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Chapter 25: Problem 60

(a) Calculate the capacitance of a capacitor consisting of two parallel plates separated by a layer of paraffin wax \(0.50 \mathrm{~cm}\) thick, the area of each plate being \(80 \mathrm{~cm}^{2}\). The dielectric constant for the wax is \(2.0 .(b)\) If the capacitor is connected to a \(100-V\) source, calculate the charge on the capacitor and the energy stored in the capacitor.

Short Answer

Expert verified
Capacitance: \(2.832 \times 10^{-11} \mathrm{~F}\), Charge: \(2.832 \times 10^{-9} \mathrm{~C}\), Energy: \(1.416 \times 10^{-7} \mathrm{~J}\).

Step by step solution

01

Understand the Formula for Capacitance with a Dielectric

The formula for the capacitance of a parallel plate capacitor with a dielectric is given by: \[ C = \varepsilon_0 \cdot \varepsilon_r \cdot \frac{A}{d} \]where:- \( C \) is the capacitance,- \( \varepsilon_0 \) is the permittivity of free space \((8.85 \times 10^{-12} \mathrm{~F/m})\),- \( \varepsilon_r \) is the relative permittivity (dielectric constant) of the material between the plates,- \( A \) is the area of one of the plates, and- \( d \) is the separation between the plates.
02

Convert Units and Substitute Known Values

Ensure all measurements are in meters:- Plate area \( A \) is given as \(80 \mathrm{~cm}^2 = 80 \times 10^{-4} \mathrm{~m}^2\).- Separation \( d \) is given as \(0.50 \mathrm{~cm} = 0.005 \mathrm{~m}\).Substitute these into the formula:\[ C = 8.85 \times 10^{-12} \times 2.0 \times \frac{80 \times 10^{-4}}{0.005} \]
03

Calculate the Capacitance

Perform the calculation:\[ C = 8.85 \times 10^{-12} \times 2.0 \times \frac{80 \times 10^{-4}}{0.005} = 2.832 \times 10^{-11} \mathrm{~F} \]Thus, the capacitance is \(2.832 \times 10^{-11} \mathrm{~F}\).
04

Calculate Charge on the Capacitor

Use the formula \( Q = C \cdot V \) to find the charge \( Q \), where \( V = 100 \mathrm{~V} \) is the voltage:\[ Q = 2.832 \times 10^{-11} \mathrm{~F} \times 100 \mathrm{~V} = 2.832 \times 10^{-9} \mathrm{~C} \]The charge on the capacitor is \(2.832 \times 10^{-9} \mathrm{~C}\).
05

Calculate the Energy Stored in the Capacitor

Use the formula for energy stored in a capacitor: \( E = \frac{1}{2} C V^2 \):\[ E = \frac{1}{2} \times 2.832 \times 10^{-11} \mathrm{~F} \times (100)^2 \mathrm{~V}^2 \]Perform the calculation:\[ E = 1.416 \times 10^{-7} \mathrm{~J} \]Thus, the energy stored in the capacitor is \(1.416 \times 10^{-7} \mathrm{~J}\).

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Key Concepts

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

Parallel Plate Capacitor
A parallel plate capacitor is a simple and common type of capacitor. It consists of two conductive plates placed parallel to each other with an insulating material, called a dielectric, between them. The ability of a capacitor to store electric charge is measured by its capacitance. In a parallel plate capacitor, the capacitance is directly proportional to the area of the plates and inversely proportional to the distance between them.
  • The larger the area of the plates, the greater the capacitance, because more electric field lines can form between the larger surfaces.
  • The closer the plates are to each other, the greater the capacitance, as a stronger electric field is established when the separation is smaller.
The formula to calculate the capacitance of a parallel plate capacitor when a dielectric is present is: \[ C = \varepsilon_0 \cdot \varepsilon_r \cdot \frac{A}{d} \]where:
  • \( C \) is the capacitance,
  • \(\varepsilon_0\) is the permittivity of free space,
  • \(\varepsilon_r\) is the dielectric constant of the material,
  • \( A \) is the area of the plates,
  • \( d \) is the distance between the plates.
Dielectric Constant
The dielectric constant, also known as the relative permittivity, is a measure of a material's ability to store electrical energy in an electric field. When a dielectric material is placed between the plates of a capacitor, it affects how much charge the capacitor can hold.
  • A higher dielectric constant indicates that the material can better support electric field lines, enhancing the capacitance of the capacitor.
  • For paraffin wax, the dielectric constant is 2.0, meaning it can hold twice the amount of charge as if there were no dielectric between the plates.
Including a dielectric between the plates of a parallel plate capacitor increases its capacitance, as shown in the formula \( C = \varepsilon_0 \cdot \varepsilon_r \cdot \frac{A}{d} \). The presence of the dielectric amplifies the effect of the applied voltage, resulting in a higher charge storage capacity.
Energy Stored in a Capacitor
Energy storage in a capacitor is an important aspect, especially in electronic circuits where capacitors are used to manage power supply efficiently. The energy stored in a capacitor is given by the formula \[ E = \frac{1}{2} C V^2 \]where:
  • \( E \) is the energy stored,
  • \( C \) is the capacitance,
  • \( V \) is the voltage across the capacitor.
This relationship shows that the energy stored in a capacitor grows with both the square of the voltage and directly with the capacitance.
  • This means even a small increase in voltage can lead to a significantly larger increase in energy storage.
  • The stored energy can be released when the capacitor discharges, which can power various electric devices or smooth out power supply dips.
By storing energy effectively, capacitors play a crucial role in regulating voltage and current in electrical systems.
Charge on a Capacitor
Charge in a capacitor is a fundamental concept. The ability of a capacitor to store charge is essential to its function in circuits. The total charge \( Q \) stored in a capacitor connected to a voltage source \( V \) is determined by the formula:\[ Q = C \times V \]where:
  • \( Q \) is the charge stored in the capacitor,
  • \( C \) is the capacitance,
  • \( V \) is the voltage applied.
In essence, the charge on a capacitor is proportional to both the capacitance and the voltage:
  • Higher capacitance allows more charge storage, as more surface area and better insulation enable effective collection and retention of charge.
  • Higher voltage increases the amount of charge that can be stored for a given capacitance.
Understanding how charge is accumulated and sustained in capacitors allows for the effective design and applications in filtering, energy storage, and timing circuits.

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Most popular questions from this chapter

An electron starts from rest and falls through a potential rise of 80 V. What is its final speed? Positive charges fall through potential drops; negative charges, such as electrons, fall through potential rises. Change in \(\mathrm{PE}_{E}=V q=(80 \mathrm{~V})\left(-1.6 \times 10^{-19} \mathrm{C}\right)=-1.28 \times 10^{-17} \mathrm{~J}\) This lost \(\mathrm{PE}_{E}\) appears as KE of the electron:= and $$ \begin{array}{c} \mathrm{PE}_{\varepsilon} \text { lost }=\text { KE gained } \\\ 1.28 \times 10^{-17} \mathrm{~J}=\frac{1}{2} m v_{f}^{2} \frac{1}{2} m v_{i}^{2}=\frac{1}{2} m v_{f}^{2}-0 \\ v_{f}=\sqrt{\frac{\left(1.28 \times 10^{-17} \mathrm{~J}(2)\right.}{9.1 \times 10^{-31} \mathrm{~kg}}}=5.3 \times 10^{6} \mathrm{~m} / \mathrm{s} \end{array} $$

A \(1.2-\mu \mathrm{F}\) capacitor is charged to \(3.0 \mathrm{kV}\). Compute the energy stored in the capacitor. $$ \text { Energy }=\frac{1}{2} q V=\frac{1}{2} C V^{2}=\frac{1}{2}\left(1.2 \times 10^{-6} \mathrm{~F}\right)(3000 \mathrm{~V})^{2}=5.4 \mathrm{~J} $$

What happens to the electric potential at a point in space due to a point charge if that charge is doubled?

Determine the electric potential \(1.00 \mathrm{~cm}\) from an electron in vacuum. [Hint: \(e=-1.6022 \times 10^{-19} \mathrm{C.}\).]

A \(2.0-\mu \mathrm{F}\) capacitor is charged to \(50 \mathrm{~V}\) and then connected in parallel (positive plate to positive plate) with a \(4.0-\mu \mathrm{F}\) capacitor charged to \(100 \mathrm{~V}\). \((a)\) What are the final charges on the capacitors? (b) What is the potential difference across each?
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