## Mastering Physics Solutions: Changing Capacitance Yields a Current

Changing Capacitance Yields a Current

Part A = I = ((K-1) * r2 * ε0 * V) / (d * Δt)

Solution Below:

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.

Part A

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, Δt, and ε0, the permittivity of free space.

C = ε0 * A / d

A is just the area of each of the capacitor plates and d is the distance between plates. Since the problem gives you side lengths of “r”, area is just r2. Now, to find the current, just find the change in charge over the change in time (that’s the definition of current). The charge q, is just equal to capacitance times voltage:

Q = CV
Q = (ε0 * A / d) * V.

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) * r2 * ε0 * V) / (d * Δt)

Tagged with: