Question

Forty parallel plate capacitors are connected in series. The distance between the plates is d for the first capacitor, 2d for the second capacitor, 3d for the third capacitor, and so on. The area of the plates is the sam for all the capacitors. Expres the equivalent capacitance of the whole set in terms of C1 (the capacitance of the first capacitor)

Answer 1

The equation is supposed to look like \[C_(eq) = C_1/x \] where I fill in the expression for x. I know that the equivalent capacitance will be the inverse of the inverse sum of capacitors 1 through 40, but I'm unclear as to how to put this into the equation.

Answer 2

So does each parallel plate capacitor have a wider and wider gap?

Answer 3

Yes.

Answer 4

Do you know the equation: \[C(eq) = \sum_{n=1}^{40}{C_{1}\over n}\]

Answer 5

I don't think I've seen it before.

Answer 6

I guess you can find it here: http://en.wikipedia.org/wiki/Series_and_parallel_circuits#Capacitors

Sorry, and I messed up the equation; it should actually be:\[{1 \over C_{eq}} = \sum_{n=1}^{40}{ 1\over{C_1/n}}\]
Therefore, going past the first capacitor would give you\[1/C_1 = 1/C_1 \] (obviously). And then, for the second, \[{1\over{C_2}} = {1\over{C_1/1}} + {1\over{C_1/2}} = {1\over{C_1}} + {2\over{C_1}}\]
If you recognize the pattern, you can see that \[{1\over{C_{40}}} = {1\over{C_1}} + {2\over{C_1}} + {3\over{C_1}} + ... + {40\over{C_1}} = {{1+2+3+...+40}\over{C_1}}\]
or
\[C_{40} = {C_1\over {1+2+3+...+40}}\]

Answer 7

Ah I see. I actually got that answer, and tried putting it into webassign using x = sum(1 + 2 +...+ 40) but it wouldn't let me. I just put the actual number (820) and it said it was right. Thank you.