[MarioNintendo]

Originally Posted by Dynam0

Looking at this...I would assume this is a weighted average as a function of length of the material.

The actual lengths don't matter...just the fraction of total length of the capacitor is necessary.

k(av) = (k1*1/5 + k2*4/5) / 2
k(av) = (4*(1/5) + 6.5*(4/5)) / 2
k(av) = 6

I don't think the actual capacitance is needed here, unless you'll use this dielectric for the total capacitance then sure. If the question is worth more than 4 or 5 marks then it might be more involved, idk

I thought it might have been the case, but there doesn't seem to be any info about this online whatsoever... how strange.

I believe the initial capacitance had to be
C=(epsilon)*A^2/b
=(epislon)*(0.05^2)/0.01
=2.21*10^-12F,

therefore the final capacitance would be 6 times that quantity......?