Potential difference between plates

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Homework Help Overview

The discussion revolves around a problem involving a capacitor used in a camera's electronic flash attachment, specifically focusing on calculating the energy stored in the capacitor and the effective power of the flash. The subject area includes concepts of capacitance, potential difference, and energy calculations in electrical circuits.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the formula for energy stored in a capacitor and raise questions about the correct values for capacitance and potential difference. There is an exploration of unit conversions and the implications of potential typos in the values used.

Discussion Status

The discussion is ongoing, with participants clarifying the values of capacitance and potential difference. Some guidance has been offered regarding unit conversions, but there is no explicit consensus on the correct values or the calculations being performed.

Contextual Notes

Participants are navigating potential typos in the problem statement, specifically regarding the capacitance value and the potential difference, which may affect the calculations. There is also a focus on ensuring proper unit conversion from microfarads to farads.

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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