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Chapter 18: Problem 75

Two parallel plate capacitors have circular plates. The magnitude of the charge on these plates is the same. However, the electric field between the plates of the first capacitor is \(2.2 \times 10^{5} \mathrm{N} / \mathrm{C},\) whereas the field within the second capacitor is \(3.8 \times 10^{5} \mathrm{N} / \mathrm{C} .\) Determine the ratio \(r_{2} / r_{1}\) of the plate radius for the second capacitor to the plate radius for the first capacitor.

Short Answer

Expert verified
The ratio \( \frac{r_2}{r_1} \) is approximately 0.761.

Step by step solution

01

Understanding the relationship between electric field and capacitance

The electric field ( \(E\)) between the plates of a parallel plate capacitor is given by \( E = \frac{\sigma}{\varepsilon_0} \), where \( \sigma \) is the surface charge density (charge per unit area) and \( \varepsilon_0 \) is the permittivity of free space. Since the charge ( \(Q\)) is the same on both capacitors, \( \sigma = \frac{Q}{A} = \frac{Q}{\pi r^2} \). This implies \( E = \frac{Q}{\varepsilon_0 \pi r^2} \).
02

Setting up the equation for the ratio of radii

Since \( E = \frac{Q}{\varepsilon_0 \pi r^2} \), we can set up the equality \( \frac{E_1}{E_2} = \frac{r_2^2}{r_1^2} \) because the charges are identical. Here, \( E_1 \) and \( E_2 \) are the electric fields of the first and second capacitors respectively.
03

Substituting given values into the equation

We have \( E_1 = 2.2 \times 10^5 \, ext{N/C} \) and \( E_2 = 3.8 \times 10^5 \, ext{N/C} \). Substituting these into the equation, we get:\[ \frac{2.2 \times 10^5}{3.8 \times 10^5} = \frac{r_2^2}{r_1^2} \]
04

Solving for the ratio \( \frac{r_2}{r_1} \)

Solve the equation \( \frac{2.2}{3.8} = \frac{r_2^2}{r_1^2} \) by taking the square root on both sides:\[ \frac{r_2}{r_1} = \sqrt{\frac{2.2}{3.8}} \approx 0.761 \]

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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 device used to store electrical energy in an electric field. It consists of two conductive plates separated by a small distance, where one plate is positively charged and the other is negatively charged. This setup creates an electric field between the plates, where the field lines run from the positive to the negative plate.
  • Capacitance is influenced by the area of the plates and the distance between them.
  • The capacitor's capacity to store charge is directly proportional to the area of the plates and inversely proportional to the distance separating them.
Capacitors are widely used in various electronic circuits and have applications in tuning radios, stabilizing voltage and power flow, and storing energy.
Surface Charge Density
Surface charge density, represented by the symbol \( \sigma \), is a measure of how much electric charge resides per unit area on a surface, such as the plates of a capacitor.
  • Mathematically, it is expressed as \( \sigma = \frac{Q}{A} \), where \( Q \) is the charge and \( A \) is the area of the surface.
  • In the context of a capacitor, surface charge density is vital because it determines the electric field strength between the plates, given by the formula \( E = \frac{\sigma}{\varepsilon_0} \).
Higher surface charge density means that more charge is stored on the plates, leading to a stronger electric field, which correlates with the ability of the capacitor to store energy.
Permittivity of Free Space
The permittivity of free space, denoted as \( \varepsilon_0 \), is a fundamental physical constant that characterizes the ability of a vacuum to permit electric field lines. It appears in the equations describing electrostatic forces and fields.
  • Its approximate value is \( 8.85 \times 10^{-12} \, \text{C}^2/\text{(N m}^2) \).
  • In the context of parallel plate capacitors, it influences the relationship between surface charge density and electric field, forming the link \( E = \frac{\sigma}{\varepsilon_0} \).
This constant is crucial in calculating how much electric flux can pass through a given medium, ultimately affecting how capacitors store and handle electric charges.

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Most popular questions from this chapter

Two objects carry initial charges that are \(q_{1}\) and \(q_{2}\), respectively, where \(\left|q_{2}\right|>\left|q_{1}\right| .\) They are located \(0.200 \mathrm{m}\) apart and behave like point charges. They attract each other with a force that has a magnitude of \(1.20 \mathrm{N} .\) The objects are then brought into contact, so the net charge is shared equally, and then they are returned to their initial positions. Now it is found that the objects repel one another with a force whose magnitude is equal to the magnitude of the initial attractive force. What are the magnitudes of the initial charges on the objects? Section 18.6 The Electric Field Section 18.7 Electric Field Lines Section 18.8 The Electric Field Inside a Conductor: Shielding

ssm mmh Two tiny spheres have the same mass and carry charges of the same magnitude. The mass of each sphere is \(2.0 \times 10^{-6} \mathrm{kg} .\) The gravitational force that each sphere exerts on the other is balanced by the electric force. (a) What algebraic signs can the charges have? (b) Determine the charge magnitude.

In a vacuum, two particles have charges of \(q_{1}\) and \(q_{2}\), where \(q_{1}=+3.5 \mu \mathrm{C}\). They are separated by a distance of \(0.26 \mathrm{m}\), and particle 1 experiences an attractive force of \(3.4 \mathrm{N}\). What is \(q_{2}\) (magnitude and \(\operatorname{sign}\) )?

Four point charges have the same magnitude of \(2.4 \times 10^{-12} \mathrm{C}\) and are fixed to the corners of a square that is \(4.0 \mathrm{cm}\) on a side. Three of the charges are positive and one is negative. Determine the magnitude of the net electric field that exists at the center of the square.

\(\mathrm{mmh}\) A small object has a mass of \(3.0 \times 10^{-3} \mathrm{kg}\) and a charge of \(-34 \mu \mathrm{C}\). It is placed at a certain spot where there is an electric field. When released, the object experiences an acceleration of \(2.5 \times 10^{3} \mathrm{m} / \mathrm{s}^{2}\) in the direction of the \(+x\) axis. Determine the magnitude and direction of the electric field.
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