00:01
Three conducting plates, each of area a, are connected as shown in figure 2422.
00:07
Are the two capacitors thus formed connected in series or in parallel? well, since the middle plate is a common plate and the other two plates are connected together, so the voltage difference at both capacitors are the same, so the capacitors are connected in parallel.
00:39
For b, determine c as a function of d1, d2, and a, assuming d1, d1 plus d2 is much less than the dimensions of the plate.
00:52
So we're given d1 d2, but we have to define the value of the capacitance.
00:57
For that, we use the equation c equals epsilon not a over d.
01:01
In substating the given separation in area, then c1 would be epsilon not a over d1 and c2, a over d2.
01:15
The total capacitance of capacitors connected in parallel are just the sum of, the two capacitors and that would be equal to then epsilon not a over d1 plus epsilon not a over d2 which we could factor out epsilon not a and then that's one over d1 plus one over d2 which would be equal to epsilon not a add fractions together to common denominator or d d1 plus d2 over d1 times d2 and that could be our our formula c.
02:13
The middle plate can be moved, changing the values of d1 and 2, so as to vary the capacitance.
02:22
What are the minimum and maximum values of the net capacitance? well, if we compute the maximum capacitance of the parallel plate capacitor, based on the given figure, if the middle plate is moved to either of the outer plates and the distance between them will be zero...