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In Fig. 25-59, two parallelplate capacitors \(A\) and \(B\) are connected in parallel across a \(600 \mathrm{~V}\) battery. Each plate has area \(80.0 \mathrm{~cm}^{2}\) the plate separations are \(3.00 \mathrm{~mm}\). Capacitor \(A\) is filled with air; capacitor \(B\) is filled with a dielectric of dielectric constant \(\kappa=2.60 .\) Find the magnitude of the electric field within (a) the dielectric of capacitor \(B\) and (b) the air of capacitor \(A\). What are the free charge densities \(\sigma\) on the higher-potential plate of (c) capacitor \(A\) and (d) capacitor \(B\) ? (e) What is the induced charge density \(\sigma^{\prime}\) on the top surface of the dielectric?

Short Answer

Expert verified
(a) 76900 V/m, (b) 200000 V/m, (c) 1.77 µC/m², (d) 1.77 µC/m², (e) 0 C/m².

Step by step solution

01

Find electric field in Capacitor A

To find the electric field within capacitor \(A\), which is filled with air, use the formula for the electric field in a parallel plate capacitor: \(E = \frac{V}{d}\), where \(V\) is the potential difference across the capacitor and \(d\) is the separation between the plates. Here, \(V = 600 \text{ V}\) and \(d = 3.00 \text{ mm} = 3.00 \times 10^{-3} \text{ m}\). Thus, \(E_A = \frac{600}{3.00 \times 10^{-3}} = 2.00 \times 10^5 \text{ V/m}\).
02

Find electric field in Capacitor B

For capacitor \(B\), which is filled with a dielectric, the electric field \(E_B\) is reduced by the dielectric constant \(\kappa\). The formula becomes \(E_B = \frac{E_0}{\kappa}\), where \(E_0 = \frac{V}{d}\). Substituting the values, \(E_0 = 2.00 \times 10^5 \text{ V/m}\), and \(\kappa = 2.60\), we find that \(E_B = \frac{2.00 \times 10^5}{2.60} = 7.69 \times 10^4 \text{ V/m}\).
03

Calculate free charge density on Capacitor A

The free charge density \(\sigma\) on the plate of capacitor \(A\) can be found using \(\sigma = \varepsilon_0 E_A\), where \(\varepsilon_0 = 8.85 \times 10^{-12} \text{ C/Vm}\). Thus, \(\sigma_A = 8.85 \times 10^{-12} \times 2.00 \times 10^5 = 1.77 \times 10^{-6} \text{ C/m}^2\).
04

Calculate free charge density on Capacitor B

The free charge density for capacitor \(B\), \(\sigma_B\), is given by the formula \(\sigma_B = \varepsilon_0 E_0\), because \(E_0\) is the original electric field in absence of the dielectric. Thus, \(\sigma_B = 1.77 \times 10^{-6} \text{ C/m}^2\), which is the same as for capacitor A due to the same potential difference and geometry.
05

Calculate induced charge density on dielectric

The induced charge density \(\sigma^\prime\) on the dielectric surface is given by \(\sigma^\prime = \sigma_B - \kappa \varepsilon_0 E_B\). We know \(\sigma_B = 1.77 \times 10^{-6} \text{ C/m}^2\) and calculate \(\kappa \varepsilon_0 E_B = 2.60 \times 8.85 \times 10^{-12} \times 7.69 \times 10^4 = 1.77 \times 10^{-6} \text{ C/m}^2\). Therefore, \(\sigma^\prime = 1.77 \times 10^{-6} - 1.77 \times 10^{-6} = 0\).

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Key Concepts

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

Electric Field
An electric field represents the force that a charged particle would feel in a given region of space. In the context of a parallel plate capacitor, the electric field (\(E\)) is directly related to the voltage (\(V\)) applied across the plates and the distance (\(d\)) between them. This relationship can be expressed with the formula:
\[E = \frac{V}{d}\]
This formula indicates that the electric field is stronger when the potential difference is larger or when the plates are closer together. In Capacitor A, which is filled with air, the electric field is straightforward to calculate using the above formula. However, in Capacitor B, where a dielectric is present, the electric field is reduced because the dielectric material opposes the field's effects. This reduction is quantified by the dielectric constant (\(\kappa\)), resulting in a weaker electric field within dielectric-filled capacitors:
\[E_B = \frac{E_0}{\kappa}\]
Understanding the electric field is fundamental when exploring the behavior of charges in capacitors.
Dielectric Constant
The dielectric constant or relative permittivity ( \( \kappa \)) is an indicator of how much a material can reduce the electric field within it compared to a vacuum. In simple terms, dielectrics like those in Capacitor B allow capacitors to store more charge without changing voltage, by partially neutralizing the electric field.Dielectric constants are dimensionless and specific to each material:
  • A higher dielectric constant implies greater ability to polarize, thus reducing the electric field more effectively.
  • Air, for example, has a dielectric constant close to 1.
  • The dielectric in Capacitor B is given a constant of 2.60, illustrating a significant reduction in electric field strength inside the capacitor.
The presence of a dielectric in capacitor circuits enables devices to operate efficiently at lower voltages while retaining high capacitance levels. The role of the dielectric constant is crucial in designing capacitors for various electrical and electronic applications.
Charge Density
Charge density on a capacitor's plates is a measure of the amount of charge per unit area. It can be divided into two types:
  • Free Charge Density (\(\sigma\)): The charge that lies on the capacitor plates and contributes to the capacitor's overall ability to store electrical energy.
  • Induced Charge Density (\(\sigma'\)): Seen in dielectric systems, where the dielectric's molecules align to reduce the effect of an applied electric field, creating an additional apparent charge layer.
The free charge density in an air-filled capacitor is simply the product of the permittivity of free space (\(\varepsilon_0\)) and the electric field (\(E\)). In the context of the problem, this is calculated as:
\[\sigma_A = \varepsilon_0 E_A\]
Similarly, without the dielectric, Capacitor B would have the same free charge density:
\[\sigma_B = \varepsilon_0 E_0\]
However, due to its dielectric, the effective electric field (\(E_B\)) is weakened, altering charge dynamics. Notably, in this case, the induced charge density does not contribute straightforwardly to net charge storage without specific configurations, as observed. The study of charge density is vital for understanding how capacitors function in their diverse roles.
Potential Difference
The potential difference or voltage across a capacitor's plates is the cause of the electric field within the capacitor. It's vital for pushing electrons through the circuit, thereby storing energy as an electric field.In a capacitor:
  • The applied potential difference determines how much charge can be effectively stored on the plates.
  • This relationship can be described via the formula for capacitance (\(C\)), given by \(C = \frac{Q}{V}\), demonstrating how capacitance is influenced by the amount of charge and voltage.
For devices connected in parallel, like Capacitors A and B in the given problem, the potential difference is the same across each device, ensuring uniform storage capacity strategy for varying internal structures (air vs. dielectric-filled). By understanding potential difference, electrical engineers can design components that handle specified voltage levels, ensuring expected operations and safety.

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Most popular questions from this chapter

What capacitance is required to store an energy of \(10 \mathrm{~kW} \cdot \mathrm{h}\) at a potential difference of \(1000 \mathrm{~V}\) ?

As a safety engineer, you must evaluate the practice of storing flammable conducting liquids in nonconducting containers. The company supplying a certain liquid has been using a squat, cylindrical plastic container of radius \(r=0.20 \mathrm{~m}\) and filling it to height \(h=10 \mathrm{~cm},\) which is not the container's full interior height (Fig. 25 -44). Your investigation reveals that during handling at the company, the exterior surface of the container commonly acquires a negative charge density of magnitude \(2.0 \mu \mathrm{C} / \mathrm{m}^{2}\) (approximately uniform). Because the liquid is a conducting material, the charge on the container induces charge separation within the liquid. (a) How much negative charge is induced in the center of the liquid's bulk? (b) Assume the capacitance of the central portion of the liquid relative to ground is \(35 \mathrm{pF}\). What is the potential energy associated with the negative charge in that effective capacitor? (c) If a spark occurs between the ground and the central portion of the liquid (through the venting port), the potential energy can be fed into the spark. The minimum spark energy needed to ignite the liquid is \(10 \mathrm{~mJ} .\) In this situation, can a spark ignite the liquid?

How many \(1.00 \mu \mathrm{F}\) capacitors must be connected in parallel to store a charge of \(1.00 \mathrm{C}\) with a potential of \(110 \mathrm{~V}\) across the capacitors?

A potential difference of \(300 \mathrm{~V}\) is applied to a series connection of two capacitors of capacitances \(C_{1}=2.00 \mu \mathrm{F}\) and \(C_{2}=8.00 \mu \mathrm{F}\). What are (a) charge \(q_{1}\) and (b) potential difference \(V_{1}\) on capacitor 1 and \((\mathrm{c}) q_{2}\) and \((\mathrm{d}) V_{2}\) on capacitor \(2 ?\) The charged capacitors are then disconnected from each other and from the battery. Then the capacitors are reconnected with plates of the same signs wired together (the battery is not used). What now are (e) \(q_{1}\), (f) \(V_{1}\), (g) \(q_{2}\), and (h) \(V_{2}\) ? Suppose, instead, the capacitors charged in part (a) are reconnected with plates of opposite signs wired together. What now are (i) \(q_{1},(\mathrm{j}) V_{1},(\mathrm{k}) q_{2},\) and \((\mathrm{l}) V_{2} ?\)

A parallel-plate air-filled capacitor having area \(40 \mathrm{~cm}^{2}\) and plate spacing \(1.0 \mathrm{~mm}\) is charged to a potential difference of \(600 \mathrm{~V}\). Find (a) the capacitance, (b) the magnitude of the charge on each plate, (c) the stored energy, (d) the electric field between the plates, and (e) the energy density between the plates.

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