# Capacitance after changing plate distances

1. Sep 20, 2009

### exitwound

1. The problem statement, all variables and given/known data, from my crappy textbook

2. Relevant equations barely explained in my crappy textbook

Q=CV
$C=\epsilon A/d$

3. The attempt at a solution that should be easy but the text book is crap

Before squeezing:
Simplify the circuit by combining the two capacitors in parallel:

$C_{12}=C_1 + C_2$
$C_{12}=7x10^-6 + 7x10^-6 = 14x10^-6 F$

$Q=CV$
$Q=C_{12}V$
$Q=(14x10^-6)(24)=3.36x10^-4 C$

After Squeezing:
$C=\epsilon A/d$
$2C=\epsilon A/(d/2)$
$C=(7x10^-6)(2)=14x10^-6 F$

$C_{12}=14x10^-6 + 7x10^-6 = 21x10^-6 F$
$Q=(21x10^-6)(24)=5.04x10^-4 C$

Am I even close???

At this point, I have absolutely no idea what the problem is asking. Did I mention this book is terrible? Aren't both questions asking the exact same thing?? This is ridiculous.

2. Sep 21, 2009

### jambaugh

The way it is worded it appears as if a.) and b.) are the same question.
Your work looks right. Remember its asking for an increase so subtract before squeezing value from after squeezing value.

3. Sep 21, 2009

### exitwound

That's what I did, but the answer was wrong.

4. Sep 25, 2009

### exitwound

Anyone??? Still no go on this one.

5. Sep 26, 2009

### jambaugh

Ah Hmmm.... b.) could be a trick question... total charge (+ plus -) is of course zero.
But your calculations are correct for what they find. Parallel capacitances add. Halving the separation doubles the capacitance. That's it.

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