Total Energy of Camera Capacitor

1. The problem statement, all variables and given/known data
A flash attatchment for a professional camera stores energy in a capactior. When a picture is taken, all of the charge is converted to energy, and the capacitor is fully discharged.

A.Assume the battery charging the capacitor is a 300V battery. When the light flashes, it produces 5000W of light for a time of .005s. Find the total energy produced by the flash.

B.Find the capacitance.

C.If the battery were replaced with a battery with a potential difference of 120V, how much light power could the flash attachment produce?


2. Relevant equations

C= q/V
PE = qV
U=(1/2)CV^2

3. The attempt at a solution
I really don't know. At first I thought I'd just solve for capacitance(which you can't anyway), until I looked at part B. I'm assuming you need to solve for the potential energy of the flash, but I'm not sure how to go about doing that.
 
I am pretty sure in that in solving for the total energy, you just need to find the potential energy. Electric potential energy would be PE=qV. There is no charge stated in the problem, so I'm not sure where to go from there.
 
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For A) Power is a measure of Energy per Time. Think about what a Watt is and how it relates to the time given, 0.005s.

For B) Once you've answered part A, it should be easy to solve for Capacitance using the equation for U.
 

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