Refer to Interactive Solution 19.39 at for one approach to...

Refer to Interactive Solution $19.39$ at for one approach to this problem. The 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 an $850-\mu \mathrm{F}$ capacitor is $280 \mathrm{~V}$. (a) Determine the energy that is used to produce the flash in this unit. (b) Assuming that the flash lasts for $3.9 \times 10^{-3} \mathrm{~s}$, find the effective power or "wattage" of the flash.

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Key Concepts

Energy Stored in a Capacitor

This concept involves understanding that a capacitor stores electrical energy in the electric field between its plates. The stored energy is quantified by the formula E = 1/2 C V², where E is the energy, C is the capacitance, and V is the voltage across the capacitor. This relation is foundational for applications involving energy storage and transient processes in circuits.

Capacitance and Voltage Relationship

This key concept covers how a capacitor’s ability to store charge (its capacitance) and the potential difference (voltage) across its plates determine the stored energy. It highlights the importance of both parameters and shows that even moderate capacitances can store significant amounts of energy if the voltage is high, which is especially relevant in devices like camera flashes.

Effective Power Calculation

Effective power relates the rapid release of stored energy over a short time interval to the concept of power, defined as the rate of energy transfer. In contexts like a camera flash, the power is calculated using the formula P = Energy / time, emphasizing how brief discharge times can result in very high instantaneous power outputs despite the total energy being limited.

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