Squeezing a capacitor, how does it effected the charge?

AI Thread Summary
When two parallel-plate capacitors of 7.0 µF each are connected to a 10 V battery and one capacitor's plate separation is halved, the charge transferred to the capacitors increases. Initially, the total capacitance is 14 µF, leading to a charge of 140 µC. Squeezing one capacitor effectively doubles its capacitance, resulting in a new total capacitance of 21 µF and an increased charge of 210 µC. The additional charge transferred due to the squeezing is 70 µC. The final charge stored on the capacitors reflects this increase, demonstrating the impact of reduced plate separation on capacitance.
mr_coffee
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I can't seem to get this one to work.

Two parallel-plate capacitors, 7.0 µF each, are connected in parallel to a 10 V battery. One of the capacitors is then squeezed so that its plate separation is halved.

(a) Because of the squeezing, how much additional charge is transferred to the capacitors by the battery?
wrong check mark µC

(b) What is the increase in the total charge stored on the capacitors, due to the squeezing?
µC

I tried the following:
Since its in paraellel, i added the C's and got C = 14, Q = CV;
Q = (14)(10) = 140, but they then said, the one capacitor is then squeezed so the plate separation is halved, so I figured I would double the result i got oringally, so i'd get 140*2 = 280, which was wrong.
 
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Nevermind i figured it out! i just had to do 7*2 + 7, so i got (a) 70;
But for part b, i thought it would just be 210, but it was wrong, any ideas? Thanks
 
How much MORE is 210 than 140?
 
ohhh! thanks lightgrav :) 70.
 
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