Question 41

[yfangl097543]

I'm curious, I went about doing this problem thinking that the plate in the middle acted kind of like a dieletric, so the capacitance would increase. Why is this reasoning wrong?

[admin]

A plate, which is presumably a metal, cannot act as a dielectric. A dielectric would have to be specified and it would be an insulating material which, as you suggest, would increase the capacitance.

[harrington5609]

I took this question as the capacitor was added in series. The way I remember this is "capacitors in series are like resistors in parallel" so that I remember how to set up the equation to find the equivalent capacitance. Capacitors in series therefore lower the capacitance. Why is the answer not choice C?

[jellywing_2058]

Capacitors in series are like Resistors in parallel. However this combination does not result in a lower capacitance. Why? The capacitance of each capacitor in series now has twice the capacitance of the earlier capacitance.

This is because capacitance of a parallel plate capacitor is given by C = ε0 A/d,
and because by inserting a plate in the middle makes the separation of the capacitor plates = d/2, we have the capacitance of each of the capacitor doubling to:

C’ = ε0 A/(d/2)
C' = 2 ε0 A/d
C' = 2 C.

In this case
C’ = 2 C1. The series combination of the 2 capacitors will have an equivalent capacitance of C2 where

1/ C2 = 1/ C’ + 1/ C’
1/ C2 = 1/ (2 C1) + 1/ (2 C1)
1/ C2 = 2 x (1/ (2 C1))
1/ C2 = 1/ C1
C2 = C1
Therefore, Choice A

[harrington5609]

That was another one of my reasonings..I saw that the distance went from d to d/2, which would double capacitance. So you're saying that because we had the distance cut in half (doubles capacitance) and what seemed to be capacitors in series (halves capacitance) that they will cancel each other out and remain unchanged, which is choice A?