interct with this field. If chrge is present, the chrge intercts with this field nd hs electric potentil energy. The electric potentil, E, is defined by the reltionship: kq E r where q is the source chrge, the electric chrge tht cretes the field, nd r is the distnce between the source chrge nd the point of interest. This leds to n expression for the electric potentil energy of: U E q E where q is the chrge on the prticle of interest, the chrge tht is intercting with the field, nd E is the net electric potentil t the loction of the prticle of interest (creted by ll of the other chrged prticles in the universe).

3 We will typiclly leve the subscript off the electric potentil nd electric potentil energy unless the possibility of confusion with the grvittionl potentil nd potentil energy re present. Relting the Electric Field nd the Electric Potentil The electric field nd the electric potentil re not two, independent fields. They re two independent wys of conceptulizing the effect tht n electric chrge hs on the spce surrounding it. Just s problems in mechnics cn be nlyzed using force-pproch or n energy-pproch, problems deling with electricl phenomenon cn be nlyzed by focusing on the electric field or on the electric potentil. Additionlly, just s it is sometimes necessry in mechnics to trnsfer between force nd energy representtions, it is sometimes necessry to trnsfer between the electric field nd electric potentil representtions. The reltionship between the two fields cn be understood by exmining the expression for work, which reltes force to trnsfer of energy. Utilizing the dot product, the work done in moving prticle from n initil point, i, to finl point, f, cn be written s: W i f f F dl i where dl is n infinitesiml portion of the pth long which the prticle moves. You my lso recll tht the difference in potentil energy between initil nd finl loctions is defined s the opposite of the work needed to move the prticle between the two points: U U f U i W i f Putting these two ides together yields: U f F dl i We cn now use this result to relte the electric potentil nd the electric field. Substituting in expressions for potentil energy nd force in terms of the fields tht convey them leds to: ( q ) f i f i ( qe ) dl E dl In English, this finl result sttes tht the electric potentil difference between ny two points is defined s the negtive of the integrl of the electric field long pth connected the two points. (I m sure tht doesn t seem like prticulrly cler English, but this ide will become more tngible once you get to work on some of the fun ctivities in this chpter.) The bottom line is tht the electric potentil cn be determined by integrting the electric field, nd, conversely, the electric field cn be determined by differentiting the electric potentil. 3

4 . Electric Potentil Anlysis Tools Point Chrges Find the electric potentil t the indicted point. The chrges re seprted by distnce 4. +q -q The electric potentil t the point specified will be the sum of the electric potentil from the left chrge ( L ) nd the electric potentil from the right chrge ( R ). L k( q) (3) R ( ) kq kq 10 1 kq ( ) 10 kq k( q) ( ) ( ) Notice tht since the electric potentil is sclr, clculting the electric potentil is often much esier thn clculting the electric field. 4

5 Continuous Chrge Distribution The plstic rod of length L t right hs uniform chrge density. Find the electric potentil t ll points to the right of the rod on the x-xis. Since electric chrge is discrete, the electric potentil cn lwys be clculted by summing the electric potentil from ech of the electrons nd protons tht mke up n object. However, mcroscopic objects contin lot of electrons nd protons, so this summtion hs mny, mny terms: every electron nd proton kq r As described erlier, we will replce this summtion over very lrge number of discrete chrges with n integrl over hypotheticlly continuous distribution of chrge. This leds to reltionship for the electric potentil t prticulr point in spce, from continuous distribution of chrge, of: k( dq) r where dq is the chrge on infinitesimlly smll portion of the object, nd the integrl is over the entire physicl object. The steps for finding the electric potentil from continuous distribution of chrge re: 1. Crefully identify nd lbel the loction of the differentil element on digrm of the sitution.. Crefully identify nd lbel the loction of the point of interest on digrm of the sitution. 3. Write n expression for dq, the chrge on the differentil element. 4. Write n expression for r, the distnce between the differentil element nd the point of interest. 5. Insert your expressions into the integrl for the electric potentil. 6. Crefully choose the limits of integrtion. 7. Evlute the integrl. I ll demonstrte below ech of these steps for the scenrio under investigtion. 5

6 1. Crefully identify nd lbel the loction of the differentil element on digrm of the sitution. The differentil element is smll (infinitesiml) piece of the object tht we will tret like point chrge. The loction of this differentil element must be rbitrry, mening it is not t specil loction like the top, middle, or bottom of the rod. Its loction must be represented by vrible, where this vrible is the vrible of integrtion nd determines the limits of the integrl. For this exmple, select the differentil element to be locted distnce s bove the center of the rod. The length of this element is ds. (Lter, you will select the limits of integrtion to go from L/ to +L/ to llow this rbitrry element to cover the entire rod.) s. Crefully identify nd lbel the loction of the point of interest on digrm of the sitution. You re interested in the electric potentil t ll points long the x-xis to the right of the rod. Therefore, select n rbitrry loction long the x-xis nd lbel it with its loction. s x 3. Write n expression for dq, the chrge on the differentil element. The rod hs uniform chrge density, mening the mount of chrge per unit length long the rod is constnt. Since the differentil element hs length ds, the totl chrge on this element (dq) is the product of the density nd the length: dq ds 4. Write n expression for r, the distnce between the differentil element nd the point of interest. By Pythgors theorem, the distnce between the differentil element nd the point of interest is: r x s 6. Insert your expressions into the integrl for the electric potentil. k( dq) r kds x s 6

7 7 7. Crefully choose the limits of integrtion. The limits of integrtion re determined by the rnge over which the differentil element must be moved to cover the entire object. The loction of the element must vry between the bottom of the rod (-L/) nd the top of the rod (+L/) in order to include every prt of the rod. The two ends of the rod form the two limits of integrtion. / / L L s x ds k 8. Evlute the integrl. ) 4 4 ln( ) ln( / / / / L x L L x L k s x s k s x ds k L L L L This expression looks dunting, but its limiting behviors (s x pproches zero nd x pproches infinity) re correct. Consider the cse where x gets smller nd smller, pproching zero: ) ln( ) 4 (0) 4 (0) ln( L L L L k L L L L k This expression diverges (goes to infinity) s x becomes close to zero, s you should expect since the electric potentil directly on n electric chrge is infinite.

8 8 As x gets lrger nd lrger, the L/ terms become insignificnt: ) ln( ) ln( ) ln( ) 4 4 ln( x x k x L x L k x L x L k L x L L x L k This expression becomes zero s x becomes infinite, s you should expect since the electric potentil extremely fr from n electric chrge is zero. Potentil Difference The long, hollow plstic cylinder t right hs inner rdius, outer rdius, nd uniform chrge density. Find the electric potentil difference between the inner nd outer rdius. An lterntive method for clculting the electric potentil t point, or the electric potentil difference between two points, is by using knowledge of the electric field. The following reltion, l d E sttes tht the potentil difference between two points cn be determined by integrting the component of the electric field tht lies long the pth connecting the two points. This mens tht if you know the electric field in region of spce, you cn esily (more or less) find the potentil difference between ny two points tht lie in tht region. This reltion is prticulrly useful in conjunction with Guss Lw for situtions with cylindricl or sphericl symmetry.

9 9 To find the potentil difference between points nd b (i.e., wht voltmeter would red if connected cross points nd b), we need the electric field in this region. Guss Lw cn be used to find tht (review Guss Lw if this step is little fuzzy): r r E 0 ) ( This electric field points rdilly wy from the center of the cylinder. For simplicity, choose pth tht directly connects to, i.e., rdil pth. r r dr r r dr r r drr r r r dl E ln 1 ) ( ) ( ˆ ˆ ) ( ln 3 ln 3 ln ) ) (( 1 Notice tht this method directly clcultes the difference in electric potentil between two points, without ctully determining the vlue of the electric potentil t either point. Since electric potentil is relted to electric potentil energy, this method llows to you to find the difference in energy between two points but not the ctul vlue of the energy of n electric chrge. This should strike you s quite similr to the grvittionl cse. For grvittionl potentil energy, the choice of the zero-point is rbitrry nd only energy differences led to chnges in kinetic energy. For electricl potentil energy, the sitution is identicl. The zero-point of electric potentil energy (nd electric potentil) is typiclly tken t infinity, lthough you cn zero the potentil t more convenient point by grounding the system t tht point. The physicl ct of grounding point on electricl device is mthemticlly equivlent to setting the potentil equl to zero t tht point.

10 Electric Potentil Energy In mny pplictions, oppositely chrged prllel pltes (with smll holes cut for the bem to pss through) re used to ccelerte bems of chrged prticles. In this exmple, proton is injected t v i into the spce between the pltes. The potentil difference between the pltes is (the left plte is t higher potentil). Wht is the velocity of the proton s it exits the device? Since the electric potentil chnges s the proton moves from the left plte to the right plte, its potentil energy chnges. This chnge in potentil energy results in chnge in kinetic energy of the proton by energy conservtion. K U 1 mv 1 mv v f i i i i q e( v K i f Left Left U e( ) m f 1 mv Right f ) q 1 mv Right f 10

11 . Electric Potentil Activities 11

12 The two chrges below re seprted by distnce 4. Sketch grphs of electric potentil long the lines indicted.. At y = 0 +q +q Electric potentil is inversely proportionl to the distnce from the source. Thus, the potentil diverges s the distnce from the source goes to zero, which occurs t x = ±, nd goes to zero s the distnce becomes very lrge. Between the chrges, the potentil stys positive nd finite. x b. At y = x c. At y = 4 x 1

13 The two chrges below re seprted by distnce 4. Sketch grphs of electric potentil long the lines indicted. +q +q. At x = 0 y b. At x = y c. At x = 4 y 13

14 The two chrges below re seprted by distnce 4. Sketch grphs of electric potentil long the lines indicted.. At y = 0 +q -q The potentil must go to + t x = - nd - t x = +, nd lso go to zero t lrge distnces. At x = 0, the potentil is positive becuse of the lrger positive chrge, but must become negtive s you pproch the negtive chrge. The loction of the = 0 point is twice s fr from the positive chrge s it is from the negtive chrge. x b. At y = x c. At x = 0 y 14

15 The two chrges below re seprted by distnce 4. Sketch grphs of electric potentil long the lines indicted. +q +q. At x = 0 y b. At x = y c. At y = x 15

16 The two chrges below re seprted by distnce 6. Sketch grphs of electric potentil long the lines indicted. -q +q. At x = 0 y b. At y = 0 x c. At y = - x 16

17 The two chrges below re locted s shown. Ech grid squre hs width. Sketch grphs of electric potentil long the lines indicted. +q -q. At x = 0 y b. At y = 0 x c. At y = x 17

18 For ech of the pths described below, rnk the pths on the bsis of the chnge in electric potentil. b +q c d e f A B C D E F from c to d from d to c from d to e from d to b from e to f from b to Lrgest Positive 1. B. E 3. D 4. F 5. C 6. A Lrgest Negtive The rnking cnnot be determined bsed on the informtion provided. Explin the reson for your rnking: The only pths tht result in positive chnge in potentil (i.e., n increse in potentil) re B nd E. The chnge long pth B is much lrger since the potentil ctully doubles long this pth. Pth D results in no chnge in potentil since both points re equidistnt from the source chrge. Of the remining pths tht result in negtive chnge in potentil (i.e., decrese in potentil), hopefully it s cler tht pth A results in the lrgest mgnitude decrese, since the potentil is quite steep close to the chrge nd the potentil is cut in hlf long this pth. Since pths C nd F both begin t the sme vlue of potentil, pth C must hve lrger mgnitude since it ends frther from the source chrge, thus ensuring tht the potentil hs decresed by lrger mount. 18

19 For ech of the pths described below, rnk the pths on the bsis of the chnge in electric potentil. b -q c d e f A B C D E F from c to d from d to c from d to e from d to b from e to f from b to Lrgest Positive Lrgest Negtive The rnking cnnot be determined bsed on the informtion provided. Explin the reson for your rnking: 19

20 For ech of the loctions indicted below, rnk the loctions on the bsis of the electric potentil. c b +q -q e f d A B C D E F b c d e f Lrgest Positive Lrgest Negtive The rnking cnnot be determined bsed on the informtion provided. Explin the reson for your rnking: 0

21 For ech of the loctions indicted below, rnk the loctions on the bsis of the electric potentil. c b +q -q e f d A B C D E F b c d e f Lrgest Positive Lrgest Negtive The rnking cnnot be determined bsed on the informtion provided. Explin the reson for your rnking: 1

22 For ech of the pths described below, rnk the pths on the bsis of the chnge in electric potentil. c b +q -q e f d A B C D E F from b to from f to e from c to d from c to e from c to b from d to Lrgest Positive Lrgest Negtive The rnking cnnot be determined bsed on the informtion provided. Explin the reson for your rnking:

23 For ech of the pths described below, rnk the pths on the bsis of the chnge in electric potentil. c b +q -q e f d A B C D E F from b to from f to e from c to d from c to e from c to b from d to Lrgest Positive Lrgest Negtive The rnking cnnot be determined bsed on the informtion provided. Explin the reson for your rnking: 3

24 A region of spce is filled with the electric potentil illustrted below. rious positions within this region re indicted by the points through f. b f 80 d e c A n lph prticle (net chrge +e) is t rest t B n electron is t rest t b C proton is t rest t c D n electron is t rest t d E neutron is t rest t e F proton is t rest t f Rnk these prticles on the bsis of their electric potentil energy. Lrgest Positive Lrgest Negtive The rnking cnnot be determined bsed on the informtion provided. Explin the reson for your rnking: 4

25 A region of spce is filled with the electric potentil illustrted below. rious positions within this region re indicted by the points through f. Individul prticles re very slowly moved between these positions by n externl force. b f 80 d e c A proton is moved from to d B n electron is moved from d to C proton is moved from d to f D n electron is moved from b to c E neutron is moved from to e F n lph prticle (net chrge +e) is moved b to d. Rnk these motions on the bsis of the chnge in electric potentil energy of the prticle during the motion. Lrgest Positive Lrgest Negtive The rnking cnnot be determined bsed on the informtion provided. b. Rnk these motions on the bsis of the work done on the prticle by the externl force during the motion. Lrgest Positive Lrgest Negtive The rnking cnnot be determined bsed on the informtion provided. Explin the reson for your rnking: 5

26 Find the electric potentil t the three indicted points long the chin of chloride (Cl) - ions nd sodium (N + ) ions. The interionic seprtion is.8 x m. N + Cl - N + Cl -. b. c. Mthemticl Anlysis 6

27 Find the electric potentil t the three indicted points in the rry of sodium ions (N + ) nd chloride ions (Cl - ). The interionic seprtion is.8 x m. N + Cl -. b. c. Cl - N + Mthemticl Anlysis 7

28 Find the electric potentil t ech of the indicted points. The chrges re seprted by distnce 4.. b. c. +q -q d. Mthemticl Anlysis 8

29 Find the electric potentil t ech of the indicted points. The chrges re seprted by distnce 4.. b. c. +q +q d. Mthemticl Anlysis 9

30 The two chrges t right re seprted by distnce. Find the electric potentil t ll points on the x-xis nd sketch it below. +q +3q Mthemticl Anlysis for x > : for - < x < : for x < -: x 30

31 The two chrges t right re seprted by distnce. Find the electric potentil t ll points on the y-xis nd sketch it below. +q +3q Mthemticl Anlysis y 31

32 The plstic rod t right hs chrge +Q uniformly distributed long its length L. Find the electric potentil t ll points on the x-xis to the right of the rod. Mthemticl Anlysis Crefully identify nd lbel s the loction of the differentil element on the digrm bove. Crefully identify nd lbel x the loction of the point of interest on the digrm bove. Write n expression for dq, the chrge on the differentil element. Write n expression for r, the distnce between the differentil element nd the point of interest. Write n integrl expression for the electric potentil. Choosing limits of integrtion, clculte the integrl. Questions At x = L/, wht should equl? Does your function gree with this observtion? If x >> L, wht should equl? Does your function gree with this observtion? 3

33 The plstic rod t right hs chrge -Q uniformly distributed long its length L. Find the electric potentil t ll points on the y-xis bove the rod. Mthemticl Anlysis Crefully identify nd lbel s the loction of the differentil element on the digrm bove. Crefully identify nd lbel y the loction of the point of interest on the digrm bove. Write n expression for dq, the chrge on the differentil element. Write n expression for r, the distnce between the differentil element nd the point of interest. Write n integrl expression for the electric potentil. Choosing limits of integrtion, clculte the integrl. Questions At y = 0, wht should equl? Does your function gree with this observtion? If y >> L, wht should equl? Does your function gree with this observtion? 33

34 The plstic rod t right forms qurter-circle of rdius R nd hs uniform chrge density +. Find the electric potentil t the origin. Mthemticl Anlysis Crefully identify nd lbel the loction of the differentil element on the digrm bove. Crefully identify nd lbel the loction of the point of interest on the digrm bove. Write n expression for dq, the chrge on the differentil element. Write n expression for r, the distnce between the differentil element nd the point of interest. Write n integrl expression for the electric potentil. Choosing limits of integrtion, clculte the integrl. 34

35 The plstic rod t right forms hlf-circle nd hs chrge -Q uniformly distributed long its length L. Find the electric potentil t the origin. Mthemticl Anlysis Crefully identify nd lbel the loction of the differentil element on the digrm bove. Crefully identify nd lbel the loction of the point of interest on the digrm bove. Write n expression for dq, the chrge on the differentil element. Write n expression for r, the distnce between the differentil element nd the point of interest. Write n integrl expression for the electric potentil. Choosing limits of integrtion, clculte the integrl. 35

36 The plstic rod t right forms circle of rdius R nd hs chrge Q uniformly distributed long its length. Find the electric potentil t the origin. Mthemticl Anlysis Crefully identify nd lbel the loction of the differentil element on the digrm bove. Crefully identify nd lbel the loction of the point of interest on the digrm bove. Write n expression for dq, the chrge on the differentil element. Write n expression for r, the distnce between the differentil element nd the point of interest. Write n integrl expression for the electric potentil. Choosing limits of integrtion, clculte the integrl. 36

37 The solid plstic sphere t right hs rdius R nd chrge +Q uniformly distributed throughout its volume. Find the electric potentil t ll points in spce nd sketch it below, lbeling the vlue of the potentil t r = R. Mthemticl Anlysis for r > R: for r < R: R r 37

38 The solid metl sphere t right hs rdius R nd totl chrge Q. Find the electric potentil t ll points in spce nd sketch it below, lbeling the vlue of the potentil t r = R. Mthemticl Anlysis for r > R: for r < R: R r 38

39 The hollow plstic sphere t right hs inner rdius, outer rdius b, nd uniform chrge density. Find the electric potentil difference between nd b. Mthemticl Anlysis 39

40 The hollow metl sphere t right hs inner rdius, outer rdius b, nd totl chrge +Q. Find the electric potentil t the center of the sphere. Mthemticl Anlysis 40

41 A hollow metl sphere of rdius 3.0 cm is chrged to 0 k reltive to ground. Wht is the totl chrge nd chrge density on the sphere? Mthemticl Anlysis 41

42 The inner wire of the very long coxil cble t right hs dimeter of 1.5 mm nd the outer shell dimeter 1.5 cm. If the liner chrge density on the inner wire is.3 nc/m, wht potentil difference ws pplied between the wire nd the shell? Mthemticl Anlysis 4

43 A Geiger counter consists of thin inner wire nd concentric cylindricl conducting shell. The spce between is typiclly filled with n inert gs, but in this exmple it is filled with ir. The wire is chrged such tht the potentil difference between the wire nd shell is very ner 3 k/mm, the brekdown potentil for ir. If chrged prticle enters the device, its energy will cuse the ir to electriclly brekdown, sending sprk between wire nd shell. This is the distinctive click of the counter. If the inner wire hs dimeter of 50 m nd the outer shell dimeter.5 cm, find the liner chrge density on the inner wire. Mthemticl Anlysis 43

44 A sodium ion (N + ) nd chloride ion (Cl - ) re seprted by.8 x m. N + Cl - Mthemticl Anlysis. Find the electric potentil energy of the sodium ion. b. Find the electric potentil energy of the chloride ion. c. How much work would need to be done to brek the bond between the ions? d. Assume the chloride ion ws replced with second sodium ion. Find the new electric potentil energy of either sodium ion. e. Explin how the lgebric sign of the electric potentil energy cn be used to determine if chrge configurtion is bound or unbound. 44

45 Find the electric potentil energy of ech ion. The interionic seprtion is.8 x m. Cl - N + Cl - N + Mthemticl Anlysis 45

46 Find the electric potentil energy of ech ion. The interionic seprtion is.8 x m. Cl - N + N + Cl - Mthemticl Anlysis 46

47 In lph decy, n lph prticle ( bound stte of protons nd neutrons) escpes from hevy nucleus nd is propelled wy due to electric repulsion. For exmple, rdon nucleus (tomic number 86) will spontneously trnsform into polonium nucleus (tomic number 84) nd n lph prticle. Immeditely following this trnsformtion, both the polonium nucleus nd the lph prticle cn be ssumed to be t rest, nd seprted by pproximtely 50 x m. Mthemticl Anlysis. Wht is the initil electric potentil energy of the lph prticle? b. Wht is the velocity of the lph prticle when it is fr from the polonium nucleus? Since the polonium is much more mssive thn the lph, ssume the polonium stys pproximtely t rest. c. How fr from the polonium nucleus is the lph prticle when it chieves 90% of the velocity clculted in (b)? 47

48 In n ttempt to chieve nucler fusion, proton is fired t fixed deuterium nucleus. (Deuterium is n isotope of hydrogen with one proton nd one neutron in the nucleus.) If the distnce of closest pproch between the proton nd the deuterium nucleus is bout 1.0 x m, the proton nd deuterium hve good chnce of fusing. Mthemticl Anlysis. With wht speed must the proton be fired in order to hve good chnce of fusing with the deuterium? Assume the proton trvels directly towrd the deuterium nd momentrily stops t the distnce of closest pproch. b. If deuterium ws lunched t fixed trget of protons, rther thn vice-vers, wht lunch velocity would be needed for the deuterium nd protons to undergo fusion? c. If the proton is lunched with only one-hlf the speed clculted in (), how close will it get to the deuterium nucleus? 48

49 The two prticles t right, with the sme lgebric sign of chrge, re held t rest with seprtion r 0. The prticles re then relesed, nd t lter time hve seprtion, r, nd velocities, v 1 nd v. q 1 q m 1 m Mthemticl Anlysis. Apply conservtion of energy to crete reltionship between v 1 nd v. b. Apply conservtion of momentum to crete reltionship between v 1 nd v. c. Eliminte v from your system of two equtions nd determine v 1 s function of the prmeters defined bove. c. Determine the mximum speed of two protons relesed from initil seprtion 1.0 nm. 49

50 In the Bohr model of the hydrogen tom, the ground stte of hydrogen consists of proton encircled by n electron t rdius 0. All llowed orbits exist t rdii given by n 0, where n is the orbit number (1,, 3, etc.). Mthemticl Anlysis. Determine the totl energy of the electron s function of the electrosttic constnt (k), the elementry chrge (e), the orbit number (n), nd the ground stte rdius ( 0 ). (Hint: You will need to pply Newton s Second Lw to the circulr motion of the electron to determine its velocity (nd kinetic energy).) b. With 0 = 5.9 x m, numericlly evlute your expression in (), leving only the orbit number (n) unspecified. c. Determine the energy needed to excite the electron from the ground stte to the first excited stte. d. Determine the energy needed to ionize hydrogen. 50

51 In mny pplictions, oppositely chrged prllel pltes (with smll holes cut for the bem to pss through) re used to ccelerte bems of chrged prticles. In this exmple, n electron is injected t.0 x 10 7 m/s into the spce between the pltes. The pltes re 5 cm prt. Mthemticl Anlysis. If the electron exits the device t.5 x 10 7 m/s, wht is the potentil difference between the pltes? b. If proton entered the sme device t the sme speed, with wht speed would it exit the device? c. Wht is the electric field between the pltes? 51

52 In mny pplictions, oppositely chrged prllel pltes (with smll holes cut for the bem to pss through) re used to ccelerte bems of chrged prticles. In this exmple, proton is injected t.0 x 10 6 m/s into the spce between the pltes. The potentil difference between the pltes is 0 k (the left plte is t higher potentil). The pltes re 5 cm prt. Mthemticl Anlysis. Wht is the speed of the proton s it exits the device? b. If the seprtion between the pltes is incresed to 10 cm while the potentil difference is held constnt t 0 k, wht is the exit speed of the proton? c. Wht hppened to the chrge density on the pltes s the seprtion ws doubled while holding the potentil difference constnt? 5

53 A proton is ccelerted from rest through potentil difference nd then enters region of uniform mgnetic field B. Mthemticl Anlysis. Determine the rdius of the proton s pth (R) s function of the chrge (e) nd mss (m) of proton,, nd B. b. Determine the period of the proton s orbit (T) s function of the chrge (e) nd mss (m) of proton,, nd B. c. If the ccelerting potentil is doubled, wht hppens to the orbitl rdius nd period? 53

54 An electron is ccelerted from rest through potentil difference nd then enters region of uniform mgnetic field B. A trget screen is distnce D from the exit of the ccelertor. Mthemticl Anlysis. Determine the minimum ccelerting potentil () needed for the electron to strike the screen s function of the chrge (e) nd mss (m) of n electron, D, nd B. b. Evlute your expression for D = 10 cm nd B = 0.5 T. c. If the electron is ccelerted with twice the potentil clculted in (b), where on the screen does it strike? 54

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