CHAPTER 30 GAUSS S LAW

Transcription

1 CHPTER GUSS S LW. Given : E /5 E î /5 E ĵ E. N/C The pane is parae to yz-pane. Hence ony /5 E î passes perpendicuar to the pane whereas /5 E ĵ goes parae. rea.m (given Fux E /5.. Nm /c Nm /c. Given ength of rod edge of cube Portion of rod inside the cube / Tota charge. Linear charge density / of rod. We know: Fux charge encosed. Charge encosed in the rod inside the cube. / / /. s the eectric fied is uniform. Considering a perpendicuar pane to it, we find that it is an euipotentia surface. Hence there is no net current fow on that surface. Thus, net charge in that region is zero. E. Given: E î cm, a cm. E 5 N/C. From fig. We see that fux passes mainy through surface areas. BDC & EFGH. s the EFB & CHGD are paraed to the Fux. gain in BDC a ; hence the Fux ony passes through the surface are EFGH. E E c î x Fux E 5 a rea a L Fux so, Fux 5 a c 5. ccording to Fux 5 (..5 Since the charge is paced at the centre of the cube. Hence the fux passing through the six surfaces Given charge is paced o a pain surface with area a, about a/ from its centre. ssumption : et us assume that the given pain forms a surface of an imaginary cube. Then the charge is found to be at the centre of the cube. Hence fux through the surface Given: Magnitude of the two charges paced 7 c. We know: from Gauss s aw that the fux experienced by the sphere is ony due to the interna charge and not by the externa one. Now E.ds N-m /C. Y ĵ Z kˆ C D o o,a,o B o,o,a / / E F H G X î a,o,o R P R E.

2 8. We know: For a spherica surface Fux E.ds [by Gauss aw] Hence for a hemisphere tota surface area 9. Given: Voume charge density. c/m In order to find the eectric fied at a point cm m from the centre et us assume a concentric spherica surface inside the sphere. Now, E.ds But Hence / R so, / R / / 7 (. / 8.85 / 7 (. 5 N/C. Charge present in a god nuceus C Since the surface encoses a the charges we have: 79.6 (a E.ds 8.85 E ds N/C 9 9. (7 5 [area r ] (b For the midde part of the radius. Now here r 7/ 5 m Voume / r Charge encosed voume [ : voume charge density] Net charge But Net voume Net charged encosed E ds E encosed S 9. Now, Voume charge density r r c r r gain voume of sphere having radius x x N/C 9 r cm O r.

3 Now charge encosed by the sphere having radius r r r ppying Gauss s aw E E r r r r r r encosed r r r. Given: The sphere is uncharged metaic sphere. Due to induction the charge induced at the inner surface, and that outer surface. (a Hence the surface charge density at inner and outer surfaces a and a respectivey.. charge tota surface area (b gain if another charge is added to the surface. We have inner surface charge density because the added charge does not affect it. On the other hand the externa surface charge density a as the gets added up. (c For eectric fied et us assume an imaginary surface area inside the sphere at a distance x from centre. This is same in both the cases as the in ineffective. Now, E.ds So, E x x. (a Let the three orbits be considered as three concentric spheres, B & C. Now, Charge of.6 6 c Charge of B.6 6 c Charge of C.6 6 c s the point P is just inside s, so its distance from centre. m Eectric fied x (b For a point just inside the s coud 9 (.. N/C Tota charge encosed Hence, Eectric fied, E (5..65 N/C. N/C. Drawing an eectric fied around the ine charge we find a cyinder of radius m. Given: inear charge density Let the ength be 6 c/m We know E r E.d E r For, r m & 6 c/m E N/C 9 5 N/C S 5. m. m N 5 m a a S B C P -6 c/m cm,

4 5. Given : 6 c/m For the previous probem. E for a cyindrica eectricfied. r Now, For experienced by the eectron due to the eectric fied in wire centripeta force. E mv E we know,m e 9. kg, ve?, r assumed radius mv r KE / E r J. r 6. Given: Voume charge density Let the height of cyinder be h. Charge at P h For eectric fied E.ds h E ds h 7. E.d Let the area be. Uniform change distribution density is E d a C Surface charge density 6 C/m We know E due to a charge conducting sheet gain Force of attraction between partice & pate E 9. Ba mass g 6 8 Charge 6 c Thread ength cm Now from the fig, T cos mg 6.5N T sin eectric force Eectric force ( surface charge density T sin Tan, T cos mg mg mg tan C/m. cm 6 d x x P <x<d T Cos T Sin mg

5 . (a Tension in the string in Euiibrium T cos 6 mg T mg cos 6 / (b Straingtening the same figure. Now the resutant for R Induces the acceeration in the penduum. T g / g m. N. / ( /.6. s cm m, u, a? t s 6 s cceeration of the eectron, s (/ at (/ a ( 6 a The eectric fied due to charge pate Now, eectric force Now m e me acceeration c/m. Given: Surface density 9 me.5 sec. a m/s 9. (a & (c For any point to the eft & right of the dua pater, the eectric fied is zero. s there are no eectric fux outside the system. (b For a test charge put in the midde. It experiences a fore Hence net eectric fied towards the (-ve pate. 6 m E m R cm 6 E 6. (a For the surface charge density of a singe pate. Let the surface charge density at both sides be & Now, eectric fied at both ends. & Due to a net baanced eectric fied on the pate & X Y So, / Net surface charge density /.5

6 (b Eectric fied to the eft of the pates Since / Hence Eectricfied / This must be directed toward eft as X is the charged pate. (c & (d Here in both the cases the charged pate X acts as the ony source of eectric fied, with (ve in the inner side and Y attracts towards it with (-ve he in its inner side. So for the midde portion E towards right. (d Simiary for extreme right the outerside of the Y pate acts as positive and hence it repes to the right with E. Consider the Gaussian surface the induced charge be as shown in figure. The net fied at P due to a the charges is Zero. 9/ (eft 9/ (eft 9/ (right 9/ (right / charge on the right side of right most pate 9 / / / C 9 9 / - B D 9 9.6