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Chapter 21: Problem 47

A 25 pF parallel-plate capacitor with an air gap between the plates is connected to a \(100 \mathrm{V}\) battery. A Teflon slab is then inserted between the plates and completely fills the gap. What is the change in the charge on the positive plate when the Teflon is inserted?

Short Answer

Expert verified
The change in the charge on the positive plate when the Teflon is inserted is given by \(\Delta Q = Q-Q_0\).

Step by step solution

01

Calculate Initial Charge

First, calculate the initial charge present on the capacitor using the formula \(Q=C_0V\), where \(C_0=25 \, pF\) is the initial capacitance and \(V=100 \, V\) is the voltage. Thus, the initial charge \(Q_0 = C_0V\).
02

Calculate New Capacitance

Next, calculate the new capacitance after the Teflon is inserted. The dielectric constant \(k\) of Teflon is 2.1. Use the formula \(C=kC_0\) to find the new capacitance. Thus, the new capacitance \(C = kC_0\).
03

Calculate New Charge

Now, calculate the new charge on the capacitor using the formula \(Q=CV\), where \(C\) is the new capacitance and \(V=100 \, V\) is the voltage. Thus, the new charge \(Q = CV\).
04

Find the Difference in Charges

Finally, find the change in the charge on the positive plate after the Teflon is inserted. This can be done by subtracting the initial charge from the new charge, i.e., \(\Delta Q = Q-Q_0\).

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Key Concepts

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

Dielectric Constant
The dielectric constant is a fascinating concept in the field of electronics. It is a measure of a material's ability to store electrical energy in an electric field.
The dielectric constant is also known as the relative permittivity of a material.
  • In the context of capacitors, this constant plays a crucial role as it determines how much the capacitance can be increased when a dielectric material is introduced between the plates.
  • For instance, when Teflon is inserted between the plates of a capacitor, it increases the capacitance by a factor of its dielectric constant, which is 2.1 in this case.
This means the Teflon allows the capacitor to store 2.1 times more charge at the same voltage, illustrating the significance of the dielectric constant in enhancing capacitor performance.
Charge on Capacitor
The charge on a capacitor is the amount of electrical charge stored on its plates. It is directly influenced by both the capacitance of the capacitor and the voltage across it.
  • Initially, the charge on a capacitor is calculated using the formula: \( Q = C_0V \)where \( Q \) is the charge, \( C_0 \) is the initial capacitance, and \( V \) is the voltage.
  • When a dielectric is introduced, the capacitance increases, which directly affects the charge stored under the same voltage conditions.
Understanding the relationship between charge, capacitance, and voltage is critical for mastering capacitor-related problems. As the capacitance goes up with a dielectric, the potential to hold more charge increases as well.
Parallel-Plate Capacitor
A parallel-plate capacitor is a common type of capacitor used in electronic circuits. It consists of two parallel conductive plates separated by an insulating material, known as a dielectric.
  • The plates hold equal and opposite charges when connected to a voltage source, creating an electric field in the dielectric region.
  • The capacitance of a parallel-plate capacitor is given by\( C = \varepsilon_r \varepsilon_0 \frac{A}{d} \)where \( \varepsilon_r \) is the relative permittivity (dielectric constant), \( \varepsilon_0 \) is the vacuum permittivity, \( A \) is the plate area, and \( d \) is the distance between plates.
The arrangement and the properties of the dielectric material influence the capacitor's ability to store charge, as demonstrated by the increased capacitance when a material like Teflon is inserted between its plates.
Electrical Potential Difference
Electrical Potential Difference, commonly known as voltage, is the measure of the work needed to move a unit charge from one point to another in an electric field.
  • In capacitors, this potential difference determines how much electrical energy can be stored for a given amount of charge.
  • While the introduction of a dielectric doesn't change the voltage across the capacitor in a given circuit, it does change the capacitance and, consequently, the potential amount of charge stored.
Thus, when solving problems involving capacitors, understanding how potential difference relates to changes in charge and capacitance due to dielectric materials is essential for predicting and calculating capacitor behavior effectively.

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Most popular questions from this chapter

The highest magnetic fields in the world are generated when large arrays, or "banks," of capacitors are discharged through the copper coils of an electromagnet. At the National High Magnetic Field Laboratory, the total capacitance of the capacitor bank is 32 mF. These capacitors can be charged to \(16 \mathrm{kV}\). a. What is the energy stored in the capacitor bank when it is fully charged? b. When discharged, the entire energy from this bank flows through the magnet coil in 10 ms. What is the average power delivered to the coils during this time?

We've seen that bees develop a positive charge as they fly through the air. When a bee lands on a flower, charge is transferred, and an opposite charge is induced in the earth below the flower. The flower and the ground together make a capacitor; a typical value is 0.80 pF. If a flower is charged to \(30 \mathrm{V}\) relative to the ground, a bee can reliably detect the added charge and then avoids the flower in favor of flowers that have not been recently visited. Approximately how much charge must a bee transfer to the flower to create a \(30 \mathrm{V}\) potential difference?

What potential difference is needed to accelerate a \(\mathrm{He}^{+}\) ion (charge \(+e,\) mass 4 u) from rest to a speed of \(1.0 \times 10^{6} \mathrm{m} / \mathrm{s} ?\)

In 1 second, a battery charger moves 0.60 C of charge from the negative terminal to the positive terminal of a 1.5 V AA battery. a. How much work does the charger do? b. What is the power output of the charger in watts?

At one point in space, the electric potential energy of a \(15 \mathrm{nC}\) charge is \(45 \mu \mathrm{J}\) a. What is the electric potential at this point? b. If a \(25 \mathrm{nC}\) charge were placed at this point, what would its electric potential energy be?
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