There are 3 questions but all tie into each other so please help! A capacitor co

Question

There are 3 questions but all tie into each other so please help!

A capacitor consists of two closely spaced metal conductors of large area, separated by a thin insulating film. One of the many practical uses of a capacitor is to store up some energy, to be released in a sudden burst (like lighting up the flash bulb in a camera!) A decent (not too expensive) capacitor in such a camera might be 175 uF, and could release a total electrical energy of 9.5 J in a flash. What voltage difference would you need across this capacitor?
NOTE: If you locate the right formula, this is a simple "plug-n-chug" problem.
So, take a moment to think about what the formula means, and think about the actual numbers a little. They're fairly realistic! But, the answer you get will come out quite a bit more than you would have expected, given that you probably put one or two small batteries, maybe just 1.5V each, in that camera. There's an electronic trick we'll learn later this term that allows you to increase the voltage, albeit with a certain "cost", namely, you have to wait a while before the capacitor is charged up again.
This large voltage (and moderately large energy) is also why it can be dangerous to take apart electronics that have capacitors in them - even a dinky little camera flash!

How much charge is stored on the NEGATIVE PLATE of the capacitor in the previous
question??

If the two plates of the capacitor in the previous question have their separation increased by a factor of 7 while the charge on the plates remains constant, by what factor is the energy stored in the capacitor increased??

Explanation / Answer

We know that

The energy stored in a capacitor is given by

E =(1/2)CV2

9.5J =0.5*175*10-6*V2

Then the potential difference between the plates is

v =Sqrt(9.5J/0.5*175*10-6)

=329.501V

The charge across the plates is given by

Q =CV =(175*10-6F)(329.501V) =0.0576C

We know that

C =eoA/d and E =Q2/2C

Now

Capacitance is inversely proportatinal to separation then

C1 =C

C2=?

d1=d

d2 =7d

Then C2 =(1/7)C

Then

E1/E2 =C2/C1

9.5J/E2 =(1/7)C/C

E2 =66.5J

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