How does the capacitance of a parallel plate capacitor change when its plates are moved to twice their initial distance and a slab of material with dielectric constant К = 2 is placed between the plates to replace air?

Question

I think that the capacitance would be halved, is this correct?

Also if two different capacitors (different material but all other parameters are the same except the dielectric constant) are being compared, would the one with the higher dielectric constant have a higher capacitance?

Answer 1

First question: It would stay the same. The effect of doubling the gap is to cut C in half, but changing K from 1 to 2 doubles the value of C. The two effects cancel out.

Second question: Higher dielectric constant, K ,means higher capacitance.
The polarization of material between the plates lets you store more charge Q at a given voltage V. (Higher Q/V)

Try to learn the plate capacitor formula

C = K*(epsilono)*A/d

A is the plate area and d is the gap between the plates.
Epsilon-zero is a constant of nature with dimensions of farads per meter