91Ó°ÊÓ

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.

Step by step solution

01

Calculate the Area of the Square Plates

The area of the square plate is calculated using the formula \( A = L^2 \), where \( L \) is the side length of the square.

02

Calculate the Area of the Circular Plates

The area of the circular plate is calculated using the formula \( A = \pi \left(\frac{L}{2}\right)^2 = \frac{\pi L^2}{4} \), where \( L \) is the diameter of the circle, making \( \frac{L}{2} \) the radius.

03

Compare Capacitance for Identical Dielectric

The capacitance of a capacitor is given by \( C = \kappa \varepsilon_0 \frac{A}{d} \), where \( \kappa \) is the dielectric constant, \( \varepsilon_0 \) is the permittivity of free space, \( A \) is the plate area, and \( d \) is the plate separation. Since \( d \) and \( \kappa \) are constant, capacitance is proportional to the plate area. Thus, \( C_{\text{square}} > C_{\text{circle}} \) because \( L^2 > \frac{\pi L^2}{4} \).

04

Determine Which Capacitance Requires Greater Dielectric Constant

To make the capacitances equal by using different dielectrics, the one with the smaller plate area (circular) requires a higher dielectric constant to compensate. Thus, the circular plates need a higher \( \kappa \).

05

Use Equation for Equal Capacitance with Different Dielectrics

To equalize capacitance \( C_{\text{square}} \) and \( C_{\text{circle}} \), we use \( \kappa_{\text{square}} A_{\text{square}} = \kappa_{\text{circle}} A_{\text{circle}} \). Substituting the areas, we have \( 3.00 L^2 = \kappa_{\text{circle}} \left(\frac{\pi L^2}{4}\right) \).

06

Solve for Dielectric Constant of Circular Plates

Cancel \( L^2 \) from both sides to simplify to: \[ 3.00 = \kappa_{\text{circle}} \cdot \frac{\pi}{4} \]. Solving for \( \kappa_{\text{circle}} \), we get \[ \kappa_{\text{circle}} = \frac{3.00 \times 4}{\pi} \approx 3.82. \]

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Dielectric constant

The dielectric constant, often represented as \( \kappa \), is a critical factor in determining a capacitor's ability to store electrical energy. It's a measure of how much a material can increase the capacitance compared to a vacuum. Dielectric materials are placed between the plates of capacitors to enhance their capacitance.
For any capacitor, the formula incorporating the dielectric constant is given by \( C = \kappa \varepsilon_0 \frac{A}{d} \). Here, \( C \) is the capacitance, \( \varepsilon_0 \) is the permittivity of free space, \( A \) is the area of one plate, and \( d \) is the separation between the plates.
The presence of a dielectric material with a high dielectric constant significantly increases capacitance because it reduces the electric field between the plates, allowing the capacitor to store more charge at the same voltage.

Plate separation

The separation between the plates of a capacitor, denoted as \( d \), has a direct inverse relationship with the capacitance. In simpler terms, the closer the plates are to each other, the greater the capacitance will be. This is because the electric field between the plates becomes stronger, thus enhancing the capacitor's ability to store charge.
In the formula \( C = \kappa \varepsilon_0 \frac{A}{d} \), as \( d \) decreases, the fraction \( \frac{A}{d} \) increases, resulting in higher capacitance, assuming all other factors remain constant. Therefore, tight plate separation is desirable in applications requiring higher capacitance.

Plate area

The area of the capacitor plates, \( A \), directly affects the overall capacitance. The larger the plate area, the greater the capacitance. This is because a larger surface allows for more charge to be stored between the plates.
For example, in the exercise with square and circular plates, the area of the square plates is \( L^2 \), while the area of the circular plates is \( \frac{\pi L^2}{4} \). Clearly, the square plates have a larger area, resulting in a higher capacitance with the same dielectric material and plate separation.
Thus, to modify the capacitance of a capacitor, adjusting the plate area is one effective approach.

Permittivity of free space

The permittivity of free space, denoted as \( \varepsilon_0 \), is a fundamental physical constant essential in electromagnetic theory. It measures the ability of vacuum to allow electric fields to pass through it and is a key factor in the equation for capacitance.
The value of \( \varepsilon_0 \) is approximately \( 8.85 \times 10^{-12} \) farads per meter (F/m). In the formula for capacitance, \( C = \kappa \varepsilon_0 \frac{A}{d} \), \( \varepsilon_0 \) helps account for how much better a dielectric material can store energy compared to a vacuum.
Understanding \( \varepsilon_0 \) is crucial, as it provides a baseline to gauge the effectiveness of different dielectric materials in capacitors.