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Chapter 25: Problem 5

What capacitance is required to store an energy of \(10 \mathrm{~kW} \cdot \mathrm{h}\) at a potential difference of \(1700 \mathrm{~V}\) ?

Short Answer

Expert verified
The required capacitance is approximately 24.91 mF.

Step by step solution

01

Convert Energy Units

First, convert the energy from kilowatt-hours to joules. The conversion is done using the formula:\[1 \text{ kWh} = 3.6 \times 10^6 \text{ J}\]Therefore, for 10 kWh, we have:\[E = 10 \times 3.6 \times 10^6 \text{ J} = 3.6 \times 10^7 \text{ J}\]
02

Use the Energy of a Capacitor Formula

The energy stored in a capacitor is given by the formula:\[E = \frac{1}{2} C V^2\]where \(E\) is the energy in joules, \(C\) is the capacitance in farads, and \(V\) is the potential difference in volts.
03

Rearrange the Formula to Solve for Capacitance

Rearrange the energy formula to solve for the capacitance \(C\):\[C = \frac{2E}{V^2}\]
04

Substitute the Known Values

Substitute \(E = 3.6 \times 10^7 \text{ J}\) and \(V = 1700 \text{ V}\) into the rearranged formula:\[C = \frac{2 \times 3.6 \times 10^7}{1700^2}\]Calculate the value to find \(C\).
05

Calculate the Capacitance

Perform the calculation:\[C = \frac{7.2 \times 10^7}{2890000} \approx 24.91 \times 10^{-3} \text{ F}\]So, the capacitance is approximately 24.91 mF (millifarads).

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Key Concepts

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

Energy Conversion
When we talk about energy conversion, especially in physics, we often deal with changing units to make calculations easier. In this exercise, we're converting energy from kilowatt-hours (kWh) to joules (J). This is crucial since the standard unit of energy in physics is joules, making this conversion a necessary step.
  • 1 kilowatt-hour (kWh) is equivalent to 3.6 x 106 joules (J).
  • The exercise starts with 10 kWh, which translates to 36 million joules (3.6 x 107 J).
Converting these units ensures that all parts of the equation are compatible, helping you to correctly apply the capacitor energy formula.
Capacitor Energy Formula
Capacitors store energy in an electric field, a concept captured by the capacitor energy formula, which relates capacitance, voltage, and energy.The formula is:\[E = \frac{1}{2} C V^2\]Here:
  • E is the energy stored in the capacitor, measured in joules.
  • C stands for capacitance, measured in farads (F).
  • V is the potential difference across the capacitor, measured in volts (V).
Using this formula, we can understand how changing the voltage or the capacitance affects the stored energy. If the voltage is increased, the energy increases quadratically, showing the potential for capacitors to store significant energy even with small changes in voltage.
Rearranging Equations
Mathematics in physics often involves rearranging equations to solve for a desired variable. In this exercise, we need to find the capacitance (C) given the energy (E) and voltage (V).From the capacitor energy formula \(E = \frac{1}{2} C V^2\), we want to isolate and solve for \(C\). By rearranging the formula, we get:\[C = \frac{2E}{V^2}\]This rearrangement involves straightforward algebra:
  • Multiply both sides by 2 to eliminate the fraction.
  • Divide by \(V^2\) to solve for \(C\).
These steps highlight the power of algebra in solving real-world physics problems, allowing us to express one physical quantity in terms of others.
Unit Conversion in Physics
Unit conversion is essential in physics for ensuring that calculations are accurate and consistent. Converting energy from kWh to J was our first step. Physics frequently uses these conversions:
  • Distance: meters (m).
  • Time: seconds (s).
  • Energy: joules (J).
  • Voltage: volts (V).
In this problem, improper unit conversion could result in incorrect values for capacitance. It's vital to understand the conversion factors and apply them correctly when switching between units. Converting each quantity into its respective unit ensures coherence and accuracy across calculations, helping solve even complex problems with ease.

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Most popular questions from this chapter

Figure 25-29 shows a variable "air gap" capacitor for manual tuning. Alternate plates are connected together; one group of plates is fixed in position, and the other group is capable of rotation. Consider a capacitor of \(n=8\) plates of alternating polarity, each plate having area \(A=1.50 \mathrm{~cm}^{2}\) and separated from adjacent plates by distance \(d=3.40 \mathrm{~mm}\). What is the maximum capacitance of the device?

If an uncharged parallel-plate capacitor (capacitance \(C\) ) is connected to a battery, one plate becomes negatively charged as electrons move to the plate face (area \(A\) ). In Fig. \(25-32\), the depth \(d\) from which the electrons come in the plate in a particular capacitor is plotted against a range of values for the potential difference \(V\) of the battery. The density of conduction electrons in the copper plates is \(8.49 \times 10^{28}\) electrons \(/ \mathrm{m}^{3}\). The vertical scale is set by \(d_{s}=2.00 \mathrm{pm}\), and the horizontal scale is set by \(V_{s}=20.0 \mathrm{~V}\). What is the ratio \(C / A\) ?

A parallel-plate capacitor has square plates with edge length \(8.20 \mathrm{~cm}\) and \(1.30 \mathrm{~mm}\) separation. (a) Calculate the capacitance. (b) Find the charge for a potential difference of \(120 \mathrm{~V}\).

The plates of a spherical capacitor have radii \(37.0 \mathrm{~mm}\) and \(40.0 \mathrm{~mm}\). (a) Calculate the capacitance. (b) What must be the plate area of a parallel-plate capacitor with the same plate separation and capacitance?

Assume that a stationary electron is a point of charge. What is the energy density \(u\) of its electric field at radial distances (a) \(r=1.00 \mathrm{~mm}\), (b) \(r=1.00 \mu \mathrm{m}\), (c) \(r=1.00 \mathrm{~nm}\), (d) \(r=1.00 \mathrm{pm}\), and (e) \(r=1.00 \mathrm{fm}\) ? (f) What is \(u\) in the limit as \(r \rightarrow 0\) ?
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