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Chapter 16: Problem 30

If you want to store \(25 \mathrm{~J}\) of energy in a capacitor charged to \(12 \mathrm{~V}\) the required capacitance is (a) \(0.35 \mathrm{~F}\) (b) \(2.9 \mathrm{~F}\) (c) \(2.1 \mathrm{~F}\) (d) \(0.48 \mathrm{~F}\).

Short Answer

Expert verified
The required capacitance is (a) \(0.35 \mathrm{~F}\).

Step by step solution

01

Recall the formula for energy in a capacitor

The energy stored in a capacitor is given by the formula \( U = \frac{1}{2} C V^2 \), where \( U \) is the stored energy, \( C \) is the capacitance, and \( V \) is the voltage across the capacitor.
02

Plug in the given values

We need to find \( C \) when \( U = 25 \mathrm{~J} \) and \( V = 12 \mathrm{~V} \). So, substitute these values into the formula: \[ 25 = \frac{1}{2} C (12)^2 \]
03

Solve for Capacitance \( C \)

Simplify the equation from Step 2: \[ 25 = 72C \] Divide both sides by 72 to solve for \( C \): \[ C = \frac{25}{72} \approx 0.3472 \mathrm{~F} \]
04

Compare with given options

The calculated capacitance \( 0.3472 \mathrm{~F} \) is closest to option (a) \( 0.35 \mathrm{~F} \).

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Key Concepts

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

Capacitance
Capacitance is a fundamental property of capacitors, symbolized by the letter \( C \). It essentially measures a capacitor's ability to store charge per unit voltage. In simple terms, capacitance tells us how much electric charge the capacitor can hold when a certain voltage is applied across it.
  • The unit of capacitance is the farad (\( \mathrm{F} \)), which is quite large. Most capacitors you'll encounter in everyday electrical devices have capacitance values in the range of microfarads (\( \mu \mathrm{F} \)) or nanofarads (\( \mathrm{nF} \)).
  • In our exercise, we calculated the capacitance of a capacitor needed to store a specific amount of energy using the formula for energy stored: \( U = \frac{1}{2} C V^2 \).
  • Understanding capacitance is crucial for designing circuits, especially when you need to store or manage electrical energy effectively.
Voltage
Voltage, denoted by \( V \), is the electrical pressure that drives charge around a circuit. It's the force that pushes electrons through the resistance of a circuit, analogous to water pressure in a pipe.
  • The unit of voltage is the volt (\( \mathrm{V} \)).
  • In our provided example, the capacitor is charged to \( 12 \mathrm{~V} \), meaning that this is the potential difference used for calculating the stored energy.
  • Voltage influences the amount of energy stored in a capacitor, as shown in the formula for stored energy \( U = \frac{1}{2} C V^2 \). A higher voltage means more potential energy stored.
When working with capacitors, it's important to ensure the applied voltage does not exceed the capacitor's rated voltage to prevent damage.
Stored Energy
Stored energy in a capacitor is represented by the symbol \( U \). This energy is the potential energy stored due to the electric field created between the plates of the capacitor when it's charged.
  • The formula \( U = \frac{1}{2} C V^2 \) is used to calculate the stored energy, where \( U \) is the energy, \( C \) is the capacitance, and \( V \) is the voltage.
  • This formula shows that the stored energy depends on both the capacitance and the square of the voltage. Thus, even a small increase in voltage can significantly increase the storage capacity due to the \( V^2 \) term.
  • In our exercise, the goal was to determine the capacitance needed to store \( 25 \, \mathrm{J} \) of energy at \( 12 \, \mathrm{V} \).
Understanding stored energy is key for applications where capacitors are used to release energy, such as in camera flashes or for stabilizing voltage in electrical circuits.
Capacitor
A capacitor is a passive electrical component with two terminals. It's designed to store and release electrical energy. Capacitors come in various shapes, sizes, and capacitance ratings, tailored for specific applications.
  • They work by accumulating electrical charge on their plates, which are separated by an insulating material known as a dielectric.
  • The amount of charge a capacitor can hold is dictated by its capacitance and the voltage applied across it.
  • Capacitors are widely used in electronic circuits to filter out noise, stabilize voltage supplies, and store energy for short bursts.
In our practical exercise, we determined the required capacitance for a capacitor to store a given amount of energy at a specified voltage, showcasing its role in energy management within circuits.

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Most popular questions from this chapter

Two charges, one positive and one negative, are separated by a fixed distance. The sign of the potential energy of this pair (a) is always positive; (b) is always negative; (c) is sometimes positive and sometimes negative; (d) depends on the distance.

Consider a spherical capacitor, in which a spherical conductor of radius \(a\) lies inside a concentric spherical conducting shell of radius \(b\). Suppose the inner and outer shells carry charges \(\pm Q\), respectively. (a) Compute the potential difference between the two shells. (Note that the electric potential outside of a uniform sphere of charge is the same as if all the charge were concentrated at the sphere's center. Your answer should be a function of \(Q\) and the two radii.) (b) Use your answer from part (a) and the definition of capacitance to show that the capacitance of the spherical capacitor is \(C=4 \pi \varepsilon_{0} a b /(b-a) .\) (c) Compute the capacitance of a spherical capacitor with inner radius \(2.0 \mathrm{~cm}\) and outer radius \(10.0 \mathrm{~cm} .\)

Find the potential energy of charges \(q_{1}\) and \(q_{2}\) located in the \(x-y\) plane: \(q_{1}=-3.6 \times 10^{-9} \mathrm{C}\) at \((0.12 \mathrm{~m}, 0.45 \mathrm{~m})\) and \(q_{2}=1.6 \times 10^{-9} \mathrm{C}\) at \((0.36 \mathrm{~m}, 0.55 \mathrm{~m})\)

In a Millikan o1l-drop experiment used to measure the elementary charge (Chapter 15 ), two horizontal parallel plates are \(1.0 \mathrm{~cm}\) apart. With a \(180-\mathrm{V}\) potential difference between the plates, a charged droplet of mass \(8.8 \times 10^{-16} \mathrm{~kg}\) is suspended motionless. (a) If the upper plate is at the higher potential, is the charge on the droplet positive or negative? (b) What's the electric field between the plates under these conditions? (c) Compute the charge on the droplet. To how many elementary charges \((e)\) does this correspond?

In terms of the charge stored \((Q)\) and the potential difference \((V)\) the energy stored in a capacitor is (a) \(2 Q V\); (b) \(Q V\); (c) \(\frac{1}{2} Q V\) (d) \(Q / V\).
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