Physics9HW2solns

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Physics 9 Fall 2009Homework 2 - Solutions

1. Chapter 27 - Exercise 5.An electric dipole is formed from ±1.0 nC charges spread 2.0 mm apart. The dipoleis at the origin, oriented along the y−axis. What is the electric field at the points (a)(x, y) = (10 cm, 0 cm) and (b) (x, y) = (0 cm, 10 cm)?

————————————————————————————————————

Solution

(a) The dipole is oriented along the y−axis, and so the point (x, y) = (10, 0) isperpendicular to the dipole, for which the electric field is given by

E dip = −1

4π0

p

r3,

where p = qs is the dipole moment. Plugging in the numbers gives

E dip = −1

4π0

p

r3 = −

9× 109

10−9 × 0.0020.13

= −18 N/C.

(b) Now the point (x, y) = (0, 10) is along the axis of the dipole, with an electric field

E dip = 1

4π0

2 p

r3 .

The only difference between this expression and our answer to part (a) is factorof −2. Thus, we can immediately write down the electric field as E dip = 36 N/C.

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2. Chapter 27 - Exercise 7.The electric field strength 5.0 cm from a very long charged wire is 2000 N/C. What isthe electric field strength 10.0 cm from the wire?

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Solution

Recall that the electric field of a very long wire is

E = λ

2π0r,

which falls off linearly with distance. This means that if we go twice as far away fromthe wire, then the field is half as strong. See that, as r → 2r, and if the field starts atE 0 = λ/2π0, then

E →λ

2π0 (2r) =

1

2

λ

2π0r =

1

2E 0.

So, if the electric field is 2000 N/C at 5.0 cm, then at 10 cm, then the electric fieldmust be 2000/2 = 1000 N/C. Thus, E (10 cm) = 1000 N/C.

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3. Chapter 27 - Exercise 20.A 0.10 g plastic bead is charged by the addition of 1 .0× 1010 excess electrons. Whatelectric field E (strength and direction) will cause the bead to hang suspended in theair?

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Solution

The bead carries a charge Q = 1010 (−1.602× 10−19) = −1.602 × 10−9 C, due to theexcess electrons. We want to support this bead in space, against the pull of gravity.Gravity pulls the bead down, and so the electric force must push the bead up. Sincethe bead is negatively charged, in order for the force to point up, the electric field mustpoint down . So, to balance the forces we need QE = mg, and so E = mgQ . Plugging inthe values gives

E = mg

Q = 10

−3

×9.8

−1.602× 10−9 = −6.1× 105 N/C.

Expressed as a vector, E = −6.1× 105ˆ j N/C.

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4. Chapter 27 - Problem 34.Derive Equation 27.12 for the field E dipole in the plane that bisects an electric dipole.

————————————————————————————————————

Solution

Equation 27.12 in the text gives the result as

E dip = −1

4π0

p

r3,

where r is the distance from the center of thedipole. Let’s start by looking at the picture to theright. The electric field from the positive chargepoints down and to the right, while that from thenegative charge points down and to the left. The

net field points straight down.The magnitude of the net field is E = −2E pos sin θ, where E pos is the field from asingle charge, and the factor of 2 comes from the contributions of the two charges.The sine factor comes from the vertical components of the fields, while the horizontalcomponents cancel. Now,

E pos = 1

4π0

q

r2 + (s/2)2,

while sin θ = s/2√ r2+(s/2)2

. So, adding in the unit vector, ̂j, we have

E = −2 1

4π0

q

r2 + (s/2)2s/2

r2 + (s/2)2ˆ j = −

1

4π0

qsˆ jr2 + (s/2)2

3/2 .But, the dipole moment p = qsˆ j Finally, for distances r s/2, then r2 + (s/2)2 ≈ r2.So, we find

E = − 14π0

p

r3,

which is precisely Eq. 27.12.

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5. Chapter 27 - Problem 45.

(a) Show that the maximum electric field strength on the axis of a ring of chargeoccurs at z = R/

√ 2.

(b) What is the electric field strength at this point?

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Solution

(a) The electric field along the axis of a ring of charge Q is given by E (z ) =1

4π0

zQ

(z2+R2)3/2, where z is the distance along the axis, and R is the radius of the

ring. So, the field depends on z . It has a maximum when dE dz = 0. Taking thederivative gives

dE

dz =

Q

4π0

d

dz

z

(z 2 + R2)3/2

=

Q

4π0

1

(z 2 + R2)3/2 − 3

2

2z 2

(z 2 + R2)5/2

= 0

This is zero when the term inside the parenthesis vanishes. This gives z 2 + R2 =3z 2, or z = R√

2, as claimed.

(b) When z = R√ 2

, then

E

R√

2

=

1

4π0

R/√

2

Q

R/√

22 + R23/2

= 1

4π0

RQ√ 2 (3R2/2)3/2

= 2

3√

3

1

4π0

Q

R2

Thus, the maximum value of the electric field is E = 23√ 3

14π0

QR2

.

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6. Chapter 27 - Problem 54.A problem of practical interest is to make a beamof electrons turn a 90◦ corner. This can be donewith the parallel-plate capacitor shown in the fig-ure. An electron with kinetic energy 3.0 × 10−17

J enters through a small hole in the bottom plateof the capacitor.

(a) Should the bottom plate be charged posi-tive or negative relative to the top plate if you want the electron to turn to the right?Explain.

(b) What strength electric field is needed if theelectron is to emerge from an exit hole 1.0cm away from the entrance hole, traveling

at right angles to its original direction?Hint: The difficulty of this problem de-pends on how you choose your coordinatesystem.

(c) What minimum separation dmin must thecapacitor plates have?

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Solution

Because the field between the plates is constant,this problem is exactly like a gravitational projec-tile problem, only with a different acceleration!So, for a given acceleration, we want to figureout what the electric field needs to be to givethe electron a range of 1 cm. We can see theangles from the diagram to the right, orientingour coordinates as shown.

(a) We want the bottom plate to attract the electron, and so it should be positively

charged.(b) Recall the projectile equations

x (t) = x0 + v0xt + 12

axt2 = v0xt

x (t) = y0 + v0yt + 12ayt

2 = v0yt− 12ayt2,

where we have set x0 = y0 = 0, and note that ax = 0. Solving the first equationfor time gives t = x/v0x, which, upon substituting into the second equation gives

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y (x) = v0yv0x

x− ay2v2

0xx2. Now, when x hits its maximum range, R, then y = 0. So,

y (R) = 0 = v0yv0x

R− ay2v20x

R2 ⇒ ay = 2v0xv0y

R .

This is our acceleration. The electric field needed to give this acceleration is

E = F

q =

mayq

= 2mv0xv0y

qR .

Finally, note that v0x = v0 cos θ, while v0y = v0 sin θ, and so 2v0xv0y = 2v20 sin θ cos θ =

v20 sin 2θ. So, E = mv2

0

qR sin2θ. Finally, recall that the kinetic energy of the electron

is K E = 12mv20, and so

E = 2 (KE )

qR sin2θ.

Now, we plug in the numbers. The launch angle is θ = 45◦, and so

E = 2 (KE )

qR sin2θ =

2× 3.0× 1071.602× 10−19 × 0.01 sin 90

◦ ≈ 37, 500 N/C.

(c) Now, we want the minimum separation. If the plates are too close, then theelectron will collide with the top plate, and won’t continue on its full parabolicarc. The electron reaches its maximum height when x = R/2, i.e., at the halfwaypoint. So, when x = R/2, then

y (R/2) = v0y

v0x

R2 − ay

2v20x

R2

2

= v0y

v0x

R

2 − 1

2v20x2v0xv0y

R R

22

= R4v0yv0x

= R4 tan θ.

This is the minimum height. Now, θ = 45◦, and tan θ = 1, so hmin = R/4. SinceR = 1 cm, then hmin = 2.5 mm.

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7. Chapter 27 - Problem 58.In a classical model of the hydrogen atom, the electron orbits the proton in a circularorbit of radius 0.053 nm. What is the orbital frequency? The proton is so much moremassive than the electron that you can assume the proton is at rest.

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Solution

The electric force holds the electron to the proton. Since it’s moving in a circle, thereis also a centripetal force. The two forces are equal in magnitude, keeping the circularorbit stable. So, F elec =

14π0

e2

r2 = mv2

r = F cent. Now, we can write v = rω, where ω isthe angular frequency. The ordinary frequency is f = ω/2π. Thus, we can solve forthe frequency to find

1

4π0

e2

r2

= mv2

r

= m

r

(rω)2 = mrω2.

Thus, ω2 = 14π0e2

mr3, or since ω = 2πf ,

f = 1

2π

1

4π0

e2

mr3.

Plugging in the numbers gives

f = 1

2π 1

4π0

e2

mr3 =

1

2π

(9× 109) (1.602× 10

−19)2

9.11×

10−31 (0.053×

10−9) = 6.65×1015 rev/sec.

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8. Chapter 27 - Problem 61.Show that an infinite line of charge with linear charge density λ exerts an attractiveforce on an electric dipole with magnitude F = 2λp/4π0r

2. Assume that r is muchlarger than the charge separation in the dipole.

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Solution

Suppose that we hold the dipole a distancer away, oriented such that the negativeside of the dipole points towards the lineof charge, while the positive charge pointsaway. The electric field of an infinite linecharge is given by E = λ2π0d , where d isthe distance away.

Now, if the distance from the line to the center of the dipole is r, then the distance tothe negative charge is r − s/2, while it’s r + s/2 to the positive charge. So, the netforce is F net = −qE neg + qE pos, or

F = −q λ2π0(r−s/2)

+ q λ2π0(r+s/2)

= − qλ2π0

1r−s/2 − 1r+s/2

= − qλ

2π0

r+s/2−r+s/2(r−s/2)(r+s/2)

= − qsλ

2π0

1r2−(s/2)2 .

Now, the dipole moment is defined as p = qs. Furthermore, for r s/2, then r2 −(s/2)2 ≈ r2, and so we find

F ≈ − λp2π0r2

,

which is precisely what we wanted, recalling that the negative sign means that the

force is attractive.

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9. Chapter 27 - Problem 73.A positron is an elementary particle identical to an electron except that its charge is+e. An electron and a positron can rotate about their center of mass as if they were adumbbell connected by a massless rod. What is the orbital frequency for an electronand a positron 1.0 nm apart?

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Solution

This problem is very similar to the hydrogen problem that we did earlier, and we cansolve it in the same way. However, this time the electron doesn’t go around the positron- both go around each other! So, the distance between the two particles is twice theorbital radius. So, the force between the charges is

F E = 1

4π0

e2

(2r)2 ,

and since they are going around in a circle,

F cent = mv2

r .

Again, replacing v = rω = 2πrf , we find that

F cent = F E ⇒mv2

r = mr (2πf )2 =

1

4π0

e2

4r2,

which gives

f = 1

2π

e2

16π0mr3.

Now, r = 0.5 nm is half the distance between the positron and electron. So, we find

f = 1

2π

(9× 109)× (1.602× 10−19)24 (9.11× 10−31) (.5× 10−9)3 = 1.13× 10

14 Hz.

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10. Chapter 27 - Problem 74.You have a summer intern position with a company that designs and builds nanoma-chines. An engineer with the company is designing a microscopic oscillator to helpkeep time, and you’ve been assigned to help him analyze the design. He wants to placea negative charge at the center of a very small, positively charged metal loop. His

claim is that the negative charge will undergo simple harmonic motion at a frequencydetermined by the amount of charge on the loop.

(a) Consider a negative charge near the center of a positively charged ring. Showthat there is a restoring force on the charge if it moves along the z −axis but staysclose to the center. That is, show there’s a force that tires to keep the charge atz = 0.

(b) Show that for small oscillations, with amplitude R, a particle of mass m withcharge −q undergoes simple harmonic motion with frequency

f =

1

2π qQ

4π0mR3 .

R and Q are the radius and charge of the ring.

(c) Evaluate the oscillation frequency for an electron at the center of a 2.0 µm-diameter ring charged to 1.0× 10−13 C.

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Solution

(a) Recall that, along the axis the electric field of a ring of charge is

E z = 14π0

Qz (z 2 + R2)3/2

.

The force on a negative charge is F = qE , or

F = − 14π0

qQz

(z 2 + R2)3/2.

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The natural frequency of a spring with constant k is f = 12π

km

, and so we find

f = 1

2π

k

m =

1

2π

qQ

4π0mR3.

(c) Now, we have an electron of mass me = 9.11 × 10−31 kg, and charge q = e =1.602× 10−19 C, at the center of a ring of radius R = 10−6 m, carrying a chargeQ = 10−13 C. With these numbers,

f = 1

2π

qQ

4π0mR3 =

1

2π

(9× 109) 1.602× 10

−31 × 10−139.11× 10−31 × 10−6 = 2.0× 10

12 Hz.

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