Physics 9 Fall 2009Homework 3 - Solutions

1. Chapter 28 - Exercise 8.

The cube in the figure contains nonet charge. The electric field is con-stant over each face of the cube.Does the missing electric field vec-tor on the front face point in or out?What is the field strength?

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Solution

Because the electric field is constant, and because the cube contains no net charge,the total flux through the cube has to be zero. The flux into the cube is 15 × A +20× A = 35A, where A is the area of the cubes face. The flux out of the the cube is10×A+ 15×A+ 15×A = 40A. This means that more flux is leaving the cube thanentering it. Because the fluxes in have to balance the fluxes out, we see that the fluxon the front face has to point in, and have a magnitude of 5 N/C.

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2. Chapter 28 - Exercise 11.

The electric flux through the sur-face shown in the figure is 25Nm2/C. What is the electric fieldstrength?

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Solution

Since the electric field is constant, the net flux through the surface is ΦE = ~E · ~A =EA cos θ. The electric field makes an angle of 60◦ with the surface of the area, meaningthat it makes an angle of θ = 30◦ with respect to the normal of the surface, which iswhat we want. Now, ΦE = EA cos θ ⇒ E = ΦE/A cos θ. With numbers we find

E =ΦE

A cos θ=

25

(0.1)2 cos 30◦= 2890 N/C.

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3. Chapter 28 - Exercise 13.A 2.0 cm × 3.0 cm rectangle lies in the xz−plane. What is the electric flux throughthe rectangle if

(a) ~E =(

50i+ 100k)

N/C?

(b) ~E =(

50i+ 100j)

N/C?

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Solution

The rectangle lies in the xz−plane, and so its normal points along the y−direction.So, ~A = Ay j, where Ay = 0.02× 0.03 = 6× 10−4 m2. Thus, ~A = 6× 10−4j.

(a) Since the flux is ΦE = ~E · ~A, and since i · j = i · k = 0, then the net flux in part(a) is zero!

(b) Again, ΦE = ~E · ~A, and since i · j = 0, and j · j = 1, we have

ΦE = ~E · ~A = EyAy = 100× 6× 10−6 = 0.06 N/m2.

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4. Chapter 28 - Exercise 29.

The figure shows four sides of a 3.0cm × 3.0 cm × 3.0 cm cube.

(a) What are the electric fluxesΦ1 to Φ4 through sides 1 to4?

(b) What is the net flux throughthese four sides?

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Solution

(a) Again, the flux is ΦE = AE cos θ. On side 1, the angle between ~E and the normalis θ1 = 150◦. For side 2, it’s θ2 = 60◦, for side 3 it’s θ3 = 30◦, and for side 4 it’sθ4 = 120◦. Thus,

Φ1 = EA cos θ1 = (500) (.03)2 cos 150◦ = −0.39 N/Cm2.

Φ2 = EA cos θ2 = (500) (.03)2 cos 60◦ = 0.225 N/Cm2.

Φ3 = EA cos θ3 = (500) (.03)2 cos 30◦ = 0.39 N/Cm2.

Φ4 = EA cos θ4 = (500) (.03)2 cos 120◦ = −0.225 N/Cm2.

(b) The net flux through the four sides is just the sum of the individual fluxes,

Φ = Φ1 + Φ2 + Φ3 + Φ4 = 0,

which we know has to be the case, since there is no enclosed net charge.

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5. Chapter 28 - Problem 39.A hollow metal sphere has inner radius a and outer radius b. The hollow sphere hascharge +2Q. A point charge +Q sits at the center of the hollow sphere.

(a) Determine the electric fields in the three regions r ≤ a, a < r < b, and r ≥ b.

(b) How much charge is on the inside surface of the hollow sphere? On the exteriorsurface?

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Solution

The setup is seen in the figureto the right. There is a singlepoint charge, Q at the center ofthe sphere. The inner radius of thesphere is a, and the outer radius isb. We need to determine the fieldsin each of the regions using Gauss’slaw.

(a) Gauss’s law says that ∮~E · d ~A =

Qencl

ε0.

Inside the sphere, for r ≤ a, the net enclosed charge is entirely from the pointcharge at the center. So, Qencl = +Q. Now, the spherical symmetry of the chargesays that we should take a sphere of radius r < a as our Gaussian surface. Every-where on that sphere, the electric field is constant. Furthermore, the direction ofthe electric field is perpendicular to the direction of the normal to the Gaussiansurface (both point radially). So,

∮~E · d ~A =

∮E dA = E

∮dA = EA, where A

is the area of the surface. Since the surface area of sphere is 4πr2, we find thatE (r) = 1

4πε0

Qr2

, as expected.

Next, let’s look outside the sphere, where r ≥ b. In this case the net enclosedcharge is +3Q, from the center point charge, and the charged sphere. We againwant a spherical Gaussian surface. Proceeding as before we find E (r) = 1

4πε0

3Qr2

.

Finally, what happens inside the metal itself? This is a static collection of charges.The charges inside the metal sphere feel the positive charge inside the shell. Theymove around due to the electrical force until they have reached a configurationwhere all the force cancels out – in other words, until the electric field is zero.What happens? Since there is a positive charge inside, negative charges in the

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metal shell are attracted to it and move to “coat” the inner surface with a negativecharge, −Q, which is just enough to cancel the positive point charge.

This leaves a net positive charge in the shell which then pushes away from itself,and onto the surface of the sphere. So, this charge is now on the surface, inaddition to the +2Q that was there before! Thus, the net charge on the surfaceof the shell is +3Q!

So, to recap,

E (r) =

1

4πε0

Qr2

r ≤ a

0 a < r < b1

4πε0

3Qr2

r ≥ b.

(b) As discussed above, there is −Q smeared over the inside surface, and +3Q on theoutside.

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6. Chapter 28 - Problem 44.A positive point charge q sits at the center of a hollow spherical shell. The shell, withradius R and negligible thickness, has net charge −2q. Find an expression for theelectric field strength (a) inside the sphere, r < R, and (b) outside the sphere, r < R.In what direction does the electric field point in each case?

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Solution

(a) Again, let’s start by drawing a pic-ture, seen to the right. Taking aGaussian surface inside the shell,r < R, encloses only the positivecharge +q. So, by Gauss’s law,∮

~E · d ~A = E(4πr2

)=

q

ε0,

which says that ~E = 14πε0

qr2r, for

r ≤ R, and points radially outward.

(b) Now, outside the shell, a Gaussian surface encloses both charges, and so the net

enclosed charge is Qencl = −2q+q = −q. So, Gauss’s theorem says that∮~E ·d ~A =

− qε0

. Evaluating the integral as before gives ~E = − 14πε0

qr2r, for r ≥ R, and points

radially inward.

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7. Chapter 28 - Problem 45.Find the electric field inside and outside a hollow plastic ball of radius R that hascharge Q uniformly distributed on its outer surface.

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Solution

Inside the sphere the field is zero – there’s no charge inside the ball; it’s all on thesurface. So, inside E = 0, for r < R. Outside the ball, we do see a net charge ofQ. Since the ball is spherically symmetric, we choose a spherical Gaussian surface ofradius r. As before, the electric field is constant everywhere on the surface because ofthe symmetry. So, as we’ve seen before,∮

~E · d ~A = EA =Qencl

ε0=Q

ε0,

and so we just find the ordinary Coulomb’s law, E = 14πε0

Qr2

. So, the field looks justlike an ordinary point charge, as we’ve seen before.

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8. Chapter 28 - Problem 53.A spherical shell has inner radius Rin and outer radius Rout. The shell contains totalcharge Q, uniformly distributed. The interior of the shell is empty of charge andmatter.

(a) Find the electric field outside of the shell, r ≥ Rout.

(b) Find the electric field in the interior of the shell, r ≤ Rin.

(c) Find the electric field within the shell, Rin ≤ r ≤ Rout.

(d) Show that your solutions match at both the inner and outer boundaries.

(e) Draw a graph of E versus r.

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Solution

Once again, we will use Gauss’s Law to find the electric field in each region.

(a) Outside of the shell, when r ≥ Rout, the field looks like that of a point charge,

since any spherical Gaussian surface encloses the full charge. So, ~E = 14πε0

Qr2r.

(b) Inside the shell, where r ≤ Rin, there is no charge and hence nothing to source

the electric field. So, the electric field is zero, ~E = 0.

(c) Now, this gets a bit tougher. Now, we want to use a Gaussian surface of radiusRin ≤ r ≥ Rout; i.e., the surface is inside the metal itself. Since the total charge onthe shell is uniformly distributed, the Gaussian surface encloses a certain amountof charge. The charge density is constant, so ρ = Q/V , where V is the volumeof the charge distribution. The total volume of the sphere is just the differencebetween two spheres of radii Rout and Rin, V = 4

3π (R3

out −R3in). Thus, ρ =

3Q

4π(R3out−R3

in).

Now the Gaussian surface encloses a volume Vencl = 43π (r3 −R3

in), which gives an

enclosed charge of Qencl = ρVencl = Q(

r3−R3in

R3out−R3

in

). Again, using Gauss’s law we

find that∮~E · d ~A = EA = E (4πr2) = Qencl

ε0, and so

~E =1

4πε0

Q

r2

(r3 −R3

in

R3out −R3

in

)r

Again, we find the following field values

~E (r) =

0 r ≤ Rin

14πε0

Qr2

(r3−R3

in

R3out−R3

in

)r Rin ≤ r ≤ Rout

14πε0

Qr2r r ≥ Rout.

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(d) Now, the boundaries are at the values r = Rin and Rout. We want to lookat the field inside the metal, when Rin ≤ r ≤ Rout. When r = Rin we get~E (Rin) = 1

4πε0

QR2

in

(R3

in−R3in

R3out−R3

in

)r = 0. This matches the value inside the sphere. On

the other boundary, when r = Rout, then ~E = 14πε0

QR2

out

(R3

out−R3in

R3out−R3

in

)r = 1

4πε0

QR2

outr,

which exactly matches our solution for r ≥ Rout on the surface. So, we can trustour solutions!

(e)

The electric field is plottedin the graph to the right.The field starts out at zeroat the origin, and remains sountil it reaches the inner sur-face at r = Rin. Then thefield begins to rise, not quitelinearly, going as r−r−2, un-til it reaches the outer sur-face, r = Rout. This is whereit reaches its maximum, af-ter which it begins to fall offwith the usual r−2 Coulomblaw.

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9. Chapter 28 - Problem 54.An early model of the atom, proposed by Rutherford after his discovery of the atomicnucleus, had a positive point charge +Ze (the nucleus) at the center of a sphere ofradius R with uniformly distributed negative charge −Ze. Z is the atomic number, thenumber of protons in the nucleus and the number of electrons in the negative sphere.

(a) Show that the electric field inside this atom is

Ein =Ze

4πε0

(1

r2− r

R3

).

(b) What is E at the surface of the atom? Is this the expected value? Explain.

(c) A uranium atom has Z = 92 and R = 0.10 nm. What is the electric field strengthat r = 1

2R?

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Solution

The atom is seen in the figureto the right. It contains a pos-itively charged nucleus of charge+Ze, which is embedded inside auniform sphere of charge −Ze. Weare interested in the field inside thesphere, so we take a spherical gaus-sian surface of radius r.

(a) Since the electric field is radial, and everywhere constant on the Gaussian surface,

Gauss’s law gives the usual result that∮~E · d ~A = EA = E (4πr2) = Qencl

ε0.

Now, what’s Qencl? This problem is very similar to the last one, except withthe addition of +Ze at the center. So, the total enclosed charge is Qencl =+Ze− negative charge. The negative charge is just the charge density, times the

enclosed volume,(−Ze43πR3

)× 4π

3r3 = −Ze r3

R3 . Thus,

Qencl = Ze

(1− r3

R3

),

which, using Gauss’s law as discussed above, gives

Ein =Ze

4πε0

(1

r2− r

R3

).

(b) On the surface of the atom, r = R, and so the electric field is zero. This is to beexpected, since on the surface the enclosed charge is zero (+Ze−Ze), and so theelectric flux is zero from Gauss’s law. The atom looks neutral outside the surface.

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(c) When r = R/2, then

E

(R

2

)=

Ze

4πε0

(4

R2− 1

2R2

)=

7

2

Ze

4πε0R2.

Plugging in the numbers gives

E

(R

2

)=

7

2

Ze

4πε0R2=

7

2

92× 1.602× 10−19 × 9× 109

(0.1× 10−9)2 = 4.64× 1013N/C.

So, E = 4.64× 1013N/C, which is a very strong electric field!

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10. Chapter 28 - Problem 58.A sphere of radius R has total charge Q. The volume charge density (C/m3) withinthe sphere is

ρ = ρ0

(1− r

R

).

This charge density decreases linearly from ρ0 at the center to zero at the edge of thesphere.

(a) Show that ρ0 = 3Q/πR2.Hint: You’ll need to do a volume integral.

(b) Show that the electric field inside the sphere points radially outward with mag-nitude

E =Qr

4πε0R3

(4− 3

r

R

).

(c) Show that your result of part b has the expected value at r = R.

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Solution

(a) Now, the charge density is not constant! So, we can’t just say that Q = ρV .Instead, we have to integrate, Q =

∫dq =

∫ρdV . As we’ve seen before, for a

sphere dV = 4πr2dr, as you can see by imagining a series of thin shells of thicknessdr piled on top of each other. So,

Q =

∫ρdV = 4πρ0

∫ R

0

(1− r

R

)r2 dr = 4πρ0

[r3

3− r4

4R

] ∣∣∣∣R0

=π

3ρ0R

3,

and so ρ0 = 3QπR3 .

(b) Now, we take a Gaussian surface of radius r inside the sphere. From Gauss’s law,∮~E · d ~A = Qencl

ε0, the left-hand side is E (4πr2) because the field is radial inside

the sphere. What’s Qencl? Again, we integrate as before, only now out to a radiusr, corresponding to our surface.

Qencl =

∫ρdV = 4πρ0

∫ r

0

(1− r

R

)r2 dr = 4πρ0

[r3

3− r4

4R

]=

4πρ0

3

[r3 − 3r4

4R

].

Plugging for ρ0 gives

Qencl =4πρ0

3

[r3 − 3r4

4R

]= Q

r3

R3

[4− 3

r

R

].

Thus, from Gauss’s law

E (r) =Qr

4πε0R3

(4− 3

r

R

).

(c) At r = R, then E (R) = QR4πε0R3

(4− 3R

R

)= Q

4πε0R2 . So, the electric field at r = Ris that of a point charge, which is exactly we should expect!

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