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Chapter 21: Problem 46

A 1.2 nF parallel-plate capacitor has an air gap between its plates. Its capacitance increases by \(3.0 \mathrm{nF}\) when the gap is filled by a dielectric. What is the dielectric constant of that dielectric?

Short Answer

Expert verified
The dielectric constant of the dielectric is \(3.5\).

Step by step solution

01

Identify given values and the unknown

The initial capacitance without the dielectric, \(C_1\) is given as \(1.2 \, nF\). After the dielectric is introduced, the capacitance increases by \(3.0 \, nF\), making the new capacitance \(C_2 = 4.2 \, nF\). The dielectric constant (\(k\)) is the unknown to be solved for.
02

Understand the formula of capacitance with a dielectric

The capacitance of a capacitor when a dielectric is inserted between its plates is given by equation: \(C_2 = kC_1\), where \(C_1\) is the original capacitance, \(C_2\) is the new capacitance and \(k\) is the dielectric constant.
03

Substitute the values and solve for \(k\)

Substitute the given values of \(C_1\) and \(C_2\) into the equation \(C_2 = kC_1\). So \(4.2 = k \cdot 1.2\). Solving this equation for \(k\), \(k = \frac{4.2}{1.2} = 3.5\).

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Key Concepts

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

Capacitance
Capacitance is the ability of a component or circuit to store electrical charge. It is measured in farads (F), but in practice, we often use smaller units like nanofarads (nF) or picofarads (pF). Capacitance depends on several factors:
  • The area of the plates (A) 鈥� larger areas allow for more charge storage.
  • The distance (d) between the plates 鈥� closer plates increase capacitance.
  • The permittivity of the material between the plates.
The basic formula for capacitance (\(C\)) in a parallel-plate capacitor is given by:\[C = \frac{\varepsilon_0 A}{d}\]where\(\varepsilon_0\)is the permittivity of free space. When other dielectric materials are used, the permittivity value changes, thus affecting the overall capacitance.
Parallel-Plate Capacitor
A parallel-plate capacitor consists of two conductive plates separated by a gap, which can be filled with an insulating material called a dielectric. This setup allows the capacitor to store an electric charge and potential energy. In a basic air-filled setup, the capacitance is influenced by:
  • The size of the plates 鈥� larger plates can hold more charge.
  • The proximity of the plates 鈥� the closer the plates, the higher the capacitance.
  • The medium between the plates 鈥� air or vacuum initially unless altered.
A parallel-plate capacitor is ideal for understanding the relationship between physical dimensions, the dielectric material, and capacitance, making it a fundamental concept in electronics.
Dielectric Material
Dielectric materials are insulators placed between the plates of a capacitor, which increases its ability to store charge. When a dielectric is introduced, it affects the electric field and reduces it, allowing for more charge to be stored at the same voltage. Key factors about dielectric materials include:
  • Dielectric constant: This is a measure of a material's ability to increase capacitance compared to a vacuum. Higher values mean better insulation and more capacitance.
  • Polarization: Dielectric materials become polarized in an electric field, contributing to additional charge storage.
In practical terms, adding a dielectric between the capacitor plates increases capacitance according to the formula:\[C_2 = kC_1\]where\(C_2\)is the capacitance with the dielectric,\(C_1\)is the initial capacitance, and\(k\)is the dielectric constant.
Problem-Solving
Effective problem-solving in physics involves understanding and applying concepts to real-world situations. In our exercise, determining the dielectric constant requires analyzing given values and plugging them into the appropriate formula. Here's a breakdown:
  • Identify the known values and what you need to find. Here, we have initial and final capacitances and need the dielectric constant.
  • Use the correct formula connecting these quantities. In this case, \(C_2 = kC_1\)was vital.
  • Substitute known values into the equation, solve algebraically, and ensure to check your calculations.
By following these steps, you can solve a range of physics problems effectively, whether it鈥檚 related to capacitors or other concepts.

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Most popular questions from this chapter

A flying hummingbird picks up charge as it moves through the air. This creates a potential near the bird. What is the "voltage of a hummingbird"? Assume that the bird acquires a charge of \(+200 \mathrm{pC}\), a typical value, and model the bird as a sphere of radius \(3 \mathrm{cm}\).

The capacity of a battery to deliver charge, and thus power, decreases with temperature. The same is not true of capacitors. For sure starts in cold weather, a truck has a 500 F capacitor alongside a battery. The capacitor is charged to the full \(13.8 \mathrm{V}\) of the truck's battery. How much energy does the capacitor store? How does the energy density of the \(9.0 \mathrm{kg}\) capacitor system compare to the \(130,000 \mathrm{J} / \mathrm{kg}\) of the truck's battery?

What potential difference is needed to accelerate a \(\mathrm{He}^{+}\) ion (charge \(+e,\) mass 4 u) from rest to a speed of \(1.0 \times 10^{6} \mathrm{m} / \mathrm{s} ?\)

An uncharged capacitor is connected to the terminals of a \(3.0 \mathrm{V}\) battery, and \(6.0 \mu \mathrm{C}\) flows to the positive plate. The \(3.0 \mathrm{V}\) battery is then disconnected and replaced with a \(5.0 \mathrm{V}\) battery, with the positive and negative terminals connected in the same manner as before. How much additional charge flows to the positive plate?

It takes \(3.0 \mu \mathrm{J}\) of work to move a \(15 \mathrm{nC}\) charge from point \(\mathrm{A}\) to B. It takes \(-5.0 \mu \mathrm{J}\) of work to move the charge from \(\mathrm{C}\) to \(\mathrm{B}\). What is the potential difference \(V_{\mathrm{C}}-V_{\mathrm{A}} ?\)
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