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Chapter 19: Problem 42

Two capacitors are identical, except that one is empty and the other is filled with a dielectric \((\kappa=4.50) .\) The empty capacitor is connected to a \(12.0-\mathrm{V}\) battery. What must be the potential difference across the plates of the capacitor filled with a dielectric such that it stores the same amount of electrical energy as the empty capacitor?

Short Answer

Expert verified
The potential difference must be approximately 5.66 V.

Step by step solution

01

Identify the Formula for Energy Stored 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 energy, \( C \) is the capacitance, and \( V \) is the voltage across the capacitor.
02

Set the Energy Equations Equal for Both Capacitors

Let \( C_0 \) be the capacitance of the empty capacitor and \( V_0 = 12.0 \) V its voltage. For the empty capacitor, \( U_0 = \frac{1}{2} C_0 V_0^2 \). The capacitance of the capacitor with the dielectric is \( C_1 = \kappa C_0 \). We need to find \( V_1 \) such that both capacitors store the same energy: \( U_0 = \frac{1}{2} C_1 V_1^2 \).
03

Equate the Energies and Solve for Voltage

Since \( \frac{1}{2} C_0 V_0^2 = \frac{1}{2} (\kappa C_0) V_1^2 \), we can cancel \( \frac{1}{2} C_0 \) from both sides to obtain: \( V_0^2 = \kappa V_1^2 \). Substitute \( \kappa = 4.50 \) and \( V_0 = 12.0 \) V into the equation to get: \( 144 = 4.50 V_1^2 \).
04

Solve for \( V_1 \)

Divide both sides by 4.5 to get \( V_1^2 = 32 \). Taking the square root of both sides, we find \( V_1 = \sqrt{32} \approx 5.66 \) V. This is the potential difference across the plates of the capacitor filled with the dielectric.

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Key Concepts

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

Dielectric Constant
The dielectric constant, often symbolized as \( \kappa \), plays a crucial role in determining how well a material can store electrical energy within a capacitor. You can think of it as a measure of a material's ability to resist an electric field. The larger the dielectric constant, the more capacity a material has to reduce the electric field within the capacitor.

In the example exercise, one capacitor is filled with a dielectric material with a dielectric constant \( \kappa = 4.50 \). This constant tells us that the dielectric material increases the capacitor's ability to store charge by a factor of 4.50 compared to when it is empty.
  • The dielectric constant is dimensionless 鈥� it has no units.
  • Its presence helps to prevent charge leakage by polarizing within the electric field.
  • By increasing the overall capacitance, it allows the capacitor to hold more energy at the same voltage.
Thus, the dielectric constant directly affects how much energy the capacitor can store, which was evident when comparing the two capacitors鈥攐ne empty and one with a dielectric.
Capacitance
Capacitance, denoted by \( C \), is the ability of a system to store an electric charge. In a simplistic form, it can be understood as the 'size' of the capacitor to hold electrical energy. It is measured in farads (F).

For the capacitors in the exercise, the empty capacitor has a specific capacitance \( C_0 \), while the one filled with the dielectric has an enhanced capacitance \( C_1 = \kappa C_0 \). This formula indicates how much more charge the capacitor with the dielectric can store, given the dielectric constant \( \kappa \).
  • Capacitance increases with a higher dielectric constant.
  • A larger capacitance allows more electric charge to be stored at a lower voltage.
  • The formula \( C = \frac{Q}{V} \) shows that capacitance is the ratio of the charge \( Q \) stored per unit voltage \( V \).
Understanding capacitance helps predict how capacitors will behave in electrical circuits and under different conditions.
Voltage and Potential Difference
Voltage, also known as potential difference, is the force that pushes electric charges through a circuit. In this context, it indicates how much energy is imparted to the charges stored across the capacitor plates. It is measured in volts (V).

In the exercise, the empty capacitor operates at a potential difference of 12.0 V. This voltage stores a certain amount of energy. When introducing a dielectric, the problem requires finding the new voltage \( V_1 \) that allows the dielectric-filled capacitor to store the same energy.
  • The energy equation used is \( U = \frac{1}{2} C V^2 \), highlighting the importance of voltage in energy storage.
  • Changing the dielectric alters the capacitance, which consequently requires adjusting the voltage to maintain energy levels.
  • The formula \( V_0^2 = \kappa V_1^2 \) shows how voltage must decrease in a capacitor with a dielectric to preserve stored energy.
Voltage and potential difference are key factors in determining how much energy can be stored and the work done on charges within a capacitor system.

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Most popular questions from this chapter

Concept Questions Two capacitors have the same plate separation. However, one has square plates, while the other has circular plates. The square plates are a length \(L\) on each side, and the diameter of the circular plates is \(L\). (a) If the same dielectric material were between the plates in each capacitor, which one would have the greater capacitance? (b) By putting different dielectric materials between the capacitor plates, we can make the two capacitors have the same capacitance. Which capacitor should contain the dielectric material with the greater dielectric constant? Give your reasoning in each case. The capacitors have the same capacitance because they contain different dielectric materials. The dielectric constant of the material between the square plates has a value of \(\kappa_{\text {square }}=3.00 .\) What is the dielectric constant \(\kappa_{\text {circle }}\) of the material between the circular plates? Be sure that your answer is consistent with your answers to the Concept Questions.

The potential at location \(A\) is \(452 \mathrm{~V}\). A positively charged particle is released there from rest and arrives at location \(B\) with a speed \(v_{B}\). The potential at location \(C\) is \(791 \mathrm{~V},\) and when released from rest from this spot, the particle arrives at \(B\) with twice the speed it previously had, or \(2 v_{B} .\) Find the potential at \(B\).

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

The potential difference between the plates of a capacitor is \(175 \mathrm{~V}\). Midway between the plates, a proton and an electron are released. The electron is released from rest. The proton is projected perpendicularly toward the negative plate with an initial speed. The proton strikes the negative plate at the same instant that the electron strikes the positive plate. Ignore the attraction between the two particles, and find the initial speed of the proton.

Interactive LearningWare 19.2 at reviews the concepts pertinent to this problem. What is the potential difference between the plates of a 3.3-F capacitor that stores sufficient energy to operate a 75 -W light bulb for one minute?
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