Each plate of a parallel-plate capacator is a square with side length r, and the plates are separated by a distance d. The capac

Question

3 months ago

13

Each plate of a parallel-plate capacator is a square with side length r, and the plates are separated by a distance d. The capac

itor 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.

Physics

1 answer:

kicyunya [3.2K] · Answer 1 · 2026-02-25T06:45:34+03:00

Answer:

Explanation:

Prior to the insertion of the dielectric, the capacitance is Co.

Upon inserting the slab,

the capacitance now is

C=kCo.

The charge Q can be expressed as:

Q=CV.

Therefore, when C=Co,

Qo=CoV.

When C=kCo,

Q=kCoV.

The alteration in charge can be given as:

Q-Qo= kCoV - CoV.

∆Q= kCoV - CoV.

The current I is defined as

I=dQ/dt.

I= (kCoV - CoV) / dt.

I=Co(kV-V)/dt.

Here, Co is the capacitance value.

The formula for capacitance of a parallel-plate capacitor is:

Co=εoA/d.

Thus,

I=εoA(kV-V)/d•dt.

I=VεoA(k-1)/d•dt.

Let A=πr².

I = V•εo•πr²•(k-1) / d•dt.

This formula represents the current in the desired terms.