Physics 304 Problem Set 4 Due: Monday, Sept. 26
(1) Griffiths problem 3.4 (page 121).
(2) An infinite grounded conducting plane sits on the x y plane at z = 0. Above this plane a thin insulating rod with uniform charge density λ is placed along the z axis, from z = a to z = b . Note that superposition will help to solve this problem, based on the more standard image problem.
(a) What will be the image charge?
(b) Solve for the charge density on the conducting plane at a radius s from the origin.
(c) Find the force on the charged rod.
(3) Griffiths problem 3.8 (page 126).
(4) Griffiths problem 3.10 (page 126).
(5) Two very thin spherical conducting shells of radius a (smaller shell) and b (larger shell) are arranged concentrically, and charged so that shell a is at potential V a
V b
, relative to infinity.
, and shell b is at potential
(a) First solve Laplace’s equation for the general form of the solution, in spherical coordinates for the case of no angular dependence.
(b) Solve for the potential vs. r in the regions r > b , a < r < b , and r < a .
(c) For the case where V a
= – V b
, solve for the electric field between the two shells vs. r , and find the total charge on each shell that will produce this situation.
(6) A parallel plate capacitor is shown, with an uncharged, electrically isolated metal slab that is allowed to slide horizontally into the capacitor. The capacitor plate separation is d , and the rectangular plates have dimensions a × b . When the metal slab extends in by a distance x , it occupies an area x × a , and the gap above it is d /2. The slab can slide so that it does not touch the bottom plate, but consider the narrow gap below it to have negligible volume. E will be finite in the tiny lower gap, so ∆ V is thus essentially zero, and the metal slab will thus be at the same potential as the lower plate. Also neglect edge effects both at the capacitor edge and at the edge of the metal slab, assuming that the E field is locally uniform and vertical everywhere.
(a) With fixed charge Q , note that the charge densities on the surfaces in the two regions will be different, but ∆ V across the two different gaps is the same. In terms of this ∆ V , solve for the E fields, and then the surface charges, and then by setting the total charge = Q , eliminate ∆ V to obtain the E fields in the two regions.
(b) Find the total energy of this configuration by integrating the field energy.
(c) Find the force on the metal slab by using the gradient of the potential method. Will it be pulled into the gap or pushed out?
(d) The force on the slab cannot be obtained directly from the fields acting on it if we neglect edge effects. Sketch a plausible charge distribution for the surface of the metal slab, with field lines included, showing where the force would be exerted on the slab.