Physics 9 Fall 2011Midterm 1 Solutions

For the midterm, you may use one sheet of notes with whatever you want to put on it, frontand back. Please sit every other seat, and please don’t cheat ! If something isn’t clear, pleaseask. You may use calculators. All problems are weighted equally. PLEASE BOX YOURFINAL ANSWERS! You have the full length of the class. If you attach any additionalscratch work, then make sure that your name is on every sheet of your work. Good luck!

1. A solid, nonconducting sphere of radius R has a nonuniform charge density, ρ(r) = Cr,where C is a constant, such that the total charge on the sphere is Q.

(a) Determine C in terms of the total charge Q and radius R of the sphere.

(b) What is the electric field at a distance r outside the sphere? (You don’t have tocalculate this, just give the expression.)

(c) Find the electric field at a distance r inside the sphere.

(d) Show that the two fields agree on the boundary of the sphere, when r = R.

Hint: recall that the volume element of a sphere is dV = 4πr2dr.

————————————————————————————————————

Solution

(a) The total charge may be found by integrating the charge density over the totalvolume of the sphere, Q =

∫ρ dV . So,

Q =

∫ρ dV = 4πC

∫ R

0

r3 dr = 4πCR4

4= πR4C,

from which we find that C = QπR4 .

(b) Outside the sphere the field is simply that of a point charge,

Eout =Q

4πε0r2,

which can be easily found using Gauss’s law.(c) To determine the field inside the sphere we use Gauss’s

law, ∮~E · d ~A =

Qencl

ε0.

The first step is to choose a Gaussian surface over whichto integrate. Based on the symmetry of the charge dis-tribution we choose a spherical surface of radius r ≤ R.As we’ve seen many times, along this surface the elec-tric field is constant and points along the direction of thenormal to the surface.

Rr

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Thus, the left-hand side of Gauss’s law becomes∮~E · d ~A =

∮EdA cos θ =

E∮dA = EA = E (4πr2). To determine the enclosed charge we again integrate

the charge density, only this time out to the same radius r ≤ R. Then

Qencl =

∫ρ dV = 4πC

∫ r

0

r3 dr = πCr4 = Qr4

R4,

after plugging in for C. Notice that, if r = 0 we get Qencl = 0, while if r = R wehave Qencl = Q, as expected. Thus, Gauss’s law gives

E(4πr2

)=Q

ε0

r4

R4⇒ Ein =

Q

4πε0R4r2,

which grows with distance.

(d) When r = R the electric field outside then

Eout =Q

4πε0R2,

while

Ein =Q

4πε0R2,

and so the fields agree on the boundary, as we expected.

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2. An uncharged, infinitely long conducting cylinder of radius a is placed in an initiallyuniform electric field ~E = E0j, such that the cylinder’s axis lies along the z axis. Theresulting electrostatic potential is V (x, y, z) = V0 for points inside the cylinder, and

V (x, y, z) = V0 − E0y +E0a

2y

x2 + y2

for points outside the cylinder, where V0 is the (constant) electrostatic potential on the

conductor. Use this expression to determine the resulting electric field, ~E.

————————————————————————————————————

Solution

The electric field can be found from the voltage by finding the gradient, ~E = −∇V .So,

Ex = −∂V∂x

= − ∂∂x

[V0 − E0y + E0a2y

x2+y2

]= 0 + 0 + E0a

2y(

2x(x2+y2)2

)= 2E0a2xy

(x2+y2)2.

Next,Ey = −∂V

∂y

= − ∂∂y

[V0 − E0y + E0a2y

x2+y2

]= 0 + E0 − E0a

2[x2+y2−y(2y)

(x2+y2)2

]= E0 + E0a

2(

y2−x2

(x2+y2)2

).

Finally, the z-component is simplest of all,

Ez = −∂V∂z

= 0,

since nothing depends on z. Thus, the electric field is

~E =2E0a

2xy

(x2 + y2)2 i+

[E0 + E0a

2

(y2 − x2

(x2 + y2)2

)]j

This electric field is made up of the original external electric field, E0j, and the inducedfield, which is everything else

~Eind =2E0a

2xy

(x2 + y2)2 i+ E0a2

(y2 − x2

(x2 + y2)2

)j

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3. Calculate the current through and powerdissipated in each resistor in the circuitseen to the right. Enter your results in thetable below with the given resistances.

Resistor Current (A) Power (W)

R1 = 2 Ω 3.76 28.3R2 = 0.5 Ω 1.88 1.77R3 = 0.5 Ω 1.88 1.77R4 = 2 Ω 3.76 28.3

R1

R2

R4 16 V

R3

————————————————————————————————————

Solution

In order to find the current through eachresistor we need to determine the currentcoming from the battery. We begin by re-drawing the circuit, combining the two re-sistors in parallel to an equivalent circuit.Recalling that

1

Reqv

=1

R2

+1

R3

=1

0.5+

1

0.5,

we see that the equivalent resistor is 1/4= 0.25 Ω, as drawn in the diagram to theright.

2.0 Ω

0.25 Ω

2.0 Ω 16 V

Now we simply have three resistors in se-ries, for which the equivalent resistance canbe found simply by adding the individualresistances,

Reqv = 2 + 2 + .25 = 4.25 Ω,

which gives us our equivalent circuit seenat right.

4.25 Ω

16 V

Now, finding the current produced by the battery is trivial using Kirchhoff’s loop lawand Ohm’s law,

∑i ∆Vi = E − IR = 0, such that I = E/R = 16/4.25 = 3.76 amps.

This is also the current through R1 and R4 in the original circuit since these resistorsare in series with the battery. From Kirchhoff’s junction rules, the 2 amp current issplit evenly between resistors 2 and 3 since they both have the same resistance. Thepower in each resistor is just P = I2R, which gives P1 = P4 = (3.76)2 × 2 = 28.3 W,and P2 = P3 = (1.88)2 × (0.5) = 1.77 W.

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4. The most reactive isotope of uranium, 235U , which has 92 protons and 143 neutrons,undergoes fission under the influence of an additional neutron. One possible (simpli-fied) decay reaction is given below

235U + n0 → 236U → 91Kr + 141Ba+ 3n0,

with the energy released being carried away by the kinetic energy of the products.

(a) Suppose we wanted to break the uranium nucleus apart using a proton, instead.How fast would we need to fire the proton at the nucleus in order to overcomethe electrostatic repulsion so that the proton hits the 235U nucleus, which has aradius of r235U ≈ 7.4× 10−15m? (Note - you can neglect the motion of the muchheavier nucleus!)

(b) The neutron breaks the nucleus apart into a 92Kr nucleus, with 36 protons, anda 141Ba nucleus, with 56 protons (and also three additional neutrons). If thesenuclei are still separated by the same distance, 7.4 × 10−15m, then what is theelectrostatic potential energy between the nuclei?

————————————————————————————————————

Solution

(a) In order to get the proton close enough to hit the nucleus, we have to give it enoughenergy to overcome the electrostatic repulsion that it feels due to the positivelycharged nucleus. This means that the kinetic energy, 1

2mv2, of the proton has to

be equal to the potential energy between the proton and the nucleus, kQqr

. Settingthese equal and solving for the velocity gives

v =

√2kQq

mr=

√2 · 8.99× 109 · 92 (1.6× 10−19)2

1.67× 10−27 · 7.4× 10−15= 5.85× 107 m/s.

(b) The electrostatic potential energy is just the potential energy between the Krand Ba nuclei, PE = kQ1Q2

r, where Q1 = 36e, and Q2 = 56e. So,

PE =kQ1Q2

r=

8.99× 109 · 36 · 56 · e2

7.4× 10−15= 6.3× 10−11 J.

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Extra Credit Question!!

The following is worth 10 extra credit points!

Embodied in Kirchhoff’s Laws are two conservation laws. Explain what they are.

————————————————————————————————————

Solution

Kirchhoff’s loop rule, which says that the net change in voltage along a closed circuitis zero, ∑

i

∆Vi = 0,

expresses the law of conservation of energy. The change in voltage that a charge qpasses through changes its potential energy by an amount ∆PE = q∆V . So, we couldwrite

q∑i

∆Vi =∑i

q∆Vi =∑i

∆PEi = 0.

Thus, the loop law says that the net change in potential energy around the loop iszero. Since ∆E = 0 around the loop for the closed system, then ∆KE = −∆PE = 0,and so energy is conserved.

The Kirchhoff’s junction rule, which says that the sum of the currents in to a junctionis equal to the sum of the currents out of the junction,∑

i

Iin =∑i

Iout,

expresses the conservation of charge. This is because the current is made up of movingcharges, and since the charges don’t get lost in the junction, whatever charges comein to the junction have to go back out again. So, charge in equals charge out, whichmeans that current in also equals current out.

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Some Possibly Useful Information

Some Useful Constants.

Coulomb’s Law constant k ≡ 14πε0

= 8.99× 109 Nm2

C2 .

The magnetic permeability constant µ0 = 4π × 10−7 NA2 .

Speed of Light c = 2.99× 108m/s.

Newton’s Gravitational Constant G = 6.672× 10−11 Nm2

kg2.

The charge on the proton e = 1.602× 10−19C

The mass of the electron, me = 9.11× 10−31 kg.

The mass of the proton, mp = 1.673× 10−27 kg.

Boltzmann’s constant, kB = 1.381× 10−23J/K.

1 eV = 1.602× 10−19 Joules⇒ 1 MeV = 106 eV .

1 A = 10−10 meters.

Planck’s constant, h = 6.63× 10−34 J s = 4.14× 10−15 eV s.

The reduced Planck’s constant, ~ ≡ h2π

= 1.05× 10−34 J s = 6.58× 10−16 eV s.

Some Useful Mathematical Ideas.

∫xndx =

xn+1

n+1n 6= −1,

ln (x) n = −1.∫dx√a2+x2 = ln

(x+√a2 + x2

).∫

x dx√a2+x2 =

√a2 + x2.

Other Useful Stuff.

The force on an object moving in a circle is F = mv2

r.

Kinetmatic equations x(t) = x0 + v0xt+ 12axt

2, y(t) = y0 + v0yt+ 12ayt

2.

The binomial expansion, (1 + x)n ≈ 1 + nx, if x 1.

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