Question

Each plate of a parallel-plate capacator is a square with side length r, and the plates are separated by a distance d. The capacitor is connected to a source of voltage, V. A plastic slab of thickness d and dielectric constant K is inserted slowly between the plates over the time period Δt until the slab is squarely between the plates. While the slab is being inserted, a current runs through the battery/capacitor circuit.
Assuming that the dielectric is inserted at a constant rate, find the current I as the slab is inserted.

Answer 1

Answer:

Explanation:

Before the dialectic was inserted the capacitor is Co

When the slab is inserted,

The capacitor becomes

C=kCo

The charge Q is given as

Q=CV

Then, when C=Co

Qo=CoV

Then, when C=kCo

Q=kCoV

Then, the change in charges is given as

Q-Qo= kCoV - CoV

∆Q= kCoV - CoV

Current is given as

I=dQ/dt

I= (kCoV - CoV) / dt

I=Co(kV-V)/dt

Note Co is the value capacitor

So, Capacitance of parallel plates capacitor is given as

Co=εoA/d

Then,

I=εoA(kV-V)/d•dt

I=VεoA(k-1)/d•dt

Where A=πr²

I = V•εo•πr²•(k-1) / d•dt

This is the required expression for current is in the required term