EAcos θ, where θ is the angle between the electric field and

Transcription

1 8.4. Modl: Th lctric flux flows out of a closd surfac around a rgion of spac containing a nt positiv charg and into a closd surfac surrounding a nt ngativ charg. Visualiz: Plas rfr to Figur EX8.4. Lt A b th ara in m of ach of th six facs of th cub. Solv: Th lctric flux is dfind as = E A= EAcos θ, whr θ is th angl btwn th lctric fild and a lin prpndicular to th plan of th surfac. Th lctric flux out of th closd cub surfac is out = ( 0 N/C + 0 N/C + 10 N/C) Acos0 = ( 50 A) N m /C Similarly, th lctric flux into th closd cub surfac is in = ( 15 N/C + 15 N/C + 15 N/C) Acos180 = ( 45 A) N m /C 50 N m/c 45 N m/c = 5 N m/c. Sinc th nt lctric flux is positiv (i.., outward), th closd box contains a positiv charg. Th nt lctric flux is ( A) ( A) ( A)

2 8.10. Modl: Th lctric fild is uniform ovr th ntir surfac. Visualiz: Plas rfr to Figur EX8.10. Th lctric fild vctors mak an angl of 0 blow th surfac. Bcaus th normal ˆn to th planar surfac is at an angl of 90 rlativ to th surfac, th angl btwn ˆn and E is θ = 10. Solv: Th lctric flux is ( )( ) = E A= EAcosθ = 00 N/C m cos10 =. N m /C

3 8.0. Visualiz: Plas rfr to Figur EX8.0. For any closd surfac that ncloss a total charg Q in, th nt lctric flux through th closd surfac is = Q ε. For th closd surfac of th torus, Q in 0 in includs only th 1 nc charg. So, th nt flux through th torus is du to this charg: This is inward flux = = 8.85 /Nm 11 N m /C

4 8.. Solv: For any closd surfac nclosing a total charg Q in, th nt lctric flux through th surfac is Q in 1 = Q ε ( )( ) ε0 in = 0 = 8.85 /Nm 1000 N m /C = 8.85 nc

5 8.6. Modl: Th xcss charg on a conductor rsids on th outr surfac. Visualiz: Plas rfr to Figur EX8.6. Solv: Point 1 is at th surfac of a chargd conductor, hnc E ( )( C m ) η =, prpndicular to surfac E = = 904 N/C surfac surfac 1 ε /N m At point th lctric fild strngth is zro bcaus this point lis insid th conductor. Th lctric fild strngth at point is zro bcaus thr is no xcss charg on th intrior surfac of th box. This can b quickly sn by considring a Gaussian surfac just insid th intrior surfac of th box as shown in Figur 8.1.

6 8.0. Modl: Th lctric fild ovr th fiv surfacs is uniform. Visualiz: Plas rfr to Figur P8.0. Solv: Th lctric flux through a surfac ara A is = E A= EAcosθ whr θ is th angl btwn th lctric fild and a lin prpndicular to th plan of th surfac. Th lctric fild is prpndicular to sid 1 and is paralll to sids,, and 5. Also th angl btwn E and A 4 is 60. Th lctric fluxs through ths fiv surfacs ar 1 = EA 1 1cosθ1 = ( 400 N/C)( m 4 m) cos( 180 ) = 00 N m /C = EA cos90 = = 5 = 0 N m /C = EAcosθ = 400 N/C m sin0 4 m cos60 = N m /C ( )(( ) ) Assss: Bcaus th flux into ths fiv facs is qual to th flux out of th fiv facs, th nt flux is zro, as w found.

7 8.8. Modl: Th ball is uniformly chargd. Th charg distribution is sphrically symmtric. Visualiz: Th figur shows sphrical Gaussian surfacs with radii r = 5 cm, 10 cm, and 0 cm. Ths surfacs match th symmtry of th charg distribution. So, E is prpndicular to th Gaussian surfac and th valu of th fild strngth is th sam at all points on th surfac. Solv: (a) Bcaus th ball is uniformly chargd, its charg dnsity is 4 9 ( ) Q Q 80 ρ = = = =.87 /m.4 /m V π r 4π 0.0 m ( ) (b) Onc again, bcaus th ball is uniformly chargd, 6 6 4π 4π Q= ρv = ρ r = (.87 /m ) r = ( 1.00 /m ) r 5 Whn r = 5 cm, ( )( ) 6 5 Q 5 = 1.00 /m m = 1.5 nc. Whn r = 10 cm and 0 cm, Q 10 = 10.0 nc and Q 0 = 80 nc. (c) Gauss s law is =. E A da = EA = Qin ε0 For th 5.0-cm-radius Gaussian sphrical surfac, 9 Q5 EA 5 5 ε = Q5 1.5 E5 = = = N/C 1 0 ε0a 5 ( 8.85 /Nm ) 4π ( m) Similarly, w can apply Gauss s law to th 10 cm-radius and 0 cm-radius sphrical surfacs and obtain E 10 = N/C and E 0 = N/C. Assss: W not that E 0 = E 10 = 4E 5. This rsult is consistnt with th rsult of Exampl 8.4 according to which 1 Q Einsid = r 4πε R 0

8 8.40. Modl: Th xcss charg on a conductor rsids on th outr surfac. Th charg distribution on th two sphrs is assumd to hav sphrical symmtry. Visualiz: Plas rfr to Figur P8.40. Th Gaussian surfacs with radii r = 8 cm, 10 cm, and 17 cm match th symmtry of th charg distribution. So, E is prpndicular to ths Gaussian surfacs and th fild strngth has th sam valu at all points on th Gaussian surfac. Solv: (a) Gauss s law is = E da = Q ε. A in 0 Applying it to a Gaussian surfac of radius 8 cm, Q ( )( ) π( ) 1 8 in = ε0easphr = 8.85 /N m 15,000 N/C m = 1.07 Bcaus th xcss charg on a conductor rsids on its outr surfac and bcaus w hav a solid mtal sphr insid our Gaussian surfac, Q in is th charg that is locatd on th xtrior surfac of th innr sphr. (b) In lctrostatics, th lctric fild within a conductor is zro. Applying Gauss s law to a Gaussian surfac just insid th insid surfac of th hollow sphr at r = 10 cm, Qin = AE da= Q in = 0 C ε 0 That is, thr is no nt charg. Bcaus th innr sphr has a charg of C, th insid surfac of th hollow sphr must hav a charg of C. (c) Applying Gauss s law to a Gaussian surfac at r = 17 cm, 1 ( )( ) ( ) 8 Qin ε0 E da ε0easphr = A = = C /Nm 15,000 N/C 4π 0.17 m = 4.8 This valu includs th charg on th innr sphr, th charg on th insid surfac of th hollow sphr, and th charg on th xtrior surfac of th hollow sphr du to polarization. Thus, Q xtrior hollow ( ) ( ) + + = Qxtrior hollow = 4.8