Capacitance after changing plate distances

[exitwound]

1. Homework Statement , from my crappy textbook

2. Homework 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 textbook 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.

 

[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.

 

[exitwound]

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

 

[exitwound]

Anyone? Still no go on this one.

 

[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.