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Chapter 18: Problem 29

The plates of a parallel-plate capacitor are \(3.28 \mathrm{~mm}\) apart, and each has an area of \(12.2 \mathrm{~cm}^{2}\). Each plate carries a charge of magnitude \(4.35 \times 10^{-8} \mathrm{C}\). The plates are in vacuum. (a) What is the capacitance? (b) What is the potential difference between the plates? (c) What is the magnitude of the electric field between the plates?

Short Answer

Expert verified
(a) Capacitance is \(3.287 \times 10^{-12} \text{ F}\). (b) Potential difference is \(1.323 \times 10^4 \text{ V}\). (c) Electric field is \(4.034 \times 10^6 \text{ V/m}\).

Step by step solution

01

Convert Units

Convert the area from \(\text{cm}^2\) to \(\text{m}^2\). Since 1 \(\text{cm}^2 = 10^{-4} \text{m}^2\), the area \(A\) is:\[A = 12.2 \text{ cm}^2 \times 10^{-4} \text{ m}^2/\text{cm}^2 = 1.22 \times 10^{-3} \text{ m}^2\]Also, convert the distance from mm to m: \[d = 3.28 \text{ mm} = 3.28 \times 10^{-3} \text{ m}\]
02

Calculate Capacitance

Use the formula for capacitance of a parallel-plate capacitor in a vacuum: \[C = \frac{\varepsilon_0 \cdot A}{d}\]where \(\varepsilon_0 = 8.854 \times 10^{-12} \text{ C}^2/\text{N}\cdot\text{m}^2\) (permittivity of free space).Substitute the values:\[C = \frac{8.854 \times 10^{-12} \cdot 1.22 \times 10^{-3}}{3.28 \times 10^{-3}} \approx 3.287 \times 10^{-12} \text{ F}\]
03

Calculate Potential Difference

Use the formula for potential difference:\[V = \frac{Q}{C}\]where \(Q = 4.35 \times 10^{-8} \text{ C}\).Substitute the values:\[V = \frac{4.35 \times 10^{-8}}{3.287 \times 10^{-12}} \approx 1.323 \times 10^4 \text{ V}\]
04

Calculate Electric Field

The magnitude of the electric field \(E\) between the plates is given by:\[E = \frac{V}{d}\]Substitute the values:\[E = \frac{1.323 \times 10^4}{3.28 \times 10^{-3}} \approx 4.034 \times 10^6 \text{ V/m}\]

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Key Concepts

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

Electric Field
The electric field is a fundamental concept in physics, especially when studying capacitors. In a parallel-plate capacitor, the electric field (E) is created by two plates with opposite charges.
  • The electric field describes the force per unit charge exerted on a charged particle within this region.
  • It always points from the positive to the negative charge.
For a parallel-plate capacitor, the electric field is uniform and can be calculated using the formula:\[E = \frac{V}{d}\]where V is the potential difference and d is the distance between the plates. This shows that the electric field depends directly on the potential difference and inversely on the separation distance between the plates.
Understanding how the electric field behaves helps in analyzing how capacitors store energy.
The greater the field, the stronger the force exerted on charges. This is crucial for applications requiring precise control of electric forces.
Potential Difference
Potential difference, also known as voltage, is crucial when dealing with capacitors. It refers to the energy required to move a unit charge between two points in an electric field, such as between the plates of a capacitor.
  • It is measured in volts (V).
  • A high potential difference means more energy is stored per unit charge.
The potential difference across a capacitor's plates can be determined using the formula:\[V = \frac{Q}{C}\]where Q is the charge stored on the plates and C is the capacitance of the capacitor.
This equation tells us that voltage is directly proportional to the charge and inversely proportional to capacitance.
A better understanding of potential difference aids in designing circuits to regulate the energy flow efficiently.
Vacuum
A vacuum, or empty space, is an environment devoid of matter, including air molecules. In the context of capacitors, it affects how they operate and their efficiency.
  • In a vacuum, there are no additional molecules to interfere with the electric field between capacitor plates.
  • This allows for a high-quality, uninterrupted field.
The permittivity of vacuum, denoted as \( \varepsilon_0 \), is a constant that represents how much electric field is 'allowed' to pass through a vacuum. The formula for the capacitance of a parallel-plate capacitor in a vacuum is:\[C = \frac{\varepsilon_0 \cdot A}{d}\]where A is the area of the plates and d is the distance between them.
Using a vacuum as a dielectric ensures that the capacitance is determined precisely without the distortions that other materials might introduce. Understanding this concept is key in applications requiring high precision.

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Most popular questions from this chapter

A particle with a charge of \(+4.20 \mathrm{nC}\) is in a uniform electric field \(\vec{E}\) directed in the negative \(x\) direction. It is released from rest, and after it has moved \(6.00 \mathrm{~cm},\) its kinetic energy is found to be \(+1.50 \times 10^{-6} \mathrm{~J}\). (a) What work was done by the electric force? (b) What was the change in electric potential over the distance that the charge moved? (c) What is the magnitude of \(\overrightarrow{\boldsymbol{E}} ?\) (d) What was the change in potential energy of the charge?

Neurons are the basic units of the ner- vous system. They contain long tubular structures called axons that propagate electrical signals away from the ends of the neurons. The axon contains a solution of potassium ions \(\mathrm{K}^{+}\) and large negative organic ions. The axon membrane prevents the large ions from leaking out, but the smaller \(\mathrm{K}^{+}\) ions are able to penetrate the membrane to some degree. (See Figure \(18.35 .)\) This leaves an excess of negative charge on the inner surface of the axon membrane and an excess of positive charge on the outer surface, resulting in a potential difference across the membrane that prevents further \(\mathrm{K}^{+}\) ions from leaking out. Measurements show that this potential difference is typically about \(70 \mathrm{mV}\). The thickness of the axon membrane itself varies from about 5 to \(10 \mathrm{nm}\), so we'll use an average of \(7.5 \mathrm{nm}\). We can model the membrane as a large sheet having equal and opposite charge densities on its faces. (a) Find the electric field inside the axon membrane, assuming (not too realistically) that it is filled with air. Which way does it point, into or out of the axon? (b) Which is at a higher potential, the inside surface or the outside surface of the axon membrane?

How much charge does a \(12 \mathrm{~V}\) battery have to supply to fully charge a \(2.5 \mu \mathrm{F}\) capacitor and a \(5.0 \mu \mathrm{F}\) capacitor when they're (a) in parallel, (b) in series? (c) How much energy does the battery have to supply in each case?

A \(10.0 \mu \mathrm{F}\) parallel-plate capacitor with circular plates is connected to a \(12.0 \mathrm{~V}\) battery. (a) What is the charge on each plate? (b) How much charge would be on the plates if their separation were doubled while the capacitor remained connected to the battery? (c) How much charge would be on the plates if the capacitor were connected to the \(12.0 \mathrm{~V}\) battery after the radius of each plate was doubled without changing their separation?

A parallel-plate vacuum capacitor has \(8.38 \mathrm{~J}\) of energy stored in it. The separation between the plates is \(2.30 \mathrm{~mm} .\) If the separation is decreased to \(1.15 \mathrm{~mm},\) what is the energy stored (a) if the capacitor is disconnected from the potential source so the charge on the plates remains constant, and (b) if the capacitor remains connected to the potential source so the potential difference between the plates remains constant?
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