Question

This is a Mastering Physics Question

Each plate of a parallel-plate capacator is a square with sidelength ,and the plates are separated by a distance . The capacitor is connected to a source of voltage.A plastic slab of thickness and dielectric constant is inserted slowly between the plates over the time period until the slab is squarely between the plates. While theslab is being inserted, a current runs through thebattery/capacitor circuit.

Assuming that the dielectric is inserted at aconstant rate, find the current as the slab is inserted.

Express your answer in terms of anyor all of the given variables ,,,,, and $epsilon_0$ , the permittivity of free space.

I =

Answer 1

As the slab is being inserted, the capacitor can be treated astwo capacitors in parallel one with a dielectric and one without.If the dielectric has been inserted a distance x at some time, thenwe can write:

     C = ε _o Kx r / d      +   ε_o (r - x) r / d

Or:     C = (ε_o r / d ) (   Kx  + r    -   x )

And we know the amount of the charge on the caps togetheris:

    q   = CV     so the current is   I = dq/dt = (d/dt) (CV)

Or:   I   = V dC/dt = V (ε _o r / d ) (d/dt)(   Kx   + r   -   x )

Or:    I   =    V(ε _o r / d ) (K - 1) (dx/dt)

What is dx/dt? It is the speed at which the dielectric isinserted into the plates. You are told this speed is constant, soit must equal the distance, r, over time, Δt  and

    I   =    V (ε _o r / d ) (K - 1) ( r / Δt )

Or:   I =    (K - 1) V ε_o r² / d  Δt