4. Compare the electric force holding the electron in orbit ( r = 0.53

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Frederick Williams
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1 Electostatics WS Electic Foce an Fiel. Calculate the magnitue of the foce between two 3.60-µ C point chages 9.3 cm apat.. How many electons make up a chage of 30.0 µ C? 3. Two chage ust paticles exet a foce of 3. 0 N on each othe. What will be the foce if they ae move so they ae only one-eighth as fa apat? 0 4. Compae the electic foce holing the electon in obit ( = m) aoun the poton nucleus of the hyogen atom, with the gavitational foce between the same electon an poton. What is the atio of these two foces? 5. A chage of 6.00 mc is place at each cone of a suae 0.00 m on a sie. Detemine the magnitue an iection of the foce on each chage. 6. A µc an a 3.55 µ C chage ae place 8.5 cm apat. Whee can a thi chage be place so that it expeiences no net foce? 7. What ae the magnitue an iection of the electic foce on an electon in a unifom electic fiel of stength 360 N C that points ue east? 8. What is the magnitue of the acceleation expeience by an electon in an electic fiel of 750 N C? How oes the iection of the acceleation epen on the iection of the fiel at that point? Electic Potential an Potential Enegy 9. How much wok oes the electic fiel o in moving a 7.7 µ C chage fom goun to a point whose potential is + 55 V highe? 0. An electon acuies J of kinetic enegy when it is acceleate by an electic fiel fom plate A to plate B. What is the potential iffeence between the plates, an which plate is at the highe potential?. Two paallel plates, connecte to a 00-V powe supply, ae sepaate by an ai gap. How small can the gap be if the ai is not to become conucting by exceeing its 6 beakown value of E = 3 0 V m?. What is the spee of a poton whose kinetic enegy is 3. kev? 3. A + 35 µ C point chage is place 3 cm fom an ientical + 35 µ C chage. How much wok woul be euie to move a µ C test chage fom a point miway between them to a point cm close to eithe of the chages?

2 4. An electon stats fom est 3.5 cm fom a fixe point chage with = 0.5 µ C. How fast will the electon be moving when it is vey fa away? 5. Two point chages, 3.0 µ C an.0 µ C, ae place 5.0 cm apat on the x axis. At what points along the x axis is (a) the electic fiel zeo an (b) the potential zeo? Let V = 0 at =.

3 Key. Use Coulomb s law to calculate the magnitue of the foce. ( C ) 9 ( m) F = k = N m C = 3.47 N 3 N. Use the chage pe electon to fin the numbe of electons. electon 4 ( C) =.87 0 electons C 3. Since the magnitue of the foce is invesely popotional to the suae of the sepaation istance, F, if the istance is multiplie by a facto of /8, the foce will be multiplie by a facto of F = 64F = N =.0 N 4. Take the atio of the electic foce ivie by the gavitational foce. k 9 9 F E k N m C.60 0 C = = = = F mm G Gmm ( N m kg )( 9. 0 kg)(.67 0 kg G ) The electic foce is about the given scenaio times stonge than the gavitational foce fo 5. Detemine the foce on the uppe ight chage, an then use the symmety of the configuation to etemine the foce on the othe thee chages. The foce at the uppe ight cone of the suae is the vecto sum of the foces ue to the othe thee chages. Let the vaiable epesent the 0.00 m length of a sie of the suae, an let the vaiable epesent the 6.00 mc chage at each cone. 6. Assume that the negative chage is = 8.5 cm to the ight of the positive chage, on the x- axis. To expeience no net foce, 4.7 µc 3.5 µc the thi chage must be close to the smalle magnitue + chage (the negative chage). The thi chage cannot be x between the chages, because it woul expeience a foce fom each chage in the same iection, an so the net foce coul not be zeo. An the thi chage must be on the line joining the othe two chages, so that the two foces on the thi chage ae along the same line. See the iagam. Euate the magnitues of the two foces on the thi chage, an solve fo x > 0. F = F k = k x = ( + x) x ( ) 39 3 F F F 4 x C = = = ( ) ( 8.5cm ) ( C C ) 6 cm 3

4 7. Use E. 6 3 to calculate the foce. F 9 E= F = E= (.60 0 C)( 360 N C east) = N west 8. Assuming the electic foce is the only foce on the electon, then Newton s n law may be use to fin the acceleation. 9 (.60 0 C) 4 F = ma= E a = E = net N C =.3 0 m s m 9. 0 kg Since the chage is negative, the iection of the acceleation is opposite to the fiel. 9. The wok one by the electic fiel can be foun fom E. 7-b. W 4 V = W = V = ( C)( + 55 V) = 4. 0 J 0. The kinetic enegy gaine by the electon is the wok one by the electic foce. Use E. 7-b to calculate the potential iffeence. 7 W J V = = = 466 V C The electon moves fom low potential to high potential, so plate B is at the highe potential.. Fin the istance coesponing to the maximum electic fiel, using E. 7-4b. V V 00 V 5 5 E = = = = m 7 0 m 6 E 3 0 V m. The kinetic enegy of the poton is given. Use the kinetic enegy to fin the spee. 3 9 ( )( ) KE 3. 0 ev.60 0 J ev 5 mv = KE v = = = m s 7 m.67 0 kg 3. The wok euie is the iffeence in potential enegy between the two locations. The test chage has potential enegy ue to each of the othe chages, given in Conceptual Example 7-7 as PE = k. So to fin the wok, calculate the iffeence in potential enegy between the two locations. Let epesent the 35µ C chage, let epesent the 0.50µ C test chage, an let epesent the 3 cm istance. k k k k PE = + PE = + initial final 0. m + 0. m [ ] [ ] 4. By enegy consevation, all of the initial potential enegy will change to kinetic enegy of the electon when the electon is fa away. The othe chage is fixe, an so has no kinetic enegy. When the electon is fa away, thee is no potential enegy. k( e)( ) E = E PE = KE = mv initial final initial final ( ) ( Ngm C )(.60 0 C)(.5 0 C) k e v = = m = m s 3 ( 9. 0 kg)( 0.35 m) 4

5 5. (a) Because of the invese suae natue of the x electic fiel, any location whee the fiel is zeo must be close to the weake chage ( ). Also, in between the two < 0 chages, the fiels ue to the two chages ae paallel to each othe (both to the left) an cannot cancel. Thus the only places whee the fiel can be zeo ae close to the weake chage, but not between them. In the iagam, this is the point labele as x. Take to the ight as the positive iection. E = k k = 0 ( + x) = x x + x.0 0 C x = = ( 5.0cm) = cm left of C.0 0 C > 0 (b) The potential ue to the positive chage is positive eveywhee, an the potential ue to the negative chage is negative eveywhee. Since the negative chage is smalle in magnitue than the positive V V chage, any point whee the potential is zeo must be close to the negative chage. So consie locations between the chages (position x ) an to the left of the negative chage (position x ) as shown in the iagam. (.0 0 C)( 5.0 cm) k k = + = 0 x = = =.0 cm location 6 ( x) x ( ) ( C) (.0 0 C)( 5.0 cm) k k = + = 0 x = = = 0.0 cm location 6 ( + x) x ( + ) (.0 0 C) So the two locations whee the potential is zeo ae.0 cm fom the negative chage towas the positive chage, an 0.0 cm fom the negative chage away fom the positive chage. x < 0 x > 0 5

SPH4UI 28/02/2011. Total energy = K + U is constant! Electric Potential Mr. Burns. GMm

SPH4UI 28/02/2011. Total energy = K + U is constant! Electric Potential Mr. Burns. GMm 8//11 Electicity has Enegy SPH4I Electic Potential M. Buns To sepaate negative an positive chages fom each othe, wok must be one against the foce of attaction. Theefoe sepeate chages ae in a higheenegy