A capacitor with a vacuum between its plates is connected to a battery and then the gap is filled with Mylar. By what percentage is its energy-storing capacity increased?

Short Answer

Expert verified
Answer: The energy-storing capacity of the capacitor is increased by 210% when the gap between its plates is filled with Mylar.

Step by step solution

01

Find the initial capacitance of the capacitor

To determine the percentage increase in energy-storing capacity, we first need to find the initial capacitance of the capacitor with a vacuum between its plates. The capacitance of a capacitor can be calculated using the formula: C = ε₀ * A / d where C is the capacitance, ε₀ is the vacuum permittivity (8.85 * 10^{-12} F/m), A is the area of the plates, and d is the distance between the plates. However, since we are interested in the percentage change, we do not need the specific values of A and d. We can denote the initial capacitance with vacuum as C₀.
02

Find the capacitance with Mylar

Next, we need to determine the capacitance of the capacitor when the gap between its plates is filled with Mylar. Mylar has a dielectric constant (relative permittivity) of 𝜀𝑟 = 3.1. The capacitance of the capacitor with Mylar can be determined using the formula: C' = ε₀ * 𝜀𝑟 * A / d Notice that the only difference in the formula is that we multiply ε₀ by the dielectric constant 𝜀𝑟. Since the initial capacitance is C₀ = ε₀ * A / d, we can rewrite the capacitance with Mylar as: C' = 𝜀𝑟 * C₀ Substituting the value of 𝜀𝑟 = 3.1, we get, C' = 3.1 * C₀
03

Calculate the percentage increase in capacitance

Now that we have the initial capacitance (C₀) and the capacitance with Mylar (C'), we can find the percentage increase in capacitance using the following formula: Percentage increase = [(C' - C₀) / C₀] * 100 Substituting the values, we get: Percentage increase = [(3.1 * C₀ - C₀) / C₀] * 100 Simplifying the equation, we get: Percentage increase = [2.1 * C₀ / C₀] * 100 Percentage increase = 2.1 * 100 Percentage increase = 210%
04

Calculate the percentage increase in energy-storing capacity

We know that the energy stored in a capacitor can be calculated using the formula: E = (1/2) * C * V^2, where V is the voltage across the capacitor. Since the battery is connected across the capacitor, we can assume that the voltage remains constant. Therefore, the energy stored in the capacitor is directly proportional to its capacitance. So, the percentage increase in energy-storing capacity is the same as the percentage increase in capacitance, which is 210%. The energy-storing capacity of the capacitor is increased by 210% when the gap between its plates is filled with Mylar.

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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, also known as the relative permittivity, is a dimensionless number that represents how much an electrical insulating material (dielectric) can increase the capacitance of a capacitor compared to a vacuum.
A vacuum has the lowest possible dielectric constant, valued at 1. When another material, such as Mylar as mentioned in the exercise, fills the space between the plates of a capacitor, it increases the capacitance by a factor equal to its dielectric constant. This is significant because the dielectric constant reflects the ability of the material to align polar molecules in response to an electric field, effectively reducing the electric field within the capacitor and increasing its capacity to store charge.
  • A high dielectric constant indicates a more effective dielectric.
  • Mylar's dielectric constant of 3.1 means it increases the capacitor's ability to store energy by 210%, compared to the same capacitor in a vacuum.
  • This property is utilized in various applications, from improving the efficacy of energy storage to the miniaturization of electronic components.
Capacitance Calculation
Capacitance calculation involves determining the ability of a capacitor to store an electric charge. The unit of capacitance is the farad (F), which describes the amount of electric charge in coulombs a capacitor can hold per volt of electric potential difference between its plates.
For a parallel-plate capacitor, the capacitance can be calculated using the formula:
\[\begin{equation}C = \frac{\epsilon_0 \cdot \epsilon_r \cdot A}{d}\end{equation}\]
where:
  • \(C\) is the capacitance in farads (F).
  • \(\epsilon_0\) is the vacuum permittivity.
  • \(\epsilon_r\) is the relative permittivity or dielectric constant of the material between the plates.
  • \(A\) is the area of one of the plates in square meters (\(m^2\)).
  • \(d\) is the distance between the plates in meters (\(m\)).

Understanding the Changes in Capacitance

Inserting a dielectric increases the capacitance by the factor of its relative permittivity. If the capacitor initially had a capacitance of \(C_0\) in a vacuum, the new capacitance with the dielectric becomes \(C' = \epsilon_r \cdot C_0\).
As illustrated in the exercise, this change in capacitance also corresponds to the change in energy-storing capacity of the capacitor.
Vacuum Permittivity
Vacuum permittivity, denoted as \(\epsilon_0\), is a fundamental constant that represents the capability of a vacuum to permit electric field lines. Its value, approximately \(8.85 \times 10^{-12}\) farads per meter (F/m), is crucial in calculating the capacitance of a capacitor in a vacuum.
Vacuum permittivity is the baseline value that allows comparison of the effect of different dielectrics on capacitance. It is involved in the general capacitance formula for a capacitor without any dielectric material between its plates, or it can be used in conjunction with the dielectric constant to calculate the increased capacitance when a dielectric material is present.
  • It is also a key value in many equations in electromagnetism, like Coulomb's Law.
  • In circuits, vacuum permittivity helps to predict how capacitors will behave in various configurations without a dielectric.
  • The value is defined through the speed of light in a vacuum, underscoring its fundamental nature in the fabric of physics.

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Most popular questions from this chapter

Two concentric metal spheres are found to have a potential difference of \(900 . \mathrm{V}\) when a charge of \(6.726 \cdot 10^{-8} \mathrm{C}\) is applied to them. The radius of the outer sphere is \(0.210 \mathrm{~m}\). What is the radius of the inner sphere?

A parallel plate capacitor consisting of a pair of rectangular plates, each measuring \(1.00 \mathrm{~cm}\) by \(10.0 \mathrm{~cm},\) with a separation between the plates of \(0.100 \mathrm{~mm},\) is charged by a power supply at a potential difference of \(1.00 \cdot 10^{3} \mathrm{~V}\). The power supply is then removed, and without being discharged, the capacitor is placed in a vertical position over a container holding de-ionized water, with the short sides of the plates in contact with the water, as shown in the figure. Using energy considerations, show that the water will rise between the plates. Neglecting other effects, determine the system of equations that can be used to calculate the height to which the water rises between the plates. You do not have to solve the system.

The Earth is held together by its own gravity. But it is also a charge-bearing conductor. a) The Earth can be regarded as a conducting sphere of radius \(6371 \mathrm{~km},\) with electric field \(\vec{E}=(-150 . \mathrm{V} / \mathrm{m}) \hat{r}\) at its surface, where \(\hat{r}\) is a unit vector directed radially outward. Calculate the total electrostatic potential energy associated with the Earth's electric charge and field. b) The Earth has gravitational potential energy, akin to the electrostatic potential energy. Calculate this energy, treating the Earth as a uniform solid sphere. (Hint: \(d U=-(G m / r) d m\). c) Use the results of parts (a) and (b) to address this question: To what extent do electrostatic forces affect the structure of the Earth?

A capacitor consists of two parallel plates, but one of them can move relative to the other as shown in the figure. Air fills the space between the plates, and the capacitance is \(32.0 \mathrm{pF}\) when the separation between plates is \(d=0.500 \mathrm{~cm} .\) a) A battery with potential difference \(V=9.0 \mathrm{~V}\) is connected to the plates. What is the charge distribution, \(\sigma\), on the left plate? What are the capacitance, \(C^{\prime},\) and charge distribution, \(\sigma^{\prime},\) when \(d\) is changed to \(0.250 \mathrm{~cm} ?\) b) With \(d=0.500 \mathrm{~cm}\), the battery is disconnected from the plates. The plates are then moved so that \(d=0.250 \mathrm{~cm}\) What is the potential difference \(V^{\prime},\) between the plates?

Consider a cylindrical capacitor, with outer radius \(R\) and cylinder separation \(d\). Determine what the capacitance approaches in the limit where \(d \ll R\). (Hint: Express the capacitance in terms of the ratio \(d / R\) and then examine what happens as the ratio \(d / R\) becomes very small compared to \(1 .)\) Explain why the limit on the capacitance makes sense.

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