R E - Lehman Collegelehman.edu/faculty/anchordoqui/169-P3-sol.pdfConcentric with this sphere is a conducting spherical shell with inner radius band outer radius c, and having a net

Prof. Anchordoqui

Problems set # 3 Physics 169 February 24, 2015

1. A point charge q is located at the center of a uniform ring having linear charge density λ

and radius a, as shown in Fig. 1. Determine the total electric flux through a sphere centered at the

point charge and having radius R, where R < a.

Solution Only the charge inside radius R contributes to the total flux, hence ΦE = q/ε0.

2. A point charge Q is located just above the center of the flat face of a hemisphere of radius

R as shown in Fig. 2. What is the electric flux (i) through the curved surface and (ii) through the

flat face?

Solution With δ very small, all points on the hemisphere are nearly at a distance R from

the charge, so the field everywhere on the curved surface is 14πε0

QR2 radially outward (normal to

the surface). Therefore, the flux is this field strength times the area of half a sphere Φcurved =∫~E · d ~A = 1

4πε0QR2 2πR2 = Q

2ε0. (ii) The closed surface encloses zero charge so Gauss’ law gives

Φcurved + Φflat = 0 or Φflat = −Φcurved = − Q2ε0

.

3. The line ag in Fig. 3 is a diagonal of a cube. A point charge q is located on the extension of

line ag, very close to vertex a of the cube. Determine the electric flux through each of the sides of

the cube which meet at the point a.

Solution No charge is inside the cube. The net flux through the cube is zero. Positive flux

comes out through the three faces meeting at g. These three faces together fill a solid angle equal

to one-eighth of a sphere as seen from q. The total flux passing through these faces is then 18qε0

.

Each face containing a intercepts equal flux going into the cube: 0 = ΦE,net = 3ΦE,abcd + q8ε0

.

Therefore, ΦE,abcd = − q24ε0

.

4. A sphere of radius R surrounds a point charge Q , located at its center. (i) Show that the

electric flux through a circular cap of half-angle (see Fig. 4) is ΦE = Q2ε0

(1 − cos θ). What is the

flux for (ii) θ = 90◦ and (iii) θ = 180◦.

Solution The charge creates a uniform ~E, pointing radially outward, so ΦE = EA. The

arc length is ds = Rdθ, and the circumference is 2πr = 2πR sin θ. Hence, A =∫

2πrds =∫ θ0 (2πR sin θ)Rdθ = 2πR2

∫ θ0 sin θ dθ = −2πR2 cos θ|θ0 = 2πR2(1 − cos θ) and ΦE = 1

4πε0QR2 ·

2πR2(1 − cos θ) = Q2ε0

(1 − cos θ), i.e., independent of R! (ii) For θ = 90◦ (hemisphere), ΦE =Q

2ε0(1 − cos 90◦) = Q

2ε0(iii) For θ = 180◦ (entire sphere), ΦE = Q

2ε0(1 − cos 180◦) = Q

ε0; this is a

formal derivation of Gauss’ law.

5. An insulating solid sphere of radius a has a uniform volume charge density and carries a

total positive charge Q. A spherical gaussian surface of radius r, which shares a common center

with the insulating sphere, is inflated starting from r = 0. (i) Find an expression for the electric

flux passing through the surface of the gaussian sphere as a function of r for r < a. (ii) Find an

expression for the electric flux for r > a. (iii) Plot the flux versus r.

Solution The charge density is determined by Q = 43πa

ρ ⇒ ρ = 3Q4πa3

. (i) The flux is that

created by the enclosed charge within radius r: ΦE = qinε0

= 4πr3ρ3ε0

= Qr3

ε0a3. (ii) ΦE = Q

ε0. Note that

the answers to parts (i) and (ii) agree at r = a. (iii) This is shown in Fig. 5.

6. A solid insulating sphere of radius a carries a net positive charge 3Q, uniformly distributed

throughout its volume. Concentric with this sphere is a conducting spherical shell with inner radius

b and outer radius c, and having a net charge −Q, as shown in Fig. 6. (i) Construct a spherical

gaussian surface of radius r > c and find the net charge enclosed by this surface. (ii) What is the

direction of the electric field at r > c? (iii) Find the electric field at r ≥ c. (iv) Find the electric

field in the region with radius r where b < r < c (v) Construct a spherical gaussian surface of radius

r, where b < r < c, and find the net charge enclosed by this surface. (vi) Construct a spherical

gaussian surface of radius r, where a < r < b, and find the net charge enclosed by this surface.

(vii) Find the electric field in the region a < r < b. (viii) Construct a spherical gaussian surface of

radius r < a, and find an expression for the net charge enclosed by this surface, as a function of r.

Note that the charge inside this surface is less than 3Q. (ix) Find the electric field in the region

r < a. ( x) Determine the charge on the inner surface of the conducting shell. (xi) Determine

the charge on the outer surface of the conducting shell. (xii) Make a plot of the magnitude of the

electric field versus r.

Solution (i) qin = 3Q − Q = 2Q. (ii) The charge distribution is spherically symmetric and

qin > 0. Thus, the field is directed radially outward. (iii) For r ≥ c, E = 14πε0

qinr2

= Q2πε0r2

.

(iv) Since all points within this region are located inside conducting material, E = 0, for b < r < c.

(v) ΦE =∫~E · d ~A = 0⇒ qin = ε0ΦE = 0. (vi) qin = 3Q. (vii) For a ≤ r < b, E = 1

4πε0qinr2

= 3Q4πε0r2

(radially outward). (viii) qin = ρV = 3Q43πa3

43πr

3 = 3Q r3

a3. (ix) For 0 ≤ r ≤ a, E = qin

4πε0r2= 3Qr

4πε0a3

(radially outward). (x) From part (iv), for b < r < c, E = 0.Thus, for a spherical gaussian sur-

face with b < r < c, qin = 3Q + qinner = 0 where qinner is the charge on the inner surface of the

conducting shell. This yields qinner = −3Q. (xi) Since the total charge on the conducting shell is

qnet = qouter + qinner = −Q, we have qouter = −Q− qinner = −Q− (−3Q) = 2Q. (xii) This is shown

in Fig. 6.

7.Consider a long cylindrical charge distribution of radius R with a uniform charge density ρ.

Find the electric field at distance r from the axis where r < R.

Solution If is positive, the field must be radially outward. Choose as the gaussian surface a cylin-

der of length L and radius r, contained inside the charged rod. Its volume is πr2L and it encloses

charge ρπr2L, see Fig. 7. Because the charge distribution is long, no electric flux passes through

the circular end caps; ~E · d ~A = EdA cos(π/2) = 0. The curved surface has ~E · d ~A = EdA cos 0◦,

and E must be the same strength everywhere over the curved surface. Gauss’ law,∮~E · d ~A = q

ε0,

becomes E∫curvedsurface

dA = ρπr2Lε0

. Now the lateral surface area of the cylinder is 2πrL, yielding

E2πrL = ρπr2L/ε0. Thus, ~E = ρr2ε0

radially away from the cylinder axis.

8. A solid, insulating sphere of radius a has a uniform charge density ρ and a total charge Q.

Concentric with this sphere is an uncharged, conducting hollow sphere whose inner and outer radii

are b and c, as shown in Fig. 8. (i) Find the magnitude of the electric field in the regions r < a,

a < r < b, b < r < c, and r > c. (ii) Determine the induced charge per unit area on the inner and

outer surfaces of the hollow sphere.

Solution (i) From Gauss’ law∮~E ·d ~A = E(4πr2) = qin

ε0. For r < a, qin = ρ4

3πr3, so E = ρr

3ε0. For

a < r < b and c < r, qin = Q, so E = Q4πε0r2

. For b ≤ r ≤ c, E = 0, since E = 0 inside a conductor.

(ii) Let q1 be the induced charge on the inner surface of the hollow sphere. Since E = 0 inside

the conductor, the total charge enclosed by a spherical surface of radius b ≤ r ≤ c must be zero.

Therefore, q1+Q = 0 and σ1 = q14πb2

= − Q4πb2

. Let q2 be the induced charge on the outside surface of

the hollow sphere. Since the hollow sphere is uncharged, we require q1+q2 = 0 and σ2 = q14πc2

= Q4πc2

.

9. An early (incorrect) model of the hydrogen atom, suggested by J. J. Thomson, proposed

that a positive cloud of charge e was uniformly distributed throughout the volume of a sphere of

radius R, with the electron an equal-magnitude negative point charge e at the center. (i) Using

Gauss’ law, show that the electron would be in equilibrium at the center and, if displaced from the

center a distance r < R, would experience a restoring force of the form F = −kr, where k is a con-

stant. (ii) Show that k = e2

4πε0R3 . (iii) Find an expression for the frequency f of simple harmonic

oscillations that an electron of mass me would undergo if displaced a small distance (< R) from

the center and released. (iv) Calculate a numerical value for R that would result in a frequency of

2.47×1015 Hz, the frequency of the light radiated in the most intense line in the hydrogen spectrum.

Solution First, consider the field at distance r < R from the center of a uniform sphere of positive

charge (Q = +e) with radius R. From Gauss’ law, 4πr2E = qinε0

= ρVε0

= +e43πR3

43πr2

ε0, so E = e

4πε0R3 r

directed outward. (i) The force exerted on a point charge q = −e located at distance r from the

center is then F = qE = − e2

4πε0R3 r = −kr. (ii) k = e2

4πε0R3 (iii) Fr = mear = − e2

4πε0R3 r, so ar =

− e2

4πε0meR3 r = −ω2r. Thus, the motion is simple harmonic with frequency f = ω2π = 1

2π

√e2

4πε0meR3 .

(iv) f = 2.47× 1015 Hz = 12π

√8.99×109N·m2/C2(1.60×10−19 C)2

9.11×10−31 kg R3 , which yields R3 = 1.05× 10−30 m3, or

R = 1.02× 10−10 m = 102 pm.

10. An infinitely long cylindrical insulating shell of inner radius a and outer radius b has a

uniform volume charge density ρ. A line of uniform linear charge density λ is placed along the axis

of the shell. Determine the electric field everywhere.

Solution The field direction is radially outward perpendicular to the axis. The field strength

depends on r but not on the other cylindrical coordinates θ or z. Choose a Gaussian cylinder of

radius r and length L. If r < a, ΦE = qinε0

and E2πrL = λL/ε0, so ~E = λ2πrε0

r. If a < r < b,

E2πrL = [λL+ ρπ(r2 − a2)L]/ε0 and ~E = λ+ρπ(r2−a2)2πrε0

r. If r > b, E2πrL = [λL+ ρπ(b2 − a2)L]/ε0

and ~E = λ+ρπ(b2−a2)2πrε0

r.

11. A particle of mass m and charge q moves at high speed along the x axis. It is initially

near x = −∞, and it ends up near x = +∞. A second charge Q is fixed at the point x = 0,

y = −d. As the moving charge passes the stationary charge, its x component of velocity does not

change appreciably, but it acquires a small velocity in the y direction. Determine the angle through

which the moving charge is deflected. [Hint: The integral you encounter in determining vy can be

evaluated by applying Gauss’ law to a long cylinder of radius d, centered on the stationary charge.]

Solution The vertical velocity component of the moving charge increases according to mdvydt =

Fy → mdvydx

dxdt = qEy, see Fig. 9. Now dx

dt = vx, has a nearly constant value v. Hence dvy = qmvEydx,

yielding vy =∫ vy0 dv′y = q

mv

∫ +∞−∞ Eydx. The radially outward compnent of the electric field varies

along the x axis; it is described by∫ +∞−∞ EydA =

∫ +∞−∞ Ey2πd dx = Q/ε0. Putting all this together∫ +∞

−∞ Eydx = Q2πdε0

and vy = qQmv2πdε0

. The angle of deflection is described by tan θ = vy/v, so

θ = tan−1 qQ2πε0dmv2

.

12. Two infinite, nonconducting sheets of charge are parallel to each other, as shown in Fig. 10.

The sheet on the left has a uniform surface charge density σ, and the one on the right has a uniform

charge density −σ. Calculate the electric field at points (i) to the left of, (ii) in between, and (iii)

to the right of the two sheets. (iv) Repeat the calculations when both sheets have positive uniform

surface charge densities of value σ.

Solution Consider the field due to a single sheet and let E+ and E− represent the fields due

to the positive and negative sheets, see Fig. 10. The field at any distance from each sheet has a

magnitude given by |E+| = |E−| = σ2ε0

. (i) To the left of the positive sheet, E+ is directed toward

the left and E− toward the right and the net field over this region is ~E = 0. (ii) In the region

between the sheets, E+ and E− are both directed toward the right and the net field is ~E = σ/ε0to the right. (iii) To the right of the negative sheet, E− and E+ are again oppositely directed and~E = 0. (iv) If both charges are positive (see Fig. 10), in the region to the left of the pair of sheets,

both fields are directed toward the left and the net field is ~E = σ/ε0 to the left; in the region

between the sheets, the fields due to the individual sheets are oppositely directed and the net field

is ~E = 0; in the region to the right of the pair of sheets, both are fields are directed toward the

right and the net field is ~E = σ/ε0 to the right.

13. A sphere of radius 2a is made of a nonconducting material that has a uniform volume charge

density ρ. (Assume that the material does not affect the electric field.) A spherical cavity of radius

a is now removed from the sphere, as shown in Fig. 11. Show that the electric field within the

cavity is uniform and is given by Ex = 0 and Ey = ρa3ε0

. [Hint: The field within the cavity is the

superposition of the field due to the original uncut sphere, plus the field due to a sphere the size of

the cavity with a uniform negative charge density −ρ].

Solution The resultant field within the cavity is the superposition of two fields, one ~E+ due to a

uniform sphere of positive charge of radius 2a, and the other ~E− due to a sphere of negative charge

of radius a centered within the cavity. From Gauss’ law we have 43πr3ρε0

= 4πr2E+, so ~E+ = ρr3ε0r =

ρ3ε0~r. Using again Gauss’ law, −4

3πr31ρε0

= 4πr21E−, so ~E− = ρr1

3ε0(−r1) = − ρ

3ε0~r1. It is easily seen in

Fig. 11 that ~r = a+ ~r1, so E= − ρ(~r−~a)3ε0

, yielding ~E = ~E+ + ~E− = ρ~r3ε0− ρ~r

3ε0+ ρ~a

3ε0= ρ~a

3ε0= 0ı+ ρa

3ε0.

Therefore Ex = 0 and Ey = ρa3ε0

at all points within the cavity.

14. A solid insulating sphere of radius R has a nonuniform charge density that varies with r

according to the expression ρ = Ar2, where A is a constant and r < R is measured from the

center of the sphere. (i) Show that the magnitude of the electric field outside (r > R) the sphere

is E = AR5

5ε0r2. (ii) Show that the magnitude of the electric field inside (r < R) the sphere is

E = Ar3

5ε0. [Hint: The total charge Q on the sphere is equal to the integral of ρdV , where r extends

from 0 to R; also, the charge q within a radius r < R is less than Q. To evaluate the integrals,

note that the volume element dV for a spherical shell of radius r and thickness dr is equal to 4r2dr.]

Solution From Gauss’ law∮~E · d ~A = E4πr2 = qin/ε0. (i) For r > R, qin =

∫ R0 Ar24πr2 dr =

4πAR5/5, and E = AR5

5ε0r2. (ii) For r < R, qin =

∫ r0 Ar

24πr2dr = 4πAr5/5, and E = Ar3

5ε0.

15. A slab of insulating material (infinite in two of its three dimensions) has a uniform positive

charge density ρ. An edge view of the slab is shown in Fig.12. (i) Show that the magnitude of

the electric field a distance x from its center and inside the slab is E = ρx/ε0. (ii) Suppose an

electron of charge −e and mass me can move freely within the slab. It is released from rest at a

distance x from the center. Show that the electron exhibits simple harmonic motion with a fre-

quency s = 12π

√ρemeε0

. (iii) A slab of insulating material has a nonuniform positive charge density

ρ = Cx2, where x is measured from the center of the slab as shown in Fig. 12, and C is a constant.

The slab is infinite in the y and z directions. Derive expressions for the electric field in the exterior

regions and the interior region of the slab (−d/2 < x < d/2).

Solution (i) Consider a cylindrical shaped gaussian surface perpendicular to the yz plane with

one end in the yz plane and the other end containing the point x: Use Gauss law:∮~E ·d ~A = qin/ε0

By symmetry, the electric field is zero in the yz plane and is perpendicular to d ~A over the wall

of the gaussian cylinder. Therefore, the only contribution to the integral is over the end cap con-

taining the point x. Hence EA = ρAx/ε0, so that at distance x from the mid-line of the slab,

E = ρx/ε0. (ii) Use Newton’s law to obtain a = Fme

= − ρemeε0

x. The acceleration of the electron

is of the form a = −ω2x with ω =√

ρemeε0

. Thus, the motion is simple harmonic with frequency

f = ω2π = 1

2π

√ρemeε0

.

Problems 755

9. On the basis of the repulsive nature of the force betweenlike charges and the freedom of motion of charge within aconductor, explain why excess charge on an isolated con-ductor must reside on its surface.

A person is placed in a large hollow metallic sphere thatis insulated from ground. If a large charge is placed onthe sphere, will the person be harmed upon touching theinside of the sphere? Explain what will happen if theperson also has an initial charge whose sign is oppositethat of the charge on the sphere.

11. Two solid spheres, both of radius R , carry identical totalcharges, Q. One sphere is a good conductor while theother is an insulator. If the charge on the insulating sphereis uniformly distributed throughout its interior volume,

10.

how do the electric fields outside these two spherescompare? Are the fields identical inside the two spheres?

A common demonstration involves charging a rubberballoon, which is an insulator, by rubbing it on your hair,and touching the balloon to a ceiling or wall, which is alsoan insulator. The electrical attraction between the chargedballoon and the neutral wall results in the balloon stickingto the wall. Imagine now that we have two infinitely largeflat sheets of insulating material. One is charged and theother is neutral. If these are brought into contact, will anattractive force exist between them, as there was for theballoon and the wall?

13. You may have heard that one of the safer places to be duringa lightning storm is inside a car. Why would this be the case?

12.

Section 24.1 Electric Flux1. An electric field with a magnitude of 3.50 kN/C is applied

along the x axis. Calculate the electric flux through a rec-tangular plane 0.350 m wide and 0.700 m long assumingthat (a) the plane is parallel to the yz plane; (b) the planeis parallel to the xy plane; (c) the plane contains the y axis,and its normal makes an angle of 40.0° with the x axis.

2. A vertical electric field of magnitude 2.00 ! 104 N/C existsabove the Earth’s surface on a day when a thunderstorm isbrewing. A car with a rectangular size of 6.00 m by 3.00 mis traveling along a roadway sloping downward at 10.0°.Determine the electric flux through the bottom of the car.

A 40.0-cm-diameter loop is rotated in a uniform electric fielduntil the position of maximum electric flux is found. Theflux in this position is measured to be 5.20 ! 105 N "m2/C.What is the magnitude of the electric field?

4. Consider a closed triangular box resting within a horizon-tal electric field of magnitude E # 7.80 ! 104 N/C asshown in Figure P24.4. Calculate the electric flux through(a) the vertical rectangular surface, (b) the slantedsurface, and (c) the entire surface of the box.

3.

5. A uniform electric field a i $ b j intersects a surface of areaA. What is the flux through this area if the surface lies(a) in the yz plane? (b) in the xz plane? (c) in the xy plane?

6. A point charge q is located at the center of a uniform ringhaving linear charge density % and radius a, as shown inFigure P24.6. Determine the total electric flux through asphere centered at the point charge and having radius R ,where R & a.

7. A pyramid with horizontal square base, 6.00 m on eachside, and a height of 4.00 m is placed in a vertical electricfield of 52.0 N/C. Calculate the total electric flux throughthe pyramid’s four slanted surfaces.

8. A cone with base radius R and height h is located on ahorizontal table. A horizontal uniform field E penetratesthe cone, as shown in Figure P24.8. Determine the electricflux that enters the left-hand side of the cone.

Section 24.2 Gauss’s LawThe following charges are located inside a submarine:5.00 'C, ( 9.00 'C, 27.0 'C, and ( 84.0 'C. (a) Calculate

9.

1, 2, 3 # straightforward, intermediate, challenging # full solution available in the Student Solutions Manual and Study Guide

# coached solution with hints available at http://www.pse6.com # computer useful in solving problem

# paired numerical and symbolic problems

P R O B L E M S

30 cm

60°10 cm

E

Figure P24.4

R

q a

λ

Figure P24.6

h

R

E

Figure P24.8

Figure 1: Problem 1.

the net electric flux through the hull of the submarine.(b) Is the number of electric field lines leaving thesubmarine greater than, equal to, or less than the numberentering it?

10. The electric field everywhere on the surface of a thinspherical shell of radius 0.750 m is measured to be890 N/C and points radially toward the center of thesphere. (a) What is the net charge within the sphere’ssurface? (b) What can you conclude about the nature anddistribution of the charge inside the spherical shell?

11. Four closed surfaces, S1 through S4, together with thecharges ! 2Q , Q , and !Q are sketched in Figure P24.11.(The colored lines are the intersections of the surfaceswith the page.) Find the electric flux through eachsurface.

12. (a) A point charge q is located a distance d from an infi-nite plane. Determine the electric flux through theplane due to the point charge. (b) What If? A pointcharge q is located a very small distance from the centerof a very large square on the line perpendicular tothe square and going through its center. Determine theapproximate electric flux through the square due to thepoint charge. (c) Explain why the answers to parts(a) and (b) are identical.

13. Calculate the total electric flux through the paraboloidalsurface due to a uniform electric field of magnitude E 0 inthe direction shown in Figure P24.13.

14. A point charge of 12.0 "C is placed at the center of aspherical shell of radius 22.0 cm. What is the total electricflux through (a) the surface of the shell and (b) any hemi-spherical surface of the shell? (c) Do the results dependon the radius? Explain.

A point charge Q is located just above the centerof the flat face of a hemisphere of radius R as shown inFigure P24.15. What is the electric flux (a) through thecurved surface and (b) through the flat face?

15.

16. In the air over a particular region at an altitude of 500 mabove the ground the electric field is 120 N/C directeddownward. At 600 m above the ground the electric field is100 N/C downward. What is the average volume chargedensity in the layer of air between these two elevations? Isit positive or negative?

17. A point charge Q # 5.00 "C is located at the center of acube of edge L # 0.100 m. In addition, six other identicalpoint charges having q # ! 1.00 "C are positioned sym-metrically around Q as shown in Figure P24.17. Determinethe electric flux through one face of the cube.

18. A positive point charge Q is located at the center of a cubeof edge L. In addition, six other identical negative pointcharges q are positioned symmetrically around Q as shownin Figure P24.17. Determine the electric flux through oneface of the cube.

756 CHAPTE R 24 • Gauss’s Law

–Q

+Q

–2Q

S2

S3

S1

S4

Figure P24.11

d

r

E0

Figure P24.13

Q0

R

δ

Figure P24.15

L

L

q

q

q

q

Qq

q

L

Figure P24.17 Problems 17 and 18.

d

RO

λ

Figure P24.19


19. An infinitely long line charge having a uniform charge perunit length ! lies a distance d from point O as shown inFigure P24.19. Determine the total electric flux throughthe surface of a sphere of radius R centered at O resultingfrom this line charge. Consider both cases, where R " dand R # d.

20. An uncharged nonconducting hollow sphere of radius10.0 cm surrounds a 10.0-$C charge located at the originof a cartesian coordinate system. A drill with a radius of1.00 mm is aligned along the z axis, and a hole is drilled inthe sphere. Calculate the electric flux through the hole.

21. A charge of 170 $C is at the center of a cube of edge80.0 cm. (a) Find the total flux through each face of thecube. (b) Find the flux through the whole surface of thecube. (c) What If? Would your answers to parts (a) or(b) change if the charge were not at the center? Explain.

22. The line ag in Figure P24.22 is a diagonal of a cube. Apoint charge q is located on the extension of line ag , veryclose to vertex a of the cube. Determine the electric fluxthrough each of the sides of the cube which meet at thepoint a.

Section 24.3 Application of Gauss’s Law to VariousCharge Distributions

23. Determine the magnitude of the electric field at thesurface of a lead-208 nucleus, which contains 82 protonsand 126 neutrons. Assume the lead nucleus has a volume208 times that of one proton, and consider a proton to bea sphere of radius 1.20 % 10&15 m.

24. A solid sphere of radius 40.0 cm has a total positive chargeof 26.0 $C uniformly distributed throughout its volume.Calculate the magnitude of the electric field (a) 0 cm,(b) 10.0 cm, (c) 40.0 cm, and (d) 60.0 cm from the centerof the sphere.

25. A 10.0-g piece of Styrofoam carries a net charge of& 0.700 $C and floats above the center of a large horizon-tal sheet of plastic that has a uniform charge density on itssurface. What is the charge per unit area on the plasticsheet?

26. A cylindrical shell of radius 7.00 cm and length 240 cm hasits charge uniformly distributed on its curved surface. Themagnitude of the electric field at a point 19.0 cm radiallyoutward from its axis (measured from the midpoint of theshell) is 36.0 kN/C. Find (a) the net charge on the shelland (b) the electric field at a point 4.00 cm from the axis,measured radially outward from the midpoint of the shell.

27. A particle with a charge of & 60.0 nC is placed at thecenter of a nonconducting spherical shell of inner radius20.0 cm and outer radius 25.0 cm. The spherical shellcarries charge with a uniform density of & 1.33 $C/m3.A proton moves in a circular orbit just outside thespherical shell. Calculate the speed of the proton.

28. A nonconducting wall carries a uniform charge density of8.60 $C/cm2. What is the electric field 7.00 cm in front ofthe wall? Does your result change as the distance from thewall is varied?

Consider a long cylindrical charge distribution ofradius R with a uniform charge density '. Find the electricfield at distance r from the axis where r " R.

30. A solid plastic sphere of radius 10.0 cm has charge with uni-form density throughout its volume. The electric field5.00 cm from the center is 86.0 kN/C radially inward. Findthe magnitude of the electric field 15.0 cm from the center.

Consider a thin spherical shell of radius 14.0 cm with atotal charge of 32.0 $C distributed uniformly on itssurface. Find the electric field (a) 10.0 cm and (b) 20.0 cmfrom the center of the charge distribution.

32. In nuclear fission, a nucleus of uranium-238, whichcontains 92 protons, can divide into two smaller spheres,each having 46 protons and a radius of 5.90 % 10&15 m.What is the magnitude of the repulsive electric forcepushing the two spheres apart?

33. Fill two rubber balloons with air. Suspend both of themfrom the same point and let them hang down on strings ofequal length. Rub each with wool or on your hair, so thatthey hang apart with a noticeable separation from eachother. Make order-of-magnitude estimates of (a) the forceon each, (b) the charge on each, (c) the field each createsat the center of the other, and (d) the total flux of electricfield created by each balloon. In your solution state thequantities you take as data and the values you measure orestimate for them.

34. An insulating solid sphere of radius a has a uniformvolume charge density and carries a total positive chargeQ. A spherical gaussian surface of radius r, which shares acommon center with the insulating sphere, is inflatedstarting from r ( 0. (a) Find an expression for the electricflux passing through the surface of the gaussian sphere asa function of r for r " a. (b) Find an expression for theelectric flux for r # a. (c) Plot the flux versus r.

A uniformly charged, straight filament 7.00 m in length hasa total positive charge of 2.00 $C. An uncharged cardboardcylinder 2.00 cm in length and 10.0 cm in radius surroundsthe filament at its center, with the filament as the axis ofthe cylinder. Using reasonable approximations, find (a) theelectric field at the surface of the cylinder and (b) the totalelectric flux through the cylinder.

36. An insulating sphere is 8.00 cm in diameter and carriesa 5.70-$C charge uniformly distributed throughout itsinterior volume. Calculate the charge enclosed by aconcentric spherical surface with radius (a) r ( 2.00 cmand (b) r ( 6.00 cm.

A large flat horizontal sheet of charge has a charge perunit area of 9.00 $C/m2. Find the electric field just abovethe middle of the sheet.

37.

35.

31.

29.

Problems 757

d c

ab

e f

gh

q

Figure P24.22Figure 3: Problem 3.

38. The charge per unit length on a long, straight filament is! 90.0 "C/m. Find the electric field (a) 10.0 cm,(b) 20.0 cm, and (c) 100 cm from the filament, wheredistances are measured perpendicular to the length ofthe filament.

Section 24.4 Conductors in Electrostatic EquilibriumA long, straight metal rod has a radius of 5.00 cm and acharge per unit length of 30.0 nC/m. Find the electricfield (a) 3.00 cm, (b) 10.0 cm, and (c) 100 cm from theaxis of the rod, where distances are measured perpendicu-lar to the rod.

40. On a clear, sunny day, a vertical electric field of about130 N/C points down over flat ground. What is the surfacecharge density on the ground for these conditions?

41. A very large, thin, flat plate of aluminum of area A has atotal charge Q uniformly distributed over its surfaces.Assuming the same charge is spread uniformly over theupper surface of an otherwise identical glass plate, comparethe electric fields just above the center of the uppersurface of each plate.

42. A solid copper sphere of radius 15.0 cm carries a charge of40.0 nC. Find the electric field (a) 12.0 cm, (b) 17.0 cm,and (c) 75.0 cm from the center of the sphere. (d) WhatIf? How would your answers change if the sphere werehollow?

43. A square plate of copper with 50.0-cm sides has no netcharge and is placed in a region of uniform electric fieldof 80.0 kN/C directed perpendicularly to the plate. Find(a) the charge density of each face of the plate and (b) thetotal charge on each face.

44. A solid conducting sphere of radius 2.00 cm has a chargeof 8.00 "C. A conducting spherical shell of inner radius4.00 cm and outer radius 5.00 cm is concentric with thesolid sphere and has a total charge of !4.00 "C. Find theelectric field at (a) r # 1.00 cm, (b) r # 3.00 cm, (c) r #4.50 cm, and (d) r # 7.00 cm from the center of thischarge configuration.

45. Two identical conducting spheres each having a radius of0.500 cm are connected by a light 2.00-m-long conduct-ing wire. A charge of 60.0 "C is placed on one of the con-ductors. Assume that the surface distribution of chargeon each sphere is uniform. Determine the tension in thewire.

46. The electric field on the surface of an irregularly shapedconductor varies from 56.0 kN/C to 28.0 kN/C. Calculatethe local surface charge density at the point on the surfacewhere the radius of curvature of the surface is (a) greatestand (b) smallest.

A long, straight wire is surrounded by a hollow metal cylin-der whose axis coincides with that of the wire. The wirehas a charge per unit length of $, and the cylinder has anet charge per unit length of 2$. From this information,use Gauss’s law to find (a) the charge per unit length onthe inner and outer surfaces of the cylinder and (b) theelectric field outside the cylinder, a distance r from theaxis.

47.

39.

48. A conducting spherical shell of radius 15.0 cm carries anet charge of ! 6.40 "C uniformly distributed on itssurface. Find the electric field at points (a) just outside theshell and (b) inside the shell.

A thin square conducting plate 50.0 cm on a side liesin the xy plane. A total charge of 4.00 % 10!8 C is placedon the plate. Find (a) the charge density on the plate,(b) the electric field just above the plate, and (c) theelectric field just below the plate. You may assume that thecharge density is uniform.

50. A conducting spherical shell of inner radius a and outerradius b carries a net charge Q. A point charge q is placedat the center of this shell. Determine the surface chargedensity on (a) the inner surface of the shell and (b) theouter surface of the shell.

51. A hollow conducting sphere is surrounded by a largerconcentric spherical conducting shell. The inner spherehas charge !Q , and the outer shell has net charge & 3Q .The charges are in electrostatic equilibrium. UsingGauss’s law, find the charges and the electric fieldseverywhere.

52. A positive point charge is at a distance R/2 from the centerof an uncharged thin conducting spherical shell of radiusR. Sketch the electric field lines set up by this arrangementboth inside and outside the shell.

Section 24.5 Formal Derivation of Gauss’s LawA sphere of radius R surrounds a point charge Q , locatedat its center. (a) Show that the electric flux through a cir-cular cap of half-angle ' (Fig. P24.53) is

What is the flux for (b) ' # 90° and (c) ' # 180°?

() #Q

2* 0 (1 ! cos ')

53.

49.

Additional Problems54. A nonuniform electric field is given by the expression

E # ay i & bz j & cx k , where a, b, and c are constants. Determine the electric flux through a rectangular surfacein the xy plane, extending from x # 0 to x # w and fromy # 0 to y # h.


R

Q

θ

Figure P24.53

Chapter 24 41

P24.50 (a) The charge �q at the center induces charge �q on the inner surface of the conductor,where its surface density is:

VSa

qa

�

4 2 .

(b) The outer surface carries charge Q q� with density

VSb

Q qb

�

4 2 .

P24.51 Use Gauss’s Law to evaluate the electric field in each region, recalling that the electric field is zeroeverywhere within conducting materials. The results are:

E 0 inside the sphere and within the material of the shell

E kQre 2 between the sphere and shell, directed radially inward

E kQ

re 2

2 outside the shell, directed radially outward .

Charge �Q is on the outer surface of the sphere .

Charge �Q is on the inner surface of the shell ,

and �2Q is on the outer surface of the shell.

P24.52 An approximate sketch is given at the right. Note that the electric field linesshould be perpendicular to the conductor both inside and outside.

FIG. P24.52

Section 24.5 Formal Derivation of Gauss‘s Law

P24.53 (a) Uniform E, pointing radially outward, so )E EA . The arc length is ds Rd T ,and the circumference is 2 2S S Tr R sin

A rds R Rd R d R R � �z zz 2 2 2 2 2 10

2

0

20

2S S T T S T T S T S TT T

Tsin sin cos cosb g a f b g

)EQR

RQ

�

� � �

�1

42 1

21

02

2

0SS T Tcos cosa f a f [independent of R!]

FIG. P24.53

(b) For T q90 0. (hemisphere): )EQ Q

�

� q �2

1 9020 0

cosa f .

(c) For T q180 (entire sphere): )EQ Q

�

� q �2

1 1800 0

cosa f [Gauss’s Law].


Chapter 24 37

P24.33 Consider two balloons of diameter 0.2 m, each with mass 1 g, hanging apart with a0.05 m separation on the ends of strings making angles of 10° with the vertical.

(a) F T mg Tmg

y¦ q� � q

coscos

10 010

F T F F Tx e e¦ q� � qsin sin10 0 10 , so

Fmg

mg

F

e

e

q

FHG

IKJ q q q

| u � �

cossin tan . . tan

1010 10 0 001 9 8 10

2 10 3 3

kg m s

N ~10 N or 1 mN

2b ge jFIG. P24.33

(b) Fk qree

2

2

2 108 99 10

0 25

1 2 10 10

39 2

2

7 7

u |u �

| u

�

� �

N N m C

m

C ~ C or 100 nC

2 2.

.

.

e ja f

q

q

(c) Ek qre |

u � u| u

�

2

9 7

24

8 99 10 1 2 10

0 251 7 10 10

. .

.. ~

N m C C

m N C kN C

2 2e je ja f

(d) )Eq

�

|u

u � u � �

�

�0

7

1241 2 10

1 4 10 10.

. ~ C

8.85 10 C N m N m C kN m C2 2

2 2

*P24.34 The charge density is determined by Q a 43

3S U US

3

4 3Qa

(a) The flux is that created by the enclosed charge within radius r:

)Eq r r Q

aQra

�

�

�

�

in

0

3

0

3

03

3

03

43

4 33 4

S U SS

(b) )EQ

�0

. Note that the answers to parts (a) and (b) agree at r a .

(c)

ar

)E

Q�0

00

FIG. P24.34(c)Figure 5: Problem 5.

55. A solid insulating sphere of radius a carries a net positivecharge 3Q , uniformly distributed throughout its volume.Concentric with this sphere is a conducting spherical shellwith inner radius b and outer radius c, and having a netcharge !Q , as shown in Figure P24.55. (a) Construct aspherical gaussian surface of radius r " c and find the netcharge enclosed by this surface. (b) What is the directionof the electric field at r " c? (c) Find the electric field atr " c. (d) Find the electric field in the region with radius rwhere c " r " b. (e) Construct a spherical gaussian surfaceof radius r, where c " r " b, and find the net chargeenclosed by this surface. (f) Construct a spherical gaussiansurface of radius r, where b " r " a, and find the netcharge enclosed by this surface. (g) Find the electric fieldin the region b " r " a. (h) Construct a spherical gaussiansurface of radius r # a, and find an expression for the netcharge enclosed by this surface, as a function of r. Notethat the charge inside this surface is less than 3Q. (i) Findthe electric field in the region r # a. ( j) Determine thecharge on the inner surface of the conducting shell.(k) Determine the charge on the outer surface of theconducting shell. (l) Make a plot of the magnitude of theelectric field versus r.

56. Consider two identical conducting spheres whose surfacesare separated by a small distance. One sphere is given alarge net positive charge while the other is given a smallnet positive charge. It is found that the force betweenthem is attractive even though both spheres have netcharges of the same sign. Explain how this is possible.

A solid, insulating sphere of radius a has a uniformcharge density $ and a total charge Q. Concentric with thissphere is an uncharged, conducting hollow sphere whoseinner and outer radii are b and c, as shown in FigureP24.57. (a) Find the magnitude of the electric field in theregions r # a, a # r # b, b # r # c, and r " c. (b) Deter-mine the induced charge per unit area on the inner andouter surfaces of the hollow sphere.

57.

58. For the configuration shown in Figure P24.57, supposethat a % 5.00 cm, b % 20.0 cm, and c % 25.0 cm. Further-more, suppose that the electric field at a point 10.0 cmfrom the center is measured to be 3.60 & 103 N/C radiallyinward while the electric field at a point 50.0 cm from thecenter is 2.00 & 102 N/C radially outward. From this infor-mation, find (a) the charge on the insulating sphere,(b) the net charge on the hollow conducting sphere, and(c) the charges on the inner and outer surfaces of thehollow conducting sphere.

59. A particle of mass m and charge q moves at high speedalong the x axis. It is initially near x % ! ', and it endsup near x % ( '. A second charge Q is fixed at the pointx % 0, y % !d. As the moving charge passes the stationarycharge, its x component of velocity does not change appre-ciably, but it acquires a small velocity in the y direction.Determine the angle through which the moving charge isdeflected. Suggestion: The integral you encounter in deter-mining vy can be evaluated by applying Gauss’s law to along cylinder of radius d, centered on the stationarycharge.

60. Review problem. An early (incorrect) model of the hydro-gen atom, suggested by J. J. Thomson, proposed that a pos-itive cloud of charge ( e was uniformly distributedthroughout the volume of a sphere of radius R, with theelectron an equal-magnitude negative point charge ! e atthe center. (a) Using Gauss’s law, show that the electronwould be in equilibrium at the center and, if displacedfrom the center a distance r # R , would experience arestoring force of the form F % !Kr, where K is a constant.(b) Show that K % ke e 2/R 3. (c) Find an expression for thefrequency f of simple harmonic oscillations that anelectron of mass me would undergo if displaced a smalldistance (#R) from the center and released. (d) Calculatea numerical value for R that would result in a frequency of2.47 & 1015 Hz, the frequency of the light radiated in themost intense line in the hydrogen spectrum.

61. An infinitely long cylindrical insulating shell of innerradius a and outer radius b has a uniform volume chargedensity $. A line of uniform linear charge density ) isplaced along the axis of the shell. Determine the electricfield everywhere.

62. Two infinite, nonconducting sheets of charge are parallelto each other, as shown in Figure P24.62. The sheet on theleft has a uniform surface charge density *, and the one

Problems 759

a

b

c

3Q

–Q

Figure P24.55

Insulator

Conductor

a

cb


σ–σ

Figure P24.62

42 Gauss’s Law

Additional Problems

P24.54 In general, E i j k � �ay bz cx� � �

In the xy plane, z 0 and E i k �ay cx� �

)

)

E

Ex

w

x

w

d ay cx dA

ch xdx chx chw

� � �

z zz

E A i k k� � �e j

0

2

0

2

2 2x

y

z

x = 0

x = w

y = 0 y = h

dA = hdx

FIG. P24.54

P24.55 (a) q Q Q Qin � � �3 2

(b) The charge distribution is spherically symmetric and qin ! 0 . Thus, the field is directedradially outward .

(c) Ek qr

k Qr

e e in2 2

2 for r ct .

(d) Since all points within this region are located inside conducting material, E 0 for

b r c� � .

(e) ) )E Ed q � � � z E A 0 00in

(f) q Qin �3

(g) Ek qr

k Qr

e e in2 2

3 (radially outward) for a r bd � .

(h) q VQa

r Qrain

�FHGIKJFHGIKJ �U

SS

3 43

343

33

3

3

(i) Ek qr

kr

Qra

k Qra

e ee �

FHG

IKJ

in2 2

3

3 33 3 (radially outward) for 0 d dr a .

(j) From part (d), E 0 for b r c� � . Thus, for aspherical gaussian surface with b r c� � ,q Q qin inner � � 3 0 where qinner is thecharge on the inner surface of theconducting shell. This yields q Qinner �3 .

(k) Since the total charge on the conductingshell is q q q Qnet outer inner � � , we have

q Q q Q Q Qouter inner � � � � � �3 2b g .

(l) This is shown in the figure to the right.

E

ra b c

FIG. P24.55(l)Figure 6: Problem 6.

36 Gauss’s Law

P24.28 V u FHG

IKJ u� �8 60 10

1008 60 106

22. . C cm

cmm

C m2 2e j

E �

u

u u

�

�

V2

8 60 10

2 8 85 104 86 10

0

2

129.

..

e j N C away from the wall

The field is essentially uniform as long as the distance from the center of the wall to the field point ismuch less than the dimensions of the wall.

P24.29 If U is positive, the field must be radially outward. Choose as thegaussian surface a cylinder of length L and radius r, contained insidethe charged rod. Its volume is S r L2 and it encloses charge US r L2 .Because the charge distribution is long, no electric flux passesthrough the circular end caps; E A� q d EdA cos .90 0 0 . The curvedsurface has E A� qd EdA cos0 , and E must be the same strengtheverywhere over the curved surface. FIG. P24.29

Gauss’s law, E A� �z dq

0, becomes E dA

r L

CurvedSurface

z �

US 2

0.

Now the lateral surface area of the cylinder is 2S rL :

E r Lr L

22

0S

USb g �

. Thus, E �Ur

2 0 radially away from the cylinder axis .

*P24.30 Let U represent the charge density. For the field inside the sphere at r1 5 cm we have

E rq r

1 12

0

13

04

43

SS U

�

�

inside Er

11

03

�U

U �

u � u

� u�

�3 3 8 85 10 86 10

0 054 57 100 1

1

12 35E

r C

.

..

C N

m Nm C m

2

23e je j

.

Now for the field outside at r3 15 cm

E rr

Ekre

3 32 2

3

0

332

3 5 9 74

3

443

43

0 10 4 57 10 8 99 10 1 91 107 64 10

76 4

SS U

S

�

� u

u � u

� u

� �. . . ..

.

m C

m

Nm C

0.15 m C N C

kN C radially inward

3

2

2 2

a f e j e ja f

E

P24.31 (a) E 0

(b) Ek Qre

u u

�

2

9 6

2

8 99 10 32 0 10

0 2007 19

. .

..

e je ja f MN C E 7 19. MN C radially outward

P24.32 The distance between centers is 2 5 90 10 15u u �. m . Each produces a field as if it were a point chargeat its center, and each feels a force as if all its charge were a point at its center.

Fk q qre u �

u

u u u

�

�

1 22

92 19 2

15 238 99 10

46 1 60 10

2 5 90 103 50 10 3 50.

.

.. . N m C

C

m N kN2 2e j

a f e je j





57.






Problems 759

a

b

c

3Q

–Q

Figure P24.55

Insulator

Conductor

a

cb


σ–σ

Figure P24.62Figure 8: Problem 8.

44 Gauss’s Law

*P24.59 The vertical velocity component of the moving chargeincreases according to

mdv

dtFy

y mdv

dxdxdt

qEyy .

Now dxdt

vx has the nearly constant value v. So

dvq

mvE dxy y v dv

qmv

E dxy y

v

y

y

z z�f

f

0

.

v x v y

x

y

d q

v 0

T

Q

FIG. P24.59

The radially outward compnent of the electric field varies along the x axis, but is described by

E dA E d dxQ

y y�f

f

�f

f

z z �

20

Sb g .

So E dxQdy

�f

fz �2 0S

and vqQ

mv dy �2 0S. The angle of deflection is described by

tanTS

�

v

vqQ

dmvy

2 02 T

S

��tan 1

022

qQdmv

.

P24.60 First, consider the field at distance r R� from the center of a uniform sphere of positive chargeQ e �b g with radius R.

4 2

0 043

3

43

3

0S

US

Sr E

q V eR

re j �

�

�FHG

IKJ �

in so Ee

R

�

FHG

IKJ4 0

3S r directed outward

(a) The force exerted on a point charge q e � located at distance r from the center is then

F qE ee

Rr

eR

r Kr ��

FHG

IKJ �

�

FHG

IKJ �

4 403

2

03S S

.

(b) Ke

Rk eRe

�

2

03

2

34S

(c) F m ak eR

rr e re �FHGIKJ

2

3 , so ak e

m Rr rr

e

e

�FHGIKJ �

2

32Z

Thus, the motion is simple harmonic with frequency fk e

m Re

e

ZS S2

12

2

3 .

(d) fR

u u � u

u

�

�2 47 10

12

8 99 10 1 60 10

9 11 1015

9 19 2

31 3.

. .

. Hz

N m C C

kg

2 2

Se je j

e jwhich yields R3 301 05 10 u �. m3 , or R u �1 02 10 10210. m pm .





57.






Problems 759

a

b

c

3Q

–Q

Figure P24.55

Insulator

Conductor

a

cb


σ–σ

Figure P24.62

Chapter 24 45

P24.61 The field direction is radially outward perpendicular to the axis. The field strength depends on r butnot on the other cylindrical coordinates T or z. Choose a Gaussian cylinder of radius r and length L.If r a� ,

)Eq

�

in

0and E rL

L2

0S

Ob g �

Er

�

OS2 0

or E r �

�O

S2 0rr a� a f .

If a r b� � , E rLL r a L

22 2

0S

O USb g e j

� �

�

E r � �

��

O US

S

r a

ra r b

2 2

02e j a f� .

If r b! , E rLL b a L

22 2

0S

O USb g e j

� �

�

E r � �

�!

O US

S

b a

rr b

2 2

02e j a f� .

P24.62 Consider the field due to a single sheet and let E� and E�represent the fields due to the positive and negative sheets. Thefield at any distance from each sheet has a magnitude given byEquation 24.8:

E E� � �V

2 0.

(a) To the left of the positive sheet, E� is directed toward theleft and E� toward the right and the net field over this

region is E 0 .

(b) In the region between the sheets, E� and E� are bothdirected toward the right and the net field is

E �V

0 to the right .

FIG. P24.62

(c) To the right of the negative sheet, E� and E� are again oppositely directed and E 0 .

46 Gauss’s Law

P24.63 The magnitude of the field due to the each sheet given byEquation 24.8 is

E �V

2 0 directed perpendicular to the sheet.

(a) In the region to the left of the pair of sheets, both fields aredirected toward the left and the net field is

FIG. P24.63

E �V

0 to the left .

(b) In the region between the sheets, the fields due to the individual sheets are oppositelydirected and the net field is

E 0 .

(c) In the region to the right of the pair of sheets, both are fields are directed toward the rightand the net field is

E �V

0 to the right .

P24.64 The resultant field within the cavity is the superposition of twofields, one E� due to a uniform sphere of positive charge of radius2a, and the other E� due to a sphere of negative charge of radius acentered within the cavity.

43

43

0

2S US

rr E

�

FHGIKJ � so E r

r� �

�

U Ur3 30 0

�

– ��

FHGIKJ �

43

413

012S U

Sr

r E so E r r� ��

��

U Ur10

10

13 3�b g .

Since r a r � 1 , Er a

� � �

�Ua f3 0

FIG. P24.64

E E Er r a a

i j � �

��

��

�

��

U U U U U3 3 3 3

030 0 0 0 0

� �a.

Thus, Ex 0

and Ea

y �U

3 0 at all points within the cavity.

*P24.65 Consider the charge distribution to be an unbroken charged spherical shell with uniform chargedensity V and a circular disk with charge per area �V . The total field is that due to the whole sphere,

QR

RR4

440

2

2

02

0SS VSH

V�

�

outward plus the field of the disk ��

�

V V2 20 0

radially inward. The total

field is V V V�

��

�0 0 02 2

outward .


on the right has a uniform charge density !". Calculatethe electric field at points (a) to the left of, (b) in between,and (c) to the right of the two sheets.

What If? Repeat the calculations for Problem 62when both sheets have positive uniform surface charge den-sities of value ".

64. A sphere of radius 2a is made of a nonconducting materialthat has a uniform volume charge density #. (Assume thatthe material does not affect the electric field.) A sphericalcavity of radius a is now removed from the sphere, asshown in Figure P24.64. Show that the electric field withinthe cavity is uniform and is given by Ex $ 0 and Ey $#a/3%0. (Suggestion: The field within the cavity is the super-position of the field due to the original uncut sphere, plusthe field due to a sphere the size of the cavity with a uni-form negative charge density !#.)

63.

65. A uniformly charged spherical shell with surface chargedensity " contains a circular hole in its surface. The radiusof the hole is small compared with the radius of thesphere. What is the electric field at the center of the hole?(Suggestion: This problem, like Problem 64, can be solvedby using the idea of superposition.)

66. A closed surface with dimensions a $ b $ 0.400 m andc $ 0.600 m is located as in Figure P24.66. The left edge ofthe closed surface is located at position x $ a. The electricfield throughout the region is nonuniform and given byE $ (3.0 & 2.0x2)i N/C, where x is in meters. Calculatethe net electric flux leaving the closed surface. What netcharge is enclosed by the surface?

A solid insulating sphere of radius R has a nonuniformcharge density that varies with r according to the expression# $ Ar 2, where A is a constant and r ' R is measured fromthe center of the sphere. (a) Show that the magnitude of theelectric field outside (r ( R) the sphere is E $ AR 5/5%0r 2.(b) Show that the magnitude of the electric field inside(r ' R) the sphere is E $ Ar 3/5%0. (Suggestion: The totalcharge Q on the sphere is equal to the integral of # dV, wherer extends from 0 to R ; also, the charge q within a radiusr ' R is less than Q. To evaluate the integrals, note that thevolume element dV for a spherical shell of radius r andthickness dr is equal to 4)r 2dr.)

68. A point charge Q is located on the axis of a disk of radiusR at a distance b from the plane of the disk (Fig. P24.68).Show that if one fourth of the electric flux from thecharge passes through the disk, then R $ √3b.

67.

69. A spherically symmetric charge distribution has a chargedensity given by # $ a/r, where a is constant. Find theelectric field as a function of r. (Suggestion: The chargewithin a sphere of radius R is equal to the integral of # dV,where r extends from 0 to R. To evaluate the integral, notethat the volume element dV for a spherical shell of radius rand thickness dr is equal to 4)r 2dr.)

70. An infinitely long insulating cylinder of radius R has avolume charge density that varies with the radius as

where #0, a, and b are positive constants and r is thedistance from the axis of the cylinder. Use Gauss’s law todetermine the magnitude of the electric field at radialdistances (a) r ' R and (b) r ( R.

Review problem. A slab of insulating material (infinite intwo of its three dimensions) has a uniform positive chargedensity #. An edge view of the slab is shown in FigureP24.71. (a) Show that the magnitude of the electric field adistance x from its center and inside the slab is E $ #x/%0.(b) What If? Suppose an electron of charge !e and massme can move freely within the slab. It is released from restat a distance x from the center. Show that the electronexhibits simple harmonic motion with a frequency

f $1

2) √ #e

me %0

71.

# $ #0 !a !rb "


y

x

2a

a

Figure P24.64

y

x

E

a

c

z

x = a

b

Figure P24.66

R

Q

b

Figure P24.68

46 Gauss’s Law

P24.63 The magnitude of the field due to the each sheet given byEquation 24.8 is

E �V

2 0 directed perpendicular to the sheet.

(a) In the region to the left of the pair of sheets, both fields aredirected toward the left and the net field is

FIG. P24.63

E �V

0 to the left .

(b) In the region between the sheets, the fields due to the individual sheets are oppositelydirected and the net field is

E 0 .

(c) In the region to the right of the pair of sheets, both are fields are directed toward the rightand the net field is

E �V

0 to the right .

P24.64 The resultant field within the cavity is the superposition of twofields, one E� due to a uniform sphere of positive charge of radius2a, and the other E� due to a sphere of negative charge of radius acentered within the cavity.

43

43

0

2S US

rr E

�

FHGIKJ � so E r

r� �

�

U Ur3 30 0

�

– ��

FHGIKJ �

43

413

012S U

Sr

r E so E r r� ��

��

U Ur10

10

13 3�b g .

Since r a r � 1 , Er a

� � �

�Ua f3 0

FIG. P24.64

E E Er r a a

i j � �

��

��

�

��

U U U U U3 3 3 3

030 0 0 0 0

� �a.

Thus, Ex 0

and Ea

y �U

3 0 at all points within the cavity.

*P24.65 Consider the charge distribution to be an unbroken charged spherical shell with uniform chargedensity V and a circular disk with charge per area �V . The total field is that due to the whole sphere,

QR

RR4

440

2

2

02

0SS VSH

V�

�

outward plus the field of the disk ��

�

V V2 20 0

radially inward. The total

field is V V V�

��

�0 0 02 2

outward .


72. A slab of insulating material has a nonuniform positivecharge density ! " Cx 2, where x is measured from thecenter of the slab as shown in Figure P24.71, and C is aconstant. The slab is infinite in the y and z directions.Derive expressions for the electric field in (a) the exteriorregions and (b) the interior region of the slab(#d/2 $ x $ d/2).

73. (a) Using the mathematical similarity between Coulomb’slaw and Newton’s law of universal gravitation, show thatGauss’s law for gravitation can be written as

where m in is the net mass inside the gaussian surface andg " Fg/m represents the gravitational field at any point on

! g%dA " #4&Gm in

the gaussian surface. (b) Determine the gravitational fieldat a distance r from the center of the Earth where r $ RE ,assuming that the Earth’s mass density is uniform.

Answers to Quick Quizzes24.1 (e). The same number of field lines pass through a

sphere of any size. Because points on the surface of thesphere are closer to the charge, the field is stronger.

24.2 (d). All field lines that enter the container also leave thecontainer so that the total flux is zero, regardless of thenature of the field or the container.

24.3 (b) and (d). Statement (a) is not necessarily true becausean equal number of positive and negative charges couldbe present inside the surface. Statement (c) is not neces-sarily true, as can be seen from Figure 24.8: a nonzeroelectric field exists everywhere on the surface, but thecharge is not enclosed within the surface; thus, the netflux is zero.

24.4 (c). The charges q1 and q4 are outside the surface andcontribute zero net flux through S '.

24.5 (d). We don’t need the surfaces to realize that any givenpoint in space will experience an electric field due to alllocal source charges.

24.6 (a). Charges added to the metal cylinder by your brotherwill reside on the outer surface of the conducting cylin-der. If you are on the inside, these charges cannottransfer to you from the inner surface. For this samereason, you are safe in a metal automobile during alightning storm.

Answers to Quick Quizzes 761

x

y

O

d


48 Gauss’s Law

P24.69 E A� �

�z zd

q ar

r drr

in

0 0

2

0

14S

E ra

rdra r

Ea

r

44 4

2

2

2

0 0 0

2

0

SS S

�

�

�

zconstant magnitude

(The direction is radially outward from center for positive a; radially inward for negative a.)

P24.70 In this case the charge density is not uniform, and Gauss’s law is written as E A� �z zd dV1

0U . We

use a gaussian surface which is a cylinder of radius r, length A , and is coaxial with the chargedistribution.

(a) When r R� , this becomes E r arb

dVr

2 0

0 0

SUAb g �

�FHGIKJz . The element of volume is a cylindrical

shell of radius r, length A , and thickness dr so that dV r dr 2S A .

E rr a r

b2

22 3

20

0S

S UA

Ab g �

FHG

IKJ �FHG

IKJ so inside the cylinder, E

ra

rb

�

�FHGIKJ

U0

0223

.

(b) When r R! , Gauss’s law becomes

E r arb

r drR

2 20

0 0

SU

SA Ab g b g �

�FHGIKJz or outside the cylinder, E

Rr

aRb

�

�FHGIKJ

U02

0223

.

P24.71 (a) Consider a cylindrical shaped gaussian surface perpendicularto the yz plane with one end in the yz plane and the other endcontaining the point x :

Use Gauss’s law: E A� �z dqin

0

By symmetry, the electric field is zero in the yz plane and isperpendicular to dA over the wall of the gaussian cylinder.Therefore, the only contribution to the integral is over the endcap containing the point x :

E A� �z dqin

0 or EA

Ax

�Ua f

0

so that at distance x from the mid-line of the slab, Ex

�U

0.

(b) aF

me E

me

mx

e e e

� �

�FHGIKJ

a f U

0

x

y

z x

gaussian surface

FIG. P24.71

The acceleration of the electron is of the form a x �Z 2 with ZU

�e

me 0.

Thus, the motion is simple harmonic with frequency fe

me

�ZS S

U2

12 0

.


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R E - Lehman Collegelehman.edu/faculty/anchordoqui/169-P3-sol.pdfConcentric with this sphere is a conducting spherical shell with inner radius band outer radius c, and having a net