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PHYSICS 110A MIDTERM
Prof. C. F. McKee Oct. 20, 2006
Partial credit will be given, so show your reasoning carefully. The number of points for each problem is listed at the left. You are permitted 1 sheet of notes, written on two sides.
- Vector math
(5) a) Evaluate ˆr at r = (1, 2 , 2)
(10) b) Give the general solution for r > 1 for the function φ(r) that satisfies
∇^2 φ = 2, r ≤ 1 , ∇^2 φ = 0, r > 1. (1)
In other words, you do not need to give the solution for r ≤ 1. You will receive partial credit if you give your answer in terms of a volume integral over a specified volume.
Solution: a)
ˆr = r r
r^2 = 1 + 4 + 4 = 9 ⇒ r = 3 (3) ˆr = 1 3
b) The formal solution is φ(r) = −
4 π
r′< 1
rsf^ dτ^
where rsf ≡ |r − r′|. This is equivalent to a charge distribution with ρ/ 0 = −2 for r ≤ 1 ⇒ the total charge is Q = 4πρ/3 = − 8 π 0 /3. Thus, the solution for r > 1 is
φ = 1 4 π 0
Q
r
3 r
- Electrostatic potential
(20) Find the potential inside a uniformly charged sphere (i.e., the charge fills the volume of the sphere) with a radius R and a total charge Q. Set the potential at infinity equal to 0.
Solution: For r ≥ R, E =
4 π 0
Q
r^2 ˆr^ ⇒^ V^ =^
4 π 0
Q
r.^ (7)
For r < R, the charge inside r is q(r) = (r/R)^3 Q, so that
E = 1 4 π 0
q(r) r^2
ˆr = 1 4 π 0
Qr R^3
ˆr
⇒ V (r) = V (R) − 1 4 π 0
∫ (^) r
R
Qr R^3
dr
4 π 0
Q
R
2 −^
r^2 2 R^2
- Force between an atom and an ion
(25) Find the force between an ion of charge q and an atom of polarizability α that are separated by a distance r. Is the force attractive or repulsive?
Solution: The ion induces a dipole p = αE in the atom. The force that the ion exerts on the dipole is F = (p · ∇)E = α(E · ∇)E, (8) where E =
4 π 0
q r^2 ˆr.^ (9) Note that since p ∝ E, it is not constant, so one cannot use the expression F = ∇(p · E). Set up a Cartesion coordinate system with ˆx in the direction of the atom from the ion; then
F = α
4 π 0
q x^2
∂x
q x^2 (10)
= − 2 α
q 4 π 0
r^5
where we replaced x by the separation r in the final step.
- Hall effect
A current I flows to the right through a rectangular bar of conducting material in the presence of a uniform magnetic field B 0 perpendicular to the current (see figure). The magnetic field deflects the charge, leading to an accumulation of charge on two opposite sides of the bar. This charge produces an electric field that exactly counteracts the magnetic force, so that the current can flow parallel to the bar. Assume that the charge carriers are electrons, with a density n and charge −e.
(5) a) Which surfaces do the charges accumulate on?
(10) b) What is the potential difference across the bar (the Hall voltage)?
(5) c) What is the magnitude of the surface charge?
By symmetry, |B 1 | = |B 2 |; furthermore, the right-hand rule implies that the field above the bar is directed in the opposite direction to that below the bar, B 1 = −B 2. It follows that
B 1 = μ^0 K 2
xˆ. (20)
Now, the surface current is
K = I w
= 1 A m−^1 , (21)
so that B 1 =^12 × 4 π × 10 −^7 T, (22)
which is far smaller than the 1 T field in the bar.
Constants: 1 4 π 0 = 9^ ×^10
(^9) N m (^2) /C 2
μ 0 = 4π × 10 −^7 N/A^2
