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.

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1 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 the angle between the two is 8 A Acos 5 N C R cos8 5 N C 8 m 77 N m C (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 A (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 The electic field can be calculated by -a, and that can be solved fo the magnitude of the chage k k 5 N C 5 m 8988 N m C 9 This coesponds to about be negative Thus 85 C 8 5 electons Since the field points towad the ball, the chage must 85 C 5 The electic field due to a long thin wie is given in xample - as (a) R 7 C m N m C N C R R 5 m The negative sign indicates the electic field is pointed towads the wie (b) 7 C m N m C 8 N C R R 5m The negative sign indicates the electic field is pointed towads the wie 8 See xample - fo a detailed discussion elated to this poblem (a) Inside a solid metal sphee the electic field is (b) Inside a solid metal sphee the electic field is (c) Outside a solid metal sphee the electic field is the same as if all the chage wee concentated at the cente as a point chage 55 C N m C 5 N C m 8 Peason ducation, Inc, Uppe Saddle Rive, NJ All ights eseved This mateial is potected unde all copyight laws as they cuently exist No potion of this mateial may be epoduced, in any fom o by any means, without pemission in witing fom the publishe 7

2 Physics fo Scientists & nginees with Moden Physics, th dition Instucto Solutions Manual The field would point towads the cente of the sphee (d) Same easoning as in pat (c) 55 C N m C 77 N C 8m The field would point towads the cente of the sphee (e) The answes would be no diffeent fo a thin metal shell (f) The solid sphee of chage is dealt with in xample - We see fom that xample that the field inside the sphee is given by Outside the sphee the field is no diffeent So we have these esults fo the solid sphee 9 55 C 5 m 8988 N m C 5 m 58 N C m 9 55 C 9 m 8988 N m C 9 m 5 N C m 9 55 C m 8988 N m C 5 N C m 9 55 C 8 m 8988 N m C 77 N C All point towads the cente of the sphee m (a) Conside a spheical gaussian suface at a adius of cm It oses all of the chage da 55 C 8988 N m C 59 N C, adially outwad 9 7 m (b) A adius of cm is inside the conducting mateial, and so the field must be Note that thee must be an induced chage of 55 C on the suface at = 5 cm, and 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 da 55 C 8988 N m C 59 N C, adially outwad 9 5 m 7 (a) In the egion, a gaussian suface would ose no chage Thus, due to the spheical symmety, we have the following da (b) In the egion, only the chage on the inne shell will be osed 8 Peason ducation, Inc, Uppe Saddle Rive, NJ All ights eseved This mateial is potected unde all copyight laws as they cuently exist No potion of this mateial may be epoduced, in any fom o by any means, without pemission in witing fom the publishe 8

3 Chapte Gauss s Law da (c) In the egion, the chage on both shells will be osed da (d) To make fo, we must have opposite chage This implies that the shells ae of (e) To make fo we must have O, if a chage, at the cente of the shells, that would also make wee placed The geomety of this poblem is simila to Poblem, and so we use the same development, following xample - See the solution of Poblem fo details V V da Rl Rl (a) Fo R R, the osed volume of the shell is V R l R R V R, adially outwad Rl R (b) Fo R R, the osed volume of the shell is V R l V R, adially outwad Rl Due to the spheical symmety of the poblem, Gauss s law using a sphee of adius leads to the following da (a) Fo the egion, the osed chage is (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 We must integate to find the total chage We follow the pocedue given in xample -5 We divide the sphee up into concentic thin shells of thickness d, as shown in Fig - We then integate to find the chage 8 Peason ducation, Inc, Uppe Saddle Rive, NJ All ights eseved This mateial is potected unde all copyight laws as they cuently exist No potion of this mateial may be epoduced, in any fom o by any means, without pemission in witing fom the publishe 9

4 Physics fo Scientists & nginees with Moden Physics, th dition Instucto Solutions Manual dv d d (c) Fo the egion, the osed chage is the total chage, found by integation in a simila fashion to pat (b) dv d d (d) See the attached gaph We have chosen Let The speadsheet used fo this poblem can be found on the Media Manage, with filename PS_ISM_CHXLS, on tab Poblem d / / 5 (a) We use Gauss s law fo a spheically symmetic chage distibution, and assume that all the chage is on the suface of the ath Note that the field is pointing adially inwad, and so the dot poduct intoduces a negative sign da 5 N C 8 m 5 5 R ath 9 79 C 8 C 8988 N m C (b) Find the suface density of electons Let n be the total numbe of electons ne A A n R 885 C N m 5 N C ath 9 ath A ea e R e C 8 electons m 9 8 Peason ducation, Inc, Uppe Saddle Rive, NJ All ights eseved This mateial is potected unde all copyight laws as they cuently exist No potion of this mateial may be epoduced, in any fom o by any means, without pemission in witing fom the publishe

5 Chapte Gauss s Law Conside this sphee as a combination of two sphees Sphee is a solid sphee of adius and chage density centeed at A and sphee is a second sphee of adius / centeed at C (a) The electic field at A will have zeo contibution fom sphee due to its symmety about point A The electic field is then calculated by ceating a gaussian suface centeed at point C with adius / enc da Since the electic field points into the gaussian suface (negative) the electic field at point A points to the ight (b) At point B the electic field will be the sum of the electic fields fom each sphee The electic field fom sphee is calculated using a gaussian suface of adius centeed at A enc da At point B the field fom sphee points towad the left The electic field fom sphee is calculated using a gaussian suface centeed at C of adius / enc d A 5 At point B, the electic field fom sphee points towad the ight The net electic field is the sum of these two fields The net field points to the left We assume the chage is unifomly distibuted, and so the field of the pea is that of a point chage R R 9 R N C 885 C N m 75 m 5 C 8 Peason ducation, Inc, Uppe Saddle Rive, NJ All ights eseved This mateial is potected unde all copyight laws as they cuently exist No potion of this mateial may be epoduced, in any fom o by any means, without pemission in witing fom the publishe

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