A point charge is held at height above the upper face of a slab of linear dielectric…
PHYSICS · Multi-part · Fermi University Physics · Electricity, Magnetism & Electromagnetism · Conductors, capacitance and boundary-value electrostatics · Electric fields in matter
The Boundary-Value Problem, Ch Linear dielectrics and the boundary-value problem with matter in it
Problem
A point charge q=5.0nC is held at height d=1.0cm above the upper face of a slab of linear dielectric (εr=4.0) of thickness t=4.0cm. The lower face of the slab rests on a large, grounded, perfectly conducting plane. The lateral extent of the slab and the conducting plane is much larger than d and t, so every interface may be treated as an infinite plane; there is no free charge anywhere except on q itself and whatever charge the conductor supplies by conduction. Take the conducting plane as z=0, the slab as 0≤z≤t, and vacuum as z>t; the charge sits at z=t+d.
((a)) Name the three regions and state, at each of the two interfaces (z=t and z=0), which quantities are fixed or continuous, and why.
((b)) The exact potential above the slab is built from an infinite sequence of image charges produced by repeated reflection between the two interfaces. Construct only the first two of these images: (i) the image that would be needed if the slab extended to z→−∞ with no conductor present, read directly from the dielectric half-space result; (ii) the further image needed to make V=0 on the actual conducting plane, obtained by first replacing q with the equivalent charge q′′ that describes the field just inside a dielectric half-space (again ignoring the finite thickness for this step), then reflecting q′′ in the conducting plane and carrying that reflection back through the top face, using the half-space result with the roles of the two media exchanged, to find its strength as an image in the potential above the slab. State the magnitude and the z-coordinate of each of these two images.
((c)) Without calculating it, say what produces the next image in the sequence, and why the whole series converges.