3rd of September, 2008

  1. a b, so a b = 0 and v a = α. Multiplying the first equation by a from the left (using the cross product), we get a × (v ×a) = a ×b a2v -a(v a) = a ×b. Using that v a = α,
    v = a ×b + αa a2

    1. ((b r)a) = b a×((b r)a) = b ×a

    2. r = 3×r = 0

    3. (rr) = 4r× (rr) = 0

      (Note: The curl of any radially symmetric vector field is 0.)

    4. (a × (b ×r)) = -2a b×(a × (b ×r)) = a ×b

    5. ((ar)(b×r)) = (a×b)r×((ar)(b×r)) = 3b(ar)-r(ab)

    6. (r × (a ×r)) = -2a r(r × (a ×r)) = 3r ×a

    The curl of an electrostatic field is 0, so if a and b are not parallel, then only the fields in points (b) and (c) may be electrostatic fields.

    1. The charge density is
      ρ = ε0E = ε0Ae-br r3 r = ε0A(e-br)r r3 + ε0Ae-br r r3

      Let us calculate the two terms of this sum separately:

      e-br = -be-brr^,

      where r^ = rr, and

      r r3 = r^ r2 = 4πδ(r).

      Thus the complete charge density is

      ρ(r) = 4πε0Aδ(r) - ε0Abe-br r2 .

    2. The total charge enclosed by a spherical surface S of radius r centred around the origin is
      Q(r) = ε0 SE S = ε0Ae-br 1 r24πr2 = 4πε 0Ae-br.

      Thus as r , the total charge enclosed by the surface goes to 0.

  2. The work needed to add another layer of charge dq to a sphere of radius r and total charge q is
    dW = 1 4πε0 qdq r

    The sphere has a homogeneous charge distribution, so q = 4π 3 r3ρ and dq = 4πρr2dr.

    Thus the energy needed to assemble a charged sphere of radius R and charge Q is

    0RdW = 4πρ2R 05 15ε0 = 3 20πε0 Q2 R .


    Figure 1: The potential and electric field as a function of the distance from the centre of the sphere.

    The electric field is

    E(r) = Q 4πε0×rR3   if r R 1r2    if r R ,

    and the potential is

    V (r) = Q 4πε0×(3R2 - r2)(2R3)  if r R 1r   if r R .
  3. Let us find the gravitational potential energy W(x) of the two half-spheres as a function of their separation x. The force keeping them together is simply F = dWdx.

    Creating a gap by separating the halves is equivalent to removing (“mining out”) all the matter from the gap, and distributing it on the surface (i.e. it causes the same change ΔW in energy). Provided that the gap between the halves is very narrow compared to the planet’s radius (x R), the change in the gravitational field is negligible.

    Using Gauß’s law, it is easily shown that the gravitational potential below the surface of the spherical planet is

    V (r) = γ M 2R3r2 +  const.,

    where γ is the gravitational constant, M is the mass of the planet, and r is the distance from the centre.

    When moving a piece of mass dm up to the surface, the increase in energy is (V (R) - V (r))dm. To get the total change in energy, we must integrate over the volume of the gap:

    ΔW(x) = (gap)(V (R) - V (r))dm = xγM 2R3 M 4 3πR30R(R2 - r2)2πrdr = x3γM2 16R2 . (1)

    The force keeping the halves together is

    F = dW(x) dx = dW(x) dx -dW(0) dx 0 = dΔW dx = 3γM2 16R2 .

    (I.e. the same as the weight of an object of mass 3 16M on the surface of the planetoid.)

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