Potential difference between plates

AI Thread Summary
The discussion revolves around calculating the energy stored in a 705 µF capacitor with a potential difference of 320 V, using the formula Energy = 1/2 CV^2. Initial calculations yielded incorrect results due to confusion over the capacitor's value and potential difference. Participants clarified that the correct capacitance is 705 µF and confirmed the voltage as 320 V. A subsequent calculation corrected the energy to approximately 36.096 J. The conversation highlights the importance of accurate unit conversions and careful attention to numerical values in physics problems.
kdrobey
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Homework Statement


he electronic flash attachment for a camera contains a capacitor for storing the energy used to produce the flash. In one such unit, the potential difference between the plates of a 705 µF capacitor is 320 V.

(a) Determine the energy that is used to produce the flash in this unit. _____J
(b) Assuming that the flash lasts for 5.0 multiplied by 10-3 s, find the effective power or "wattage" of the flash. _______W

Homework Equations


Energy=1/2CV^2

The Attempt at a Solution


.5(7.05 x 10^-6 F)(320V)^2. this gave me .3606, which was not the correcct answer for part A
 
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kdrobey said:
.5(7.05 x 10^-6 F)(320V)^2. this gave me .3606, which was not the correcct answer for part A
Is the capacitance 705 µF or 7.05 µF?
 
it's 705 uF. i thought i was supposed to convert uF to F. (1F=10^-6uf?)
 
kdrobey said:
it's 705 uF. i thought i was supposed to convert uF to F. (1F=10^-6uf?)
Right. But you also "converted" 705 to 7.05 for some reason. :wink:
 
ohhh i see. so (705 x 10^-6)*(302)^2=36.096?
 
Yet another typo. Is it 302 V or 320 V?
 

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