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.

Explanation:

Since a slab is inserted between the plates, so we consider them as two capacitors attached in Parallel. One with dielectric and one without dielectric.

Equation will become:

Total Capacitance=Capacitance With dielectric+ Capacitance without Dielectric

Where:

a is the distance at which slab is added between the plates.

Rearranging the above equation:

Charge on the capacitor Q is given by:

Q=CV

Current "I" is given by:

Now,

Taking derivative:

      dr/dt=0 (r is constant)

In the above equation, d(a)/dt is the speed which is constant.

Speed= Distance/time

d(a)/dt= r/Δt

Final Equation will become:

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

which piece of furniture will have the most inertia and give the furniture movers the most difficulty in moving?  

answer

a 75kg dining table