Emple 1: Two point chge e locted on the i, q 1 = e t = 0 nd q 2 = e t =.. Find the wok tht mut be done b n etenl foce to bing thid point chge q 3 = e fom infinit to = 2. b. Find the totl potentil eneg of the tem of thee chge. Emple 1:. W foce =? = 0 = = 2 q 1 = e q 2 = e q 3 = e W field = "U = " U 2 "U 1 = " U 13 U 23 13 = 23 = " 1 q W field = " 1 q 3 1 q 2 q 3 = " 1 "e e e e 4 13 4 4 2 23 W field = " 1 e 2 4 2 = " e2 8 e2 W foce = "W field = 8 Electic otentil 1 Emple 1: b. U totl =? q 1 = e q 2 = e = 0 = U = 1 q i q j 4" i< j ij q 3 = e = 2 Emple 2: A poiton h m of 9.11 10 31 kg nd chge of e = 1.60 10 19 C. Suppoe poiton moe in the icinit of ttion lph pticle, which h chge of 2e = 3.20 10 19 C. When the poiton i 1.00 10 10 m fom the lph pticle, it i moing diectl w fom the lph pticle t peed of 3.00 10 6 m/. U totl = U 12 U 13 U 23 U totl = 1 q 1 q 2 q 1q 3 q 2q 3 4" 12 13 23 = 1 e e e e e e 4" 2. Wht i the poiton peed when the two pticle e 2.00 10 10 m pt? b. Wht i the poiton peed when it i e f w fom the lph pticle? c. Suppoe the moing pticle wee n electon heding w the lph pticle. Wht i the futhet ditnce w fom the lph pticle tht the electon eche? e2 U totl = " 8 Electic otentil 4 Emple 2: 1 2 =? q 1 = 2e q 2 = e m = 9.11"10 31 kg 2 = 2.00 "10 10 m Emple 2b: 1 2 =? q 1 = 2e q 2 = e m = 9.11"10 31 kg 2 = " 1 2 m 1 2 k q 1 q 2 1 = 1 2 m 2 2 k q 1 q 2 2 2 = 1 2 2kq 1q 2 m 1 = 1.00 "10 10 m 1 = 3.00 "10 6 m K 1 U 1 = K 2 U 2 1 " 1 1 2 = 1 2 4ke2 m 1 " 1 1 2 1 2 m 1 2 k q 1 q 2 1 = 1 2 m 2 2 k q 1 q 2 2 = 1 2 2kq 1q 2 m 1 1 = 1.00 "10 10 m 1 = 3.00 "10 6 m K 1 U 1 = K 2 U 2 2 2 = " = 1 2 4ke2 m 1 2 = 3" 10 6 m 4 9" 10 9 N m2 2 C 2 1.6 " 10 19 C 9.11" 10 31 kg 2 2 = 3.75"10 6 m 1 1" 10 10 m 1 2 " 10 10 m 2 = 3" 10 6 m 4 9 " 10 9 N m2 2 C 2 1.6 " 10 19 C 9.11" 10 31 kg 2 1" 10 10 m 2 = 4.37 "10 6 m
Emple 2c: q 1 = 2e q 2 = e m = 9.11"10 31 kg 1 = 1.00 "10 10 m 1 = 3.00 "10 6 m K 1 U 1 = K 2 U 2 1 2 m 1 2 k q 1 q 2 = 1 1 2 m 2 2 k q 1 q 2 2 2 = 0 kq 2 = 1 q 2 2kq 1 2 m 1 2 k q 1 q = 1 q 2 "4ke 2 2 m 2 1 2k q 1 q = 2 1 m 2 1 " 4k e2 1 1 "4 9 10 9 N m2 C 2 1.6 10 "19 C 2 2 = 9.11 10 "31 kg 3 10 6 m 2 " 4 9 10 9 N m2 1.6 10 "19 C C 2 1 10 "10 m 1 2 = 9.06 "10 10 m 2 2 = 0 2 =? Emple 3: 13 cm 13 cm b q 1 q2 4 cm 6 cm 4 cm An electic dipole conit of two point chge, q 1 = 12 nc nd q 2 = 12 nc, plced 10 cm pt. Compute the potentil t point, b, nd c b dding the potentil due to eithe chge. c Electic otentil 8 Emple 3: q 1 = 12 nc q 2 = "12 nc c 1 q i = k 13 cm 13 cm 4" i i q b 1 q 2 " k q 1 q 2 1 2 4 cm 6 cm 4 cm V = 9 "10 9 N m 2 12 "10 9 C C 2 0.06 m 12 "109 C 0.04 m q i i i V = "900 V V b = 9 "10 9 N m 2 12 "10 9 C C 2 0.04 m 12 "109 C 0.14 m V b = 1930 V V c = 9 "10 9 N m 2 12 "10 9 C C 2 0.13 m 12 "109 C 0.13 m 10,000 N/C Emple 4: 8 cm 6 cm A chge q = 2 µc i moed though unifom field diected to the ight hown in the figue boe. Find the wok done b the field when the chge moe fom:. to b b. b to c c. to c d. c to c b V c = 0 Electic otentil 10 Emple 4: c 10,000 N/C 6 cm q 0 = 2 µc W field =? 8 cm b o W field = "q 0 q 0 Ed. to b b. b to c W field = q 0 Ed W field = q 0 Ed W field = 2 "10 6 C 10,000 N 0.08 m W C field = 2 "10 6 C W field = 1.6 "10 3 J W field = 0 W field = "U = "q 0 V Fo unifom E field : " Ed Whee d i the ditnce in the diection of the field. 10,000 N C 0 Emple 5: A olid conducting phee of diu h totl chge. Find the potentil eewhee, both outide nd inide the phee. c. to c W field = q 0 Ed W field = 2 "10 6 C W field = 1.6 "10 3 J 10,000 N C 0.08 m d. c to W field = q 0 Ed W field = 2 "10 6 C W field = "1.6 10 "3 J 10,000 N C 0.08 m Electic otentil 12
Emple 5: Emple 5: < E " da = q enc " EdA = q enc E " da = q enc E4" 2 = 0 > E " da = q enc " EdA = q enc E " da = q enc E4" 2 = 0 4" 2 Emple 5: Chged Spheicl Conducto Emple 5: > 2 = " V " = 0 4" 2 E V =? V " V = " E d V " 0 = " 4 2 d V = " d 4 2 0 E " 1 2 V = 1 4" = 1 4" = 1 4" 1 V = 4" > Emple 5: < Emple 5: Chged Spheicl Conducto 2 = V = 4" V =? V " V = " E d V " V = " d V " V = 0 4" V V = V V = 4" < V " 1
Emple 6: A olid inulting phee of diu h totl chge unifoml ditibuted thoughout it olume. Find the potentil eewhee, both outide nd inide the phee. Emple 6: E " da = q enc q enc = "dv " EdA = q 0 enc q 0 enc = " dv 0 E " da = q enc q enc = "V q 4 enc = " 4 3 3 E4" 2 = 3 " 3 " = 3 0 4 3 q < " enc = 3 = 3 3 4" 3 Electic otentil 19 Emple 6 Emple 6: Unifoml Chged Spheicl Inulto q enc = E " da = q enc " EdA = q enc E " da = q enc E4" 2 = 4" 2 E E " E " 1 2 > 4" 2 Emple 6: > V =? 2 = " V " = 0 V " V = " E d V " 0 = " 4 2 d V = " d 4 2 V = 1 4" = 1 4" = 1 4" 1 V = 4" > Emple 6: < V " V = " E d 2 = V " V = " V 4 3 d = 4" " V " V = 4 3 d V =? " 2 " V " V = 4 3 = 2 8 3 2 " 2 V = 8" 3 2 2 V V = 8" 3 2 2 4" V = 8" 2 3 2 <
3 8" Emple 6: Unifoml Chged Spheicl Inulto V V " 2 V " 1 Emple 7: A olid inulting phee of diu h totl chge unifoml ditibuted thoughout it olume nd i concentic with conducting hell of inne diu 2 nd oute diu 3 tht h totl chge of 2. Find the potentil fo:. > 3 b. 2 < < 3 c. < < 2 d. < Electic otentil 26 Emple 7: > 3 Emple 7: > 3 2 = " V " = 0 2 3 E " da = q enc E " da = 2 E 4" 2 = " 4 2 2 3 V =? V " V = " E d " V " 0 = " 4 2 d V = d 4" 2 V = 1 4" = 1 4" 1 V = 1 4" = " 4 Emple 7b: 2 < < 3 Emple 7b: 2 < < 3 2 3 E " da = q enc E " da = E 4" 2 = 0 0 No electic field inide conducto. 2 = 3 V =? 2 V 3 = 3 " 12 V " V 3 = " E d 3 V " V 3 = " d V " V 3 = 0 V = V 3 = 3 " 12 Conducto e equipotentil thoughout.
Emple 7c: < < 2 Emple 7c: < < 2 2 E " da = q enc E " da = E 4" 2 = 4" 2 2 = 2 2 V 2 = V 3 = V =? " 12 V " V 2 = " E d 2 V " V 2 = " 2 4 2 d V " " 12 = 1 4 2 V = 1 4" 1 2 12" V = 1 4" 5 6 Emple 7d: < E " da = q enc E " da = q enc = V E 4" 2 = E 4" 2 4" 3 3 4" 3 4" 3 3 Emple 7d: V = 1 4" 5 = 1 6 4" 6 2 = V =? < V " V = " E d V " V = " 4 3 d V " 1 4 6 = " 2 4 3 2 V " 1 4 6 = " 2 4 3 2 " 2 2 2 V = 4" 3 2 2 1 2 4" 6 4 V = 8" 3 2 2 Emple 7: E 4" 3 4" 2 Emple 7: V 4 V = 8" 3 2 2 < V = " 12 2 < < 3 2 3 0 " 4 2 2 3 V = " 4 3 < V = 1 4" 5 < < 2 6
O Emple 8: dq O = 2 2 Emple 8: A inghped conducto with diu in the z plne cie totl chge unifoml ditibuted ound it. Find the electic potentil t point tht lie on the i of the ing t ditnce fom the oigin. Electic otentil 37 " k dq = 2 2 d k dq k 2 2 = k 4" 2 2 dq " dq = 2 2 k 2 2 Emple 8: V 4" Emple 8: 4" 2 2 E = " dv d = " d d 4 2 2 E = " d 4 d 2 2 " 1 2 4" 2 2 E = 2 2 8" 32 2 E = 4" 2 2 3 2 Emple 9: oitie electic chge i ditibuted unifoml long line with length 2, ling long the i between = nd =. Find the electic potentil t point on the i t ditnce of fom the oigin. Emple 9: 0, dq = "d = 2 d = 2 2 fom integl tble, 0 d k dq k "d = k" d 2 2 2 2 = k " d = ln 2 2 2 2 "d C 2 2 Electic otentil 41
Emple 9: fom integl tble C " d = ln 2 2 2 2 ] k" d = k"ln 2 2 2 2 Emple 9: V " t = 0 k" ln 2 2 ln 2 2 2 2 k"ln 2 2 8" ln 2 2 2 2 8" ln 2 2 2 2 Emple 9: 8" ln 2 2 2 2 E = " dv d = " d d 8 ln 2 2 2 2 " E = " dv d = " 2 2 " 2 2 " 8 2 2 1 2 2 2 " 12 2 " 2 2 1 2 2 2 " 12 2 2 2 " 2 E = " 2 2 " 2 2 " 12 " 2 2 8 2 2 2 2 " 12 2 2 " E = " 2 2 " " 2 2 " 8 2 2 2 2 " 2 E = " 8 "2 2 2 2 = Emple 10: A negtie chge i unifoml ditibuted ound emicicle of diu. Find the electic potentil t the cente of cutue. Epe ou nwe in tem of,!, nd " o. 4 2 2 Electic otentil 46 Emple 10: = dq " = d k dq " = k dq " k dq = k " dq = k " "k = "" = " 4 0 4