An electronic flash unit for a camera contains a capacitor with a capacitance of $890 \mu$ F. When the unit is fully charged and ready for operation, the potential difference between the capacitor plates is $330 \mathrm{V}$. (a) What is the magnitude of the charge on each plate of the fully charged capacitor? (b) Find the energy stored in the "charged-up" flash unit.
Added by Kim B.
Step 1
Given: Capacitance, C = 890 $\mu$F = 890 x $10^{-6}$ F Potential difference, V = 330 V Use the formula: Q = C x V Substitute the values: Q = 890 x $10^{-6}$ F x 330 V Calculate: Q = 0.2937 C ** Show more…
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An electronic flash unit for a camera contains a capacitor with a capacitance of $890 \mu \mathrm{F}$. When the unit is fully charged and ready for operation, the potential difference between the capacitor plates is $330 \mathrm{~V}$. (a) What is the magnitude of the charge on each plate of the fully charged capacitor? (b) How much energy is stored in the charged-up flash unit?
An electronic flash unit for a camera contains a capacitor with a capacitance of 880 μF . When the unit is fully charged and ready for operation, the potential difference between the capacitor plates is 300 V . Part A: What is the magnitude of the charge on each plate of the fully charged capacitor? Express your answer using two significant figures. Part B: Find the energy stored in the "charged-up" flash unit. Express your answer using two significant figures.
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Electronic flash units for cameras contain a capacitor for storing the energy used to produce the flash. In one such unit, the flash lasts for $\frac{1}{675}$ s with an average light power output of $2.70 \times 10^{5} \mathrm{W}$ (a) If the conversion of electrical energy to light is 95$\%$ efficient (the rest of the energy goes to thermal energy), how much energy must be stored in the capacitor for one flash? (b) The capacitor has a potential difference between its plates of 125 $\mathrm{V}$ when the stored energy equals the value calculated in part (a). What is the capacitance?
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