Physics 9 Fall 2010 Midterm 1 Solutions For the midterm, you may use one sheet of notes with whatever you want to put on it, front and back. Please sit every other seat, and please don’t cheat! If something isn’t clear, please ask. You may use calculators. All problems are weighted equally. PLEASE BOX YOUR FINAL ANSWERS! You have the full length of the class. If you attach any additional scratch work, then make sure that your name is on every sheet of your work. Good luck! 1. A point particle that has charge +q and unknown mass m is released from rest in a ~ that is directed vertically downward. The region that has a constant electric field E √ particle hits the ground at a speed v = 2 gh, where h is the initial height of the particle. Find m in terms of E, q, and g. ———————————————————————————————————— Solution When the particle starts out at height h, it has two forces acting on it, both pointing downward: gravity, Fg = mg, and the electric force FE = qE. So, Fnet = qE + mg, . For a constant and is constant. So, the constant acceleration is a = Fnet /m = g + qE m acceleration, the particle travels a distance h in time t, such that r 1 2 2h . h = at ⇒ t = 2 a q √ So, after this time, the particle is moving at velocity v = at = a 2h = 2ha. But, a √ √ √ we’re told that it hits the ground with speed v = 2 gh, and so 2 gh = 2ha, or a = 2g. Since a = g + qE/m, we see that qE qE =g⇒m= . m g (You could figure out the form of this expression just by looking at the units.)
1
2. A nonconducting solid sphere of radius 10.0 cm has a uniform volume charge density. The magnitude of the electric field 20.0 cm from the sphere’s center is 1.88 × 103 N/C. (a) What is the sphere’s volume charge density, ρ? (b) Find the magnitude of the electric field at a distance of 5.00 cm from the sphere’s center. (Hint: use Gauss’s law to figure out the field at this position.) ———————————————————————————————————— Solution (a) As we’ve discussed many times, the field outside the sphere is just that of a point charge, E = 4πQ0 r2 . We can determine the charge on the sphere by solving for Q = 4π0 r2 E. Plugging in the numbers gives Q = 4π0 r2 E =
1 (.200)2 (1.88 × 103 ) = 9.36 × 10−9 C. 9 × 109
For a constant charge density ρ = Q/V , and for a sphere of radius R, V = 4π R3 , 3 so 3 × 9.36 × 10−9 3Q = = 2 × 10−6 C/m3 . ρ= 3 3 4πR 4π(.1 ) H ~ ·dA ~ = Qencl /0 . (b) Now we want the field inside the sphere. We need Gauss’s law, E Taking our Gaussian surface as a sphere of radius r the spherical symmetry of the charge distribution tells us that the field is constant on the Gaussian surface, and H ~ ~ = EA, where A = 4πr2 is the surface area. Now, because so, as always, E · dA the charge density is uniform, the total charge enclosed in the Gaussian surface 3 3Q 4πr3 = Qr , and so Gauss’s law gives is ρVencl = 4πR 3 × 3 R3 E(4πr2 ) =
Q Qr3 ⇒ E = r. 0 R3 4π0 R3
Plugging in the numbers gives E=
Q 9 × 109 (9.36 × 10−9 ) r = (.05) = 4212 N/C. 4π0 R3 .13
2
3. A radioactive 210 Po nucleus emits an α particle that has a charge +2e. When the α particle is a large distance from the nucleus, it has a kinetic energy of 5.30 MeV. Assume that the α particle had negligible kinetic energy as it left the surface of the nucleus. The “daughter” (or residual) nucleus 206 Pb has a charge of +82e. Determine the radius of the 206 Pb nucleus. (Neglect the radius of the α particle and assume the 206 Pb nucleus remains at rest.) Express your answer in terms of fermis (1 fm = 10−15 meters). ———————————————————————————————————— Solution The α particle starts off at the edge of the 206 Pb nucleus, say at a distance R. If the α particle starts with no kinetic energy, then the initial energy is purely potential, so . The α particle is repelled from the 206 Pb nucleus, and picks up Ei = P E = 82e×2e 4π0 R speed. When it’s far away all the energy is kinetic, since the potential goes to zero at large distances. The final energy is KE = 5.30 MeV, and so equating the two energies 164e2 gives 4π = 5.30 MeV. Solving for R gives 0R R=
164e2 4π0 × (5.30 MeV)
Now, 1 MeV = 106 ×1.602×10−19 = 1.602×10−13 joules, and so 5.30 MeV = 8.5×10−13 joules. Thus, R=
164 × (1.602 × 10−19 )2 164e2 = 9 × 109 × = 4.45 × 10−14 m, 4π0 × (5.30 MeV) 8.5 × 10−13
or about 45 fm.
3
4. Consider the circuit in the figure to the right. (a) What is the equivalent resistance between points a and b? (b) How would adding a fifth resistor that has resistance R between point c and d affect the equivalent resistance between point a and b? (Hint: what is the potential difference between points c and d?)
———————————————————————————————————— Solution
(a) This circuit is equivalent to a par allel circuit with two resistors in series on each side, as we see in the a figure to the right. The equivalent ð resistance of each side of the paral lel circuit is R + R = 2R.
Now we have an equivalent parallel circuit with just two resistors, each of resistance 2R, as seen in the next figure. Since this is a parallel circuit, the equivalent resistance is 1 Requiv
=
2R
R
R
b
ò
a
b
ò
and so the equivalent resistance of the entire circuit is Requiv = R.
R
ð
1 1 1 + = , 2R 2R R
R
2R
(b) Because all of the resistors are identical, the points c and d are at the same potential. This means that if we connected a wire between points c and d, then no current would flow along it. Thus, if we added a resistor between these two points, it would do nothing, and so the equivalent resistance of the entire circuit would still be R.
4
Extra Credit Question!! The following is worth 10 extra credit points!
A charge q sits at the back corner of a cube, as ~ through shown in the figure. What is the flux of E the shaded side? If the charge was in the center, the flux would be ΦE = 6Q0 (the total flux divided evenly into six cube faces), but the charge is in the corner. (Hint: Think of the symmetry of the problem - you don’t have to do any complicated integrals!)
———————————————————————————————————— Solution We could try to figure out the components of the electric field, and integrate it over the surface of the cube, but there’s a much easier and more clever way. Suppose that we imagine stacking other boxes around the cube, keeping the charge at the center, as seen in the figure to the right. The net flux of the electric field through a volume doesn’t depend on the shape of the volume, ~ through the whole set of and so the net flux of E boxes is just q/0 . We only want to know the flux through the shaded side. Since the charge is at the center, each face of the cube has 1/6 of the flux passing through it. The shaded side is one quarter of the face of the cube, and so the net flux passing through there is 1/4 of 1/6, or 1/24 of the total flux. So, passing through R the net flux ~ · dA ~= q . the shaded face is E 240
5
q
Some Possibly Useful Information Some Useful Constants. Coulomb’s Law constant k ≡
1 4π0
= 8.99 × 109
N m2 . C2
The magnetic permeability constant µ0 = 4π × 10−7
N . A2
Speed of Light c = 2.99 × 108 m/s. Newton’s Gravitational Constant G = 6.672 × 10−11
N m2 . kg 2
The charge on the proton e = 1.602 × 10−19 C 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 × 10−23 J/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, ~ ≡
h 2π
= 1.05 × 10−34 J s = 6.58 × 10−16 eV s.
Some Useful Mathematical Ideas. ( n+1 x R n n 6= −1, x dx = n+1 ln (x) n = −1. √ R dx 2 + x2 . √ = ln x + a 2 2 a +x R
√ x dx a2 +x2
=
√
a2 + x 2 .
Other Useful Stuff. The force on an object moving in a circle is F =
mv 2 . r
Kinetmatic equations x(t) = x0 + v0x t + 21 ax t2 , y(t) = y0 + v0y t + 12 ay t2 . The binomial expansion, (1 + x)n ≈ 1 + nx, if x 1.
6
