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Concept Questions 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

Understand the Capacitance Formula

Capacitance for a parallel plate capacitor is given by \(C = \frac{\kappa \varepsilon_0 A}{d}\), where \(A\) is the area of the plate, \(d\) is the separation between the plates, \(\varepsilon_0\) is the permittivity of free space, and \(\kappa\) is the dielectric constant.

02

Calculate the Area of the Square Plates

For square plates with side length \(L\), the area \(A_{\text{square}} = L^2\).

03

Calculate the Area of the Circular Plates

The area of circular plates with diameter \(L\) is \(A_{\text{circle}} = \pi \left(\frac{L}{2}\right)^2 = \frac{\pi L^2}{4}\).

04

Compare Capacitance with the Same Dielectric

Given that both capacitors have the same plate separation and dielectric, \(C = \frac{\kappa \varepsilon_0 A}{d}\). Since \(A_{\text{square}} = L^2 > \frac{\pi L^2}{4} = A_{\text{circle}}\), the square capacitor will have the greater capacitance.

05

Determine which Capacitor Needs Greater Dielectric Constant

To make the capacitance equal given different materials, the capacitor with smaller area \(A_{\text{circle}}\) needs a larger dielectric constant \(\kappa\). This compensates for its smaller plate area compared to the square plates.

06

Set the Capacitance Equations Equal

To find \(\kappa_{\text{circle}}\), set \(C_{\text{square}} = C_{\text{circle}}\). Thus, \(\kappa_{\text{square}} L^2 = \kappa_{\text{circle}} \frac{\pi L^2}{4}\).

07

Solve for the Dielectric Constant of the Circular Plates

Simplify and solve \(3.00 L^2 = \kappa_{\text{circle}} \frac{\pi L^2}{4}\) to find \(\kappa_{\text{circle}} = \frac{12}{\pi} \approx 3.82\).

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Key Concepts

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

Dielectric Constant

In the world of capacitors, the dielectric constant is a key player. It represents the ability of a material to store electrical energy in an electric field. In simpler terms, it's a measure of how well a material can "hold" or "trap" electric charge.
This constant, denoted by \(\kappa\), appears in the formula for capacitance: \(C = \frac{\kappa \varepsilon_0 A}{d}\). Here, \(\varepsilon_0\) is the permittivity of free space, while \(A\) and \(d\) are the area and separation of the capacitor's plates, respectively.
Students might wonder why the dielectric constant matters so much. Essentially, a higher \(\kappa\) increases capacitance. This means the capacitor can store more charge for the same voltage. In our exercise, different dielectrics compensate for different plate areas, ensuring capacitances equalize.

Parallel Plate Capacitor

A parallel plate capacitor is a device that stores electrical energy and, as the name suggests, features two parallel plates. These plates are separated by some distance and usually have a dielectric material in between. This basic structure allows the capacitor to store charge effectively.
The formula for the capacitance of a parallel plate capacitor is: \(C = \frac{\kappa \varepsilon_0 A}{d}\). Each variable here plays a vital role:

• \(\kappa\) is the dielectric constant, describing the material's influence.
• \(\varepsilon_0\) is the permittivity of free space, a fixed value.
• \(A\) is the area of one plate, impacting how much charge can be stored.
• \(d\) is the separation, influencing the capacitance inversely.

By changing these variables, particularly the dielectric constant and area, we can modify how much charge the capacitor can handle. This makes parallel plate capacitors highly versatile in different electrical applications.

Plate Area

Plate area is an essential aspect of capacitor design, directly affecting the capacitance. In the formula \(C = \frac{\kappa \varepsilon_0 A}{d}\), the area \(A\) is crucial because larger plates can store more charge. Think of it like trying to collect rainwater; a bigger surface catches more rain.
In our exercise, we dealt with square plates with area \(L^2\) and circular plates with area \(\frac{\pi L^2}{4}\). It's clear that the shape of the plates and their dimensions greatly affect the overall capacitance. The square plates, having a larger area, naturally have more capacity to store charge compared to the circular ones, when other factors like dielectric material are the same.
This leads to the conclusion that when designing a capacitor, both the shape and size of the plates must be considered, alongside other variables, to achieve the desired capacitance.