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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.

Short Answer

Expert verified
(a) 9.352 Joules, (b) approximately 2399 watts.

Step by step solution

01

Understand the Problem

We are given a capacitor with a capacitance of 850 \(\mu\mathrm{F}\) (microfarads) and a voltage of 280 V. We need to calculate the energy stored in the capacitor and then find the power of the flash during its operation time of 3.9 \(\times 10^{-3}\) seconds.
02

Convert Capacitance to Farads

First, we convert the capacitance from microfarads to farads because the standard unit of capacitance is farads. 850 \(\mu\mathrm{F} = 850 \times 10^{-6} \mathrm{F}\).
03

Use the Energy Formula for a Capacitor

The energy \(E\) stored in a capacitor is given by the formula:\[ E = \frac{1}{2} C V^2 \]where \(C\) is the capacitance in farads, and \(V\) is the voltage. Substitute \(C = 850 \times 10^{-6} \mathrm{F}\) and \(V = 280 \mathrm{V}\).
04

Calculate the Energy

Substitute the values into the formula:\[E = \frac{1}{2} \times 850 \times 10^{-6} \times (280)^2 = \frac{1}{2} \times 850 \times 10^{-6} \times 78400\]Calculate the energy to find:\[E = 9.352 \text{ Joules}\]
05

Calculate the Power Output

Power \(P\) is calculated with the formula:\[ P = \frac{E}{t} \]where \(E\) is the energy in joules and \(t\) is time in seconds. We know \(E = 9.352 \text{ J}\) and \(t = 3.9 \times 10^{-3} \text{ s}\).
06

Compute the Power

Insert the values into the power formula:\[ P = \frac{9.352}{3.9 \times 10^{-3}} \approx 2398.97 \text{ watts (W)}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Capacitance
Capacitance refers to a capacitor's ability to store charge. It's an essential concept in electronics, where a capacitor in a circuit can hold energy in an electric field. Capacitors come with a specific capacitance value, which is usually represented in microfarads (\(\mu F\)), but the standard unit is farads (\(F\)).

To convert microfarads to farads, you multiply by \(10^{-6}\). Therefore, an 850 \(\mu F\) capacitor is equivalent to \(850 \times 10^{-6}\) farads.

Understanding capacitance is key when calculating how much energy a capacitor can store. This is determined using the formula:
  • \( E = \frac{1}{2} C V^2 \)
  • Where \(E\) is energy in joules, \(C\) is capacitance in farads, and \(V\) is voltage in volts.
This equation helps us understand how efficiently a capacitor stores energy based on its capacitance and the voltage applied across it.
Electrical Power
Electrical power defines the rate at which energy is consumed or produced by a device. It is measured in watts (W), which is equivalent to joules per second. When dealing with capacitors, you calculate power by considering the energy discharged over a period.

The formula to find the power is:
  • \( P = \frac{E}{t} \)
  • Where \(P\) is power in watts, \(E\) is energy in joules, and \(t\) is time in seconds.
This calculation helps in determining the wattage of the flash in a camera capacitor as it dictates how much energy is used per second.

The higher the wattage, the more energy is being consumed quickly, which can be crucial in devices like flashlights or cameras that need instantaneous bursts of power.
Time Duration
Time duration in the context of capacitors and electrical devices refers to the length of time over which a device operates, often impacting how power is calculated. This is especially important for components like the flash in a camera, where operation times can be very brief.

In calculations involving power consumption or production, time is a critical component. It influences how you compute average power over short or extended periods.
  • A shorter time duration generally requires a higher power output for the same amount of energy to be delivered.
  • In our exercise, the flash operation lasted \(3.9 \times 10^{-3}\) seconds, necessitating an understanding of how a brief duration impacts total energy and power calculations.
Considering time duration ensures accurate understanding of how quickly energy is utilized or needs to be supplied, which is vital for electrical devices that have specific performance requirements.

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Most popular questions from this chapter

Two identical point charges \(\left(+2.4 \times 10^{-9} \mathrm{C}\right)\) are fixed in place, separated by 0.50 \(\mathrm{m}\) (see the figure). Concepts: (i) The electric field is a vector and has a direction. At the midpoint, what are the directions of the individual electric-field contributions from \(q_{\mathrm{A}}\) and \(q_{\mathrm{B}} ?\) (ii) Is the magnitude of the net electric field at the midpoint greater than, less than, or equal to zero? (iii) Is the total electric potential at the midpoint positive, negative, or zero? (iv) Does the electric potential have a direction associated with it? Explain. Calculations: Find the electric field and the electric potential at the midpoint of the line between the charges \(q_{\mathrm{A}}\) and \(q_{\mathrm{B}}\).

Two charges \(A\) and \(B\) are fixed in place, at different distances from a certain spot. At this spot the potentials due to the two charges are equal. Charge A is 0.18 \(\mathrm{m}\) from the spot, while charge \(\mathrm{B}\) is \(0.43 \mathrm{m}\) from it. Find the ratio \(q_{\mathrm{B}} / q_{\mathrm{A}}\) of the charges.

A particle with a charge of \(-1.5 \mu \mathrm{C}\) and a mass of \(2.5 \times 10^{-6} \mathrm{kg}\) is released from rest at point \(A\) and accelerates toward point \(B\), arriving there with a speed of \(42 \mathrm{m} / \mathrm{s} .\) The only force acting on the particle is the electric force. (a) Which point is at the higher potential? Give your reasoning. (b) What is the potential difference \(V_{\mathrm{B}}-V_{\mathrm{A}}\) between \(\mathrm{A}\) and \(\mathrm{B} ?\)

An electric field has a constant value of \(4.0 \times 10^{3} \mathrm{V} / \mathrm{m}\) and is directed downward. The field is the same everywhere. The potential at a point \(P\) within this region is 155 V. Find the potential at the following points: (a) \(6.0 \times 10^{-3} \mathrm{m}\) directly above \(P,(\) (b) \(3.0 \times 10^{-3} \mathrm{m}\) directly below \(P\) (c) \(8.0 \times 10^{-3} \mathrm{m}\) directly to the right of \(P\)

Two equipotential surfaces surround a \(+1.50 \times 10^{-8} \mathrm{C}\) point charge. How far is the \(190-\mathrm{V}\) surface from the \(75.0-\mathrm{V}\) surface?

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