What happens to the capacitance when a conducting slab of thickness is partially placed in between?

Answer 1

If the slab is zero thickness, nothing. The lines of electric field not intercepted by the slab are unaffected, so what goes on there is unaffected. The part where the lines are intercepted is converted to two capacitors in series, and the charges on opposite sides of the slab are equal and opposite, equal to the charge on the parts original outer plates facing it. From the outside view of the world, nothing changes on the outer plates, so the capacitance is the same. If the plate is not zero thickeness, then the capacitors on that side become higher and so the total capacitance will increase, with significant distortions of field lines.

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For a parallel-plate capacitor with small spacing, so that fringing can be neglected, with a flat slab of dielectric, the new capacitance is easily calculated by noting that the electric flux D is continuous, and in the volume occupied by the dielectric the E-field is reduced by a factor equal to the relative permittivity. Thus the voltage is reduced for a given amount of charge and the capacitance increased.

If the plate area is A and the spacing is s and the dielectric thickness d, then with a charge Q on the plates the electric flux is just D = Q/A. The electric field on a line between the plates is D/eps0 in the air and D/(eps0 x epsr) inside the dielectric. Therefore the voltage is D/eps0 x (s-d) from the two air paths plus D/(eps0 x epsr) x d from the dielectric path.

So the capacitance in this case is given by:

C = A x eps0 / [s - d(1 - 1/epsr)]

So if d=0 or epsr=1, as they are when there is no dielectric, the capacitance reverts to the standard formula for parallel plates:

C = A x eps0 / s