A horizontal parallel-plate capacitor (plate separation
PHYSICS · Multi-part · AP · Physics 2: Algebra-based · Electric Force Field and Potential · Electric Force Field and Potential
Problem
A horizontal parallel-plate capacitor (plate separation d, area A) is initially charged to total charge Q and then isolated from any external circuit. The lower plate is rigidly fixed to a table, while the upper plate is free to exert forces only in the vertical direction. A dielectric slab of thickness d, uniform mass density ρ, and dielectric constant κ>1 initially hangs just outside the gap, aligned with the plates but not yet inserted. The slab is slowly released so that it can slide into the gap under the combined influence of gravity (downward) and the electric forces exerted by the charged plates (pulling the slab upward into the capacitor).
Let y be the vertical distance that the slab has entered the region between the plates (y=0 when the slab just touches the gap and y=d when it is fully inserted). Neglect edge effects.
(a) Derive an expression for the net vertical force F(y) acting on the slab as a function of the insertion depth y.
(b) Determine the equilibrium depth yeq at which the slab comes to rest (if such a position exists). Express your result in terms of Q, A, d, κ, ρ, and g.
(c) Evaluate yeq numerically for the following values: A=5.0×10−3m2, d=2.0mm, Q=1.8×10−6C, κ=5.0, and ρ=850kgm−3. State clearly whether the slab will be completely drawn into the capacitor or only partially inserted at equilibrium.