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Chapter 24: Problem 83

The power supply for a pulsed nitrogen laser has a \(0.080-\mu \mathrm{F}\) capacitor with a maximum voltage rating of \(25 \mathrm{kV}\). ( \(a\) ) Estimate how much energy could be stored in this capacitor. (b) If \(15 \%\) of this stored electrical energy is converted to light energy in a pulse that is \(4.0-\mu\) s long, what is the power of the laser pulse?

Short Answer

Expert verified
(a) 25 Joules (b) 937.5 kW power

Step by step solution

01

Identify the formula for energy stored in 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 and \( V \) is the voltage across the capacitor.
02

Calculate the energy stored in the capacitor

Substitute \( C = 0.080 \mu F = 0.080 \times 10^{-6} F \) and \( V = 25 \times 10^3 V \) into the formula \( E = \frac{1}{2} C V^2 \). Calculate: \[ E = \frac{1}{2} \times 0.080 \times 10^{-6} \times (25 \times 10^3)^2 = \frac{1}{2} \times 0.080 \times 10^{-6} \times 625 \times 10^6 \]\[ E = 0.080 \times 312.5 \] \[ E = 25 \] Thus, the energy stored is \( 25 \) joules.
03

Determine the energy converted to light

Since only \( 15\% \) of the stored energy is converted to light energy, we calculate \( E_{light} = 0.15 \times 25 = 3.75 \) joules.
04

Calculate the power of the laser pulse

Power is defined as energy divided by time. Here, the energy converted to light is \( 3.75 \, \text{Joules} \) and the time of the pulse is \( 4.0 \, \mu s = 4.0 \times 10^{-6} \, s \). Thus, the power \( P \) is:\[ P = \frac{3.75}{4.0 \times 10^{-6}} \]\[ P = 937500 \, \text{W} \]Therefore, the power of the laser pulse is \( 937500 \, \text{watts} \) or \( 937.5 \, \text{kW} \).

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Key Concepts

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

Laser Energy Conversion
When dealing with lasers, understanding the conversion of electrical energy into light energy is key. In this exercise, a fraction of the energy stored in the capacitor is transformed into light energy. This process can be understood as laser energy conversion. Here, the efficiency rate tells us how much of the electrical energy becomes light. In our situation, only 15% of the total stored energy in the capacitor is converted for the laser pulse, showcasing typical efficiency concerns in such systems. Understanding this fraction is crucial:
  • Start with the total energy stored in a capacitor, calculated earlier as 25 joules for our exercise.
  • Multiply this total energy by 0.15 (which is 15%) to find out how much is converted into light energy. This amounts to 3.75 joules in our example.
  • This conversion efficiency influences how effective the laser is at emitting energy in light form, an important factor in laser design and application.
Remember, perfect energy conversion is theoretically attractive but practically almost impossible due to losses inherent in the process.
Power Calculation
Power, in physics, represents the rate at which energy is transferred or converted. It is usually measured in watts (W). In the scenario given, we want to calculate the power of a laser pulse based on the energy converted into light and the duration of each pulse. The formula for power is:
  • Power (P) = Energy (E) / Time (t).
In the exercise:
  • The energy converted to light, as established, is 3.75 joules.
  • The duration of the laser pulse is given as 4.0 microseconds, which translates to 4.0 x 10^{-6} seconds.
  • By substituting these values into the formula, you get the power output of the laser pulse.
    Therefore, the calculation is P = 3.75 / (4.0 x 10^{-6}), which gives a result of 937,500 watts or 937.5 kilowatts (kW).
This high power level in such a short time frame is typical for laser pulses, which underscores the intense energy output that lasers can achieve.
Capacitor Discharge
Capacitors store electrical energy in an electric field and discharge it when needed. This discharge is critical in systems requiring quick releases of stored energy, such as in the pulsed laser described in the exercise. Here's how it works:
  • Capacitors are charged with electricity until they reach their maximum voltage capacity.
  • When a circuit requires the stored energy, the capacitor discharges, releasing its energy rapidly.
  • In our example, a 0.080 渭F capacitor charged to 25 kV stores 25 joules of energy.
  • Upon discharge, 15% of this energy converts into a laser light pulse, as previously discussed.
The ability of a capacitor to deliver such fast energy bursts makes it invaluable in various applications, especially where rapid energy deployment is necessary, such as in lasers, camera flashes, and other electronic devices.

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Most popular questions from this chapter

The charge on a capacitor increases by \(26 \mu \mathrm{C}\) when the voltage across it increases from \(28 \mathrm{~V}\) to \(78 \mathrm{~V}\). What is the capacitance of the capacitor?

A multilayer film capacitor has a maximum voltage rating of \(100 \mathrm{~V}\) and a capacitance of \(1.0 \mu \mathrm{F}\). It is made from alternating sheets of metal foil connected together, separated by films of polyester dielectric. The sheets are \(12.0 \mathrm{~mm}\) by \(14.0 \mathrm{~mm}\) and the total thickness of the capacitor is \(6.0 \mathrm{~mm}\) (not counting the thickness of the insulator on the outside). The metal foil is actually a very thin layer of metal deposited directly on the dielectric, so most of the thickness of the capacitor is due to the dielectric. The dielectric strength of the polyester is about \(30 \times 10^{6} \mathrm{~V} / \mathrm{m}\). Estimate the dielectric constant of the polyester material in the capacitor.

(II) Two identical capacitors are connected in parallel and each acquires a charge \(Q_{0}\) when connected to a source of voltage \(V_{0} .\) The voltage source is disconnected and then a dielectric \((K=3.2)\) is inserted to fill the space between the plates of one of the capacitors. Determine \((a)\) the charge now on each capacitor, and \((b)\) the voltage now across each capacitor.

Consider the use of capacitors as memory cells. A charged capacitor would represent a one and an uncharged capacitor a zero. Suppose these capacitors were fabricated on a silicon \(\begin{array}{llllll}\text { chip and each has a capacitance of } 30 & \text { femto-farads }\end{array}\) \(\left(1 \mathrm{fF}=10^{-15} \mathrm{~F} .\right)\) The dielectric filling the space between the parallel plates has dielectric constant \(K=25\) and a dielectric strength of \(1.0 \times 10^{9} \mathrm{~V} / \mathrm{m} .\) ( \(a\) ) If the operating voltage is \(1.5 \mathrm{~V}\), how many electrons would be stored on one of these capacitors when charged? \((b)\) If no safety factor is allowed, how thin a dielectric layer could we use for operation at \(1.5 \mathrm{~V} ?\) (c) Using the layer thickness from your answer to part \((b),\) what would be the area of the capacitor plates?

When two capacitors are connected in parallel and then connected to a battery, the total stored energy is 5.0 times greater than when they are connected in series and then connected to the same battery. What is the ratio of the two capacitances? (Before the battery is connected in each case, the capacitors are fully discharged.)
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