Download Homework IV

Homework IV
1.) A charge of +2e, where e is the fundamental charge=1.602 10-19 C, is
approaching from a great distance away at v=5 106 m/s a fixed charge of +Q, where
Q is some integer number of positive elemental charges. If the point of closest
approach (i. e. where the +2e charge – a 4He nucleus with 2 protons – stops, turns
around, and is repelled away) is 10-14 m, and if the mass of the 4He nucleus is 6.7 1027 kg, what is Q (round to the nearest integer)?
Solution: as we discussed in class, the alpha particle reaches a distance ‘r’ away
from the Q nucleus where the potential energy (+2e)(+Q)/40r equals the incoming
kinetic energy (1/2)mv2 = 0.5 (6.7 10-27) 25 1012 J = 83.75 10-15 J = ∆U (since U is
defined as 0 at infinity) = 2Q(1.602 10-19 C)2 (8.99 109 Nm2/C2)(1/10-14 m)
the units of the potential energy turn into Nm, which is also J, so the units are good.
Solving for Q: 83.75 10-15 * 10-14/(2 * 2.566 10-38 * 8.99 109) = 1.81 ~ 2, i. e. this would
be the distance of closest approach to another alpha particle.
2.) A very large flat conducting surface is charged up to some charge density . At
the center of this large sheet of charge, the electric field is measured to be -2000 V/m.
What is , in units of number of fundamental charges/square meter? 1 C=6.24 1018
charges.
Solution: We know that the electric field outside of a conductor, via Gauss’ law, is
given by eq. 27.18 E=/0. so we have -2000 V/m = /8.85 10-12 C2/Nm2,
so  = - 2000 8.85 10-12 (V/m)(C2/Nm2). change V/m to (N/C) to get the units of C/m2
so  is 1.1 1011 electrons/m2
3.) We learned a microscopic view of Ohm’s law in class, where the resistivity, in
units of Ω-m, is given by ρ = m/ne2 where n is the density of electrons/volume, e is
the fundamental charge 1.602 10-19 C, and  is the time between collisions.
Remember also that the distance an electron goes between collisions is vaverage*
If the resistivity of some material is 250 µΩ-cm (beware units!), the distance
between collisions is 3 Å (where 3 Å = 3 10-10 m), and n=5 1021 electrons/cm3 (beware
units!), what is vaverage? melectron = 9.11 10-31 kg . The units of Ω can be expressed, as
discussed in class, from V=IR; volts = (Coulombs/sec)Ω as Ω=Vs/C
Solution:
3 10-10 m = vave =m/ne2ρ = 9.11 10-31/ (5 1027 e/m3)(1.602 10-19)2(250 10-8 Ω-m)
where units have been adjusted to kg m s (MKS) units. so =0.284 10-14 s and
vave=3*10-10m/ = 1.06 105 m/s.
5.) A 1 µF capacitor should have 100 J stored in it. What Voltage should it be
charged to?
solution: U=(1/2)CV2. If U=100 J, the V2 =100/0.5 10-6 V2 (where the ‘V’ on the
right is units of voltage), so the voltage is 14.1 kV
5.) Two capacitors are arranged in series to give a total capacitance of 10 µF. The
same two capacitors are now arranged in parallel to give a total capacitance of 45
µF. What are the values of the two capacitors?
solution: in series, Ctotal = 10 µF = C1C2/(C1+C2)
in parallel, Ctotal = 45 µF = C1 + C2. Solve these two equations for the two unknown
values to get that one capacitor is 15 µF and the other is 30 µF. (The equation I
wound up with , by substituting C1+C2 in the first equation by 45 µF from the
second equation, then solving for C1C2 = 450, saying that C1=450/C2, substituting
this back into the second equation and solving for C2 using the quadratic formula, is
C22 – 45 C2 +450 = 0, to give C2 = [45  (452 – 4*450)1/2]/2
to give C2 = either 30 or 15 µF, and then C1 is the other value.)