Homework #2. Solutions. Chapter 22. The Electric Field II: Continuous Charge Distributions

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1 Homewok #. Solutions. Chapte. The lectic Field II: Continuous Chage Distibutions 4 If the electic flux though a closed suface is zeo, must the electic field be zeo eveywhee on that suface? If not, give a specific example. Fom the given infomation can the net chage inside the suface be detemined? If so, what is it? Detemine the Concept No, this is not necessaily tue. The only conclusion that we can daw is that thee is equal positive and negative flux. Fo example, the net flux though a Gaussian suface completely enclosing a dipole is zeo. If the electic flux is zeo though the closed suface, we can conclude that the net chage inside the suface is zeo. 8 xplain why the electic field stength inceases linealy with, athe than deceases invesely with, between the cente and the suface of a unifomly chaged solid sphee. Detemine the Concept We can show that the chage inside a unifomly chaged solid sphee of adius is popotional to and that the aea of a sphee is popotional to. Using Gauss s law, it follows that the electic field must be popotional to /. Use Gauss s law to expess the electic field inside a spheical chage distibution of constant volume chage density: 4πkQinside A whee A 4π. xpess and : Q inside as a function of ρ Q inside ρ V 4 πρ Substitute fo Qinside to obtain: 4 4πk πρ 4kπρ 4π This esult shows that the electic field inceases linealy as you move out fom the cente of a spheical chage distibution. 4 Two infinite non-conducting sheets of chage ae paallel to each othe, with sheet A in the x. m plane and sheet B in the x +. m plane. Find the electic field in the egion x <. m, in the egion x > +. m, and between the sheets fo the following situations. (a) When each sheet has a unifom suface chage density equal to +. µ and (b) when sheet A has a unifom suface chage density equal to +.

2 µ and sheet B has a unifom suface chage density equal to. µ. (c) Sketch the electic field-line patten fo each case. Pictue the Poblem Let the chage densities on the two plates be and and denote the thee egions of inteest as,, and. Choose a coodinate system in which the positive x diection is to the ight. We can apply the equation fo nea an infinite plane of chage and the supeposition of fields to find the field in each of the thee egions. (a) Use the equation fo nea an infinite plane of chage to expess the field in egion when +. µ : + πk πk 4πk Substitute numeical values and evaluate : 4π µ 9 ( )( ) N m /C. (.4 N/C) Poceed as above fo egion : + πk πk πk πk Poceed as above fo egion : + 9 N m 4π C 5 (.4 N/C) πk + πk 4πk (.µ ) (b) Use the equation fo nea an + infinite plane of chage to expess and πk evaluate the field in egion when +. µ and. µ : πk πk πk

3 Poceed as above fo egion : + 9 N m 4π C 5 (.4 N/C) πk + πk 4πk (.µ ) Poceed as above fo egion : + πk πk πk πk (c) The electic field lines fo (a) and (b) ae shown below: (a) (b) (a) Show that the electic-field stength on the axis of a ing chage of adius a has maximum values at z±a/. (b) Sketch the field stength vesus z fo both positive and negative values of z. (c) Detemine the maximum value of. Pictue the Poblem The electic field on the axis of a ing chage as a function of distance z along the axis fom the cente of the ing is given by kqz z. We can show that it has its maximum and minimum values at z + a ( ) z + a and z a by setting its fist deivative equal to zeo and solving the esulting equation fo z. The gaph of z will confim that the maximum and minimum occu at these coodinates. (a) The vaiation of z with z on the kqz z z + a axis of a ing chage is given by: ( )

4 Diffeentiate this expession with espect to z to obtain: d x dz kq kq d ( z a ) z ( z a ) d x + + dz kq dz ( z + a ) ( z + a ) ( z + a ) z( )( z + a ) ( z) ( z + a ) z ( z + a ) kq ( z + a ) ( z + a ) Set this expession equal to zeo fo extema and simplify: Solving fo z yields: ( z + a ) z ( z + a ) ( z + a ) ( + a ) z ( z + a ), z, and z + a z z ± a as ou candidates fo maxima o minima. (b) A plot of the magnitude of z, in units of kq/a, vesus z/a follows. This gaph shows that the extema at z ± a ae, in fact, maxima..4. (z /a ) z/a

5 a (c) valuate z ± and simplify to obtain the maximum value of the magnitude of z : ± a kq ± 9 z,max z a a ± a + a kq a kq ( a + a ) 6 A thin hemispheical shell of adius R has a unifom suface chage. Find the electic field at the cente of the base of the hemispheical shell. Pictue the Poblem Conside the ing with its axis along the z diection shown in the diagam. Its adius is z cosθ and its width is dθ. We can use the equation fo the field on the axis of a ing chage and then integate to expess the field at the cente of the hemispheical shell. d cosθ z θ sinθ dθ dθ y x xpess the field on the axis of d the ing chage: ( sin θ + cos θ) kzdq whee z cosθ kzdq xpess the chage dq on the ing: dq da ( π sinθ) π sinθdθ Substitute to obtain: k( cosθ) dθ π sinθdθ d πk sinθ cosθdθ

6 Integating d fom θ to π/ yields: πk sinθ cosθdθ πk π π [ sin θ] πk What is the electic flux though one side of a cube that has a single point chage of. µc placed at its cente? HINT: You do not need to integate any equations to get the answe. Pictue the Poblem The flux though the cube is given by φnet Qinside, whee Q inside is the chage at the cente of the cube. The flux though one side of the cube is one-sixth of the total flux though the cube. The flux though one side of the cube is one-sixth of the total flux though the cube: φ face 6 φ tot Q 6 Substitute numeical values and evaluate φ : face φ face.µ C C N m N m C 5 An imaginay ight cicula cone (Figue -4) that has a base angle θ and a base adius R is in a chage fee egion that has a unifom electic field (field lines ae vetical and paallel to the cone s axis). What is the atio of the numbe of field lines pe unit aea penetating the base to the numbe of field lines pe unit aea penetating the conical suface of the cone? Use Gauss's law in you answe. (The field lines in the figue ae only a epesentative sample.) Pictue the Poblem Because the cone encloses no chage, we know, fom Gauss s law, that the net flux of the electic field though the cone s suface is zeo. Thus, the numbe of field lines penetating the cuved suface of the cone must equal the numbe of field lines penetating the base and the enteing flux must equal the exiting flux. R θ θ n The flux penetating the base of the φ enteing A base

7 cone is given by: The flux penetating the cuved suface of the cone is given by: quating the fluxes and simplifying yields: φ A exiting base nda S cosθ da S S cosθda ( cosθ) A cuved suface The atio of the density of field lines is: A A base cuved suface cosθ 4 A non-conducting solid sphee of adius. cm has a unifom volume chage density. The magnitude of the electic field at. cm fom the sphee s cente is.88 N/C. (a) What is the sphee s volume chage density? (b) Find the magnitude of the electic field at a distance of 5. cm fom the sphee s cente. Pictue the Poblem (a) We can use the definition of volume chage density, in conjunction with quation -8a, to find the sphee s volume chage density. (b) We can use quation -8b, in conjunction with ou esult fom Pat (a), to find the electic field at a distance of 5. cm fom the solid sphee s cente. (a) The solid sphee s volume chage density is the atio of its chage to its volume: Q Q () V inside ρ inside 4 πr Fo R, quation -8a gives the electic field at a distance fom the cente of the sphee: Qinside () 4π Solving fo Q inside yields: Qinside 4π Substitute fo Q inside in equation () and simplify to obtain: 4π ρ 4 πr R Substitute numeical values and evaluate ρ: ( C /N m )(.88 N/C)(. cm) ρ.µ (. cm).997µ

8 (b) Fo R, the electic field at a distance fom the cente of the sphee is given by: xpess Q fo R: Q () 4π R 4 Q ρv π R ρ Substituting fo Q in equation () 4 π R ρ ρ and simplifying yields: 4π R Substitute numeical values and evaluate (5. cm): ( ) (.997µ )( 5. cm). cm 5 ( C /N m ) N/C 68 An infinitely long line chage that has a unifom linea chage density equal to.5 µ lies paallel to the y axis at x. m. A positive point chage that has a magnitude equal to. µc is located at x. m, y. m. Find the electic field at x. m, y.5 m. Pictue the Poblem Let P denote the point of inteest at (. m,.5 m). The electic field at P is the sum of the electic fields due to the infinite line chage and the point chage. λ.5µ. y, m q.µc (.,.) P(.,.5).... x, m xpess the esultant electic field at P: λ + q Find the field at P due the infinite line chage: λ kλ 9 ( N m /C )(.5µ ) m ( 6.74kN/C)

9 xpess the field at P due the point chage: Refeing to the diagam above, detemine and : q kq.8m and.8944 i.447 j Substitute and evaluate (. m,.5 m) q : q (.8944 i.447 j ) ( 9.48 kn/c)(.8944 i.447 j ) 9 ( ) ( N m /C )(.µ C).m,.5m (.8 m) Substitute and simplify to obtain: ( 8.6kN/C) ( 4.8 kn/c)j (. m,.5 m) ( 6.74kN/C) + ( 8.6kN/C) ( 4.8 kn/c) (.6 kn/c) ( 4.8 kn/c)j 77 An infinite non-conducting plane sheet of chage that has a suface chage density +. µ lies in the y.6 m plane. A second infinite non-conducting plane sheet of chage that has a suface chage density of. µ lies in the x. m plane. Lastly, a non-conducting thin spheical shell of adius of. m and that has its cente in the z plane at the intesection of the two chaged planes has a suface chage density of. µ. Find the magnitude and diection of the electic field on the x axis at (a) x.4 m and (b) x.5 m. Pictue the Poblem Let the numeal efe to the infinite plane whose chage density is and the numeal to the infinite plane whose chage density is. We can find the electic fields at the two points of inteest by adding the electic fields due to the chage distibutions on the infinite planes and the sphee. j

10 y, m. µ.µ x, m.6 (.,.6).µ xpess the electic field due to the infinite planes and the sphee at any point in space: (a) Because (.4 m, ) is inside the sphee: + + () sphee sphee (.4 m,) Find the field at (.4 m, ) due to plane :.µ j ( ).4 m, j ( C /N m ) Find the field at (.4 m, ) due to plane : (.4 m,) ( i ).µ Substitute in equation () to obtain: ( 69.4 kn/c)j ( ) ( i) (.9 kn/c C /N m )i (.4 m,) + ( 69.4 kn/c) j + (.9 kn/c) (.9 kn/c) + ( 69.4 kn/c)j Find the magnitude and diection of (.4 m,) : and (.4 m,) (.9 kn/c) + ( 69.4 kn/c).6 kn/c 4 kn/c 69.4 kn/c θ tan kn/c 56.

11 (b) Because (.5 m, ) is outside the sphee: kq sphee (.4 m,) sphee whee is a unit vecto pointing fom (. m,.6 m) to (.5 m, ). valuate Q sphee : Q sphee A 4πR 4π sphee (.µ )(.m) 7.7µ C Refeing to the diagam above, detemine and :.66m and.985 i +.74 j Substitute and evaluate (.5 m,) sphee : sphee 9 ( ) ( N m /C )( 7.7µ C).5 m, (.66 m) Find the field at (.5 m, ) due to plane : ( 9.8 kn/c)(.985 i+.74 j ) (.5 kn/c) + ( 48. kn/c)j.µ j ( ).5 m, j ( C /N m ) Find the field at (.5 m, ) due to plane :. µ (.5 m,) ( C /N m ) Substitute in equation () to obtain: ( 69.4 kn/c)j (.9 kn/c) (.4 m,) (.5kN/C) + ( 48.9 kn/c) j + ( 69.4 kn/c) j + (.9 kn/c) (.5kN/C) + (. kn/c)j Find the magnitude and diection of (.5 m,) :

12 and (.5 m,) (.5 kn/c) + (. kn/c) 6kN/C. kn/c θ tan 5.5 kn/c