Each plate of a parallel-plate capacitor is a square with side length r, and the

Question

Each plate of a parallel-plate capacitor 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 Delta t until the slab is squarely between the plates. While the slab is being inserted, a current runs through the battery/capacitor circuit. (Figure 1) Assuming that the dielectric is inserted at a constant rate, find the current I as the slab is inserted. Express your answer in terms of any or all of the given variables V, K, r, d, Delta t, and epsilon_0, the permittivity of free space. I =

Explanation / Answer

here by using the formula

C = e0 * A / d

A = area of plate

d = distance between plates

e0 = vaccum permitivity constant value

C = capacitance

and then the charge is

Q = C * V

Q = e0 * A * V / d

The dielectric constant is just relative to vaccum – just subtract one from it and multiply to get the charge in different mediums. And make sure to do change in charge over change in time to get the current

I = ( (K-1) * r^2 * e0 * V) / (d * delta t)

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