2 E. on each of these two surfaces. r r r r. Q E E ε. 2 2 Qencl encl right left 0

Size: px

Start display at page:

Download "2 E. on each of these two surfaces. r r r r. Q E E ε. 2 2 Qencl encl right left 0"

Charles Rogers
5 years ago
Views:

1 Ch : 4, 9,, 9,,, 4, 9,, 4, 8 4 (a) Fom the diagam in the textbook, we see that the flux outwad though the hemispheical suface is the same as the flux inwad though the cicula suface base of the hemisphee On that suface all of the flux is pependicula to the suface O, we say that on the cicula base, A Thus Φ = A = π (b) is pependicula to the axis, then evey field line would both ente though the hemispheical suface and leave though the hemispheical suface, and so Φ = 9 The only contibutions to the flux ae fom the faces pependicula to the electic field Ove each of these two sufaces, the magnitude of the field is constant, so the flux is just A on each of these two sufaces Φ = ( A) ( A) = l l = ight left ight left ( ) l ( ) 7 = = 4 N C 56 N C 5m 885 C N m = 8 C ight left Because of the symmety of the poblem one sixth of the total flux will pass though each face Φ = Φ = = face 6 total Fo points inside the nonconducting sphees, the electic field will be detemined by the chage inside the spheical suface of adius 4 π = 4 = π The electic field fo can be calculated fom Gauss s law ( ) = 4π = = 4π 4π The electic field outside the sphee is calculated fom Gauss s law with = = = ( ) 4π 4π The speadsheet used fo this poblem can be found on the Media Manage, with filename PS4_ISM_CHXLS, on tab Poblem 9 lectic field ( 6 N/C) (cm)

2 (a) When close to the sheet, we appoximate it as an infinite sheet, and use the esult of xample -7 We assume the chage is ove both sufaces of the aluminum 9 75 C ( 5m) σ = = = 885 C N m o ( ) (b) When fa fom the sheet, we appoximate it as a point chage 9 75 C 5 5 N C, away fom the sheet 9 = = ( 8988 N m C ) = N C, away fom the sheet π ( ) 4 5m (a) Conside a spheical gaussian suface at a adius of cm It oses all of the chage d A = ( 4 π ) = 55 C = = 8988 N m C = 549 N C, adially outwad π m ( ) ( ) (b) A adius of 6 cm is inside the conducting mateial, and so the field must be Note that 6 thee must be an induced chage of 55 C on the suface at = 45 cm, and 6 then an induced chage of 55 C on the oute suface of the sphee (c) Conside a spheical gaussian suface at a adius of cm It oses all of the chage 4 Since the chages ae of opposite sign, and since the chages ae fee to move since they ae on conductos, the chages will attact each othe and move to the inside o facing edges of the plates Thee will be no chage on the outside edges of the plates And thee cannot be chage in the plates themselves, since they ae conductos All of the chage must eside on sufaces Due to the symmety of the poblem, all field lines must be pependicula to the plates, as discussed in xample -7 (a) To find the field between the plates, we choose a gaussian cylinde, pependicula to the plates, with aea A fo the ends of the cylinde We place one end inside the left plate (whee the field must be zeo), and the othe end between the plates No flux passes though the cuved suface of the cylinde d A d A d A d A = = = ends side ight end σ A σ A= = between between The field lines between the plates leave the inside suface of the left plate, and teminate on the inside suface of the ight plate A simila deivation could have been done with the ight end of the cylinde inside of the ight plate, and the left end of the cylinde in the space between the plates (b) If we now put the cylinde fom above so that the ight end is σ σ between σ σ outside

3 inside the conducting mateial, and the left end is to the left of the left plate, the only possible location fo flux is though the left end of the cylinde Note that thee is NO chage osed by the Gaussian cylinde d A= d A d A= d A= ends side left end A= = outside outside (c) If the two plates wee nonconductos, the esults would not change The chage would be distibuted ove the two plates in a diffeent fashion, and the field inside of the plates would not be zeo, but the chage in the empty egions of space would be the same as when the plates ae conductos 9 Due to the spheical symmety of the poblem, Gauss s law using a sphee of adius leads to the following d A = ( 4 π ) = = 4π (a) Fo the egion < <, the osed chage is = = 4π, (b) Fo the egion < < the osed chage is the poduct of the volume chage density times the volume of chaged mateial osed The chage density is given by ρ = = 4 4 π π 4π ( ) π π V = = = = = 4π 4π 4π 4π 4π ρ ρ π π 4π ( ) (c) Fo the egion >, the osed chage is the total chage, = 4π (a) Ceate a gaussian suface that just oses the inne suface of the spheical shell Since the electic field inside a conducto must be zeo, Gauss s law equies that the osed chage be zeo The osed chage is the sum of the chage at the cente and chage on the inne suface of the conducto = q = enc inne Theefoe = q inne

4 (b) The total chage on the conducto is the sum of the chages on the inne and oute sufaces will = = = q oute inne oute inne (c) A gaussian suface of adius < only oses the cente chage, q The electic field theefoe be the field of the single chage ( < ) = q 4π (d) A gaussian suface of adius < < is inside the conducto so = (e) A gaussian suface of adius > oses the total chage q The electic field will then be the field fom the sum of the two chages q ( > ) = 4π 4 The geomety of this poblem is simila to Poblem, and so we use the same development, following xample -6 See the solution of Poblem fo details V V d A ρ ρ = ( π l ) = = = π l (a) Fo >, the osed volume of the shell is V = π l ρ V ρ = =, adially outwad π l (b) Fo <, the osed volume of the shell is V = π l ρ V ρ = =, adially outwad π l 8 The geomety of this poblem is simila to Poblem, and so we use the same development, following xample -6 See the solution of Poblem fo details d A = ( π l ) = = π l (a) Fo < <, the osed chage is the volume of chage osed, times the chage density ρ π l ρ = = = π l π l (b) Fo < <, the osed chage is all of the chage on the inne cylinde ρ ρ

5 ρ π l ρ = = = π l π l (c) Fo < <, the osed chage is all of the chage on the inne cylinde, and the pat of the chage on the shell that is osed by the gaussian cylinde ρ π l ρ π l π l ρ = = = π l π l (d) Fo >, the osed chage is all of the chage on both the inne cylinde and the shell ρ π l ρ π l π l ρ = = = π l π l (e) See the gaph The speadsheet used fo this poblem can be found on the Media Manage, with filename PS4_ISM_CHXLS, on tab Poblem 8e ( 4 N/C) (cm)

13. The electric field can be calculated by Eq. 21-4a, and that can be solved for the magnitude of the charge N C m 8.

13. The electric field can be calculated by Eq. 21-4a, and that can be solved for the magnitude of the charge N C m 8. CHAPTR : Gauss s Law Solutions to Assigned Poblems Use -b fo the electic flux of a unifom field Note that the suface aea vecto points adially outwad, and the electic field vecto points adially inwad Thus