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Chapter 24: Problem 36

A parallel-plate capacitor has capacitance \(C_{0}=5.00 \mathrm{pF}\) when there is air between the plates. The separation between the plates is 1.50 \(\mathrm{mm}\) (a) What is the maximum magnitude of charge \(Q\) that can be placed on each plate if the electric field in the region between the plates is not to exceed \(3.00 \times 10^{4} \mathrm{V} / \mathrm{m}\) (b) A dielectric with \(K=2.70\) is inserted between the plates of the capacitor, completely filling the volume between the plates. Now what is the maximum magnitude of charge on each plate if the electric field between the plates is not to exceed \(3.00 \times 10^{4} \mathrm{V} / \mathrm{m}\) ?

Short Answer

Expert verified
(a) 2.25 nC; (b) 6.075 nC

Step by step solution

01

Understand the Given Information

We have a parallel-plate capacitor with a capacitance of \( C_0 = 5.00 \mathrm{pF} \) and a plate separation of \( d = 1.50 \mathrm{mm} \). The maximum electric field allowed is \( E = 3.00 \times 10^4 \mathrm{V}/\mathrm{m} \). A dielectric with a constant \( K = 2.70 \) is also considered.
02

Formulate the Relation between Electric Field and Potential Difference

The electric field \( E \) between the plates of a capacitor is related to the potential difference \( V \) and the plate separation \( d \) by the equation \( E = \frac{V}{d} \). Solve for \( V \): \( V = E \times d \).
03

Calculate the Maximum Potential Difference without a Dielectric

Substitute the given values: \( V = (3.00 \times 10^4) \times (1.5 \times 10^{-3}) \). Calculating gives \( V = 45 \mathrm{V} \).
04

Calculate Maximum Charge without Dielectric

For a given potential difference, the charge \( Q \) on a capacitor is \( Q = C_0 \times V \). Substitute \( C_0 = 5.00 \mathrm{pF} = 5.00 \times 10^{-12} \mathrm{F} \) and \( V = 45 \mathrm{V} \) to get \( Q = 5.00 \times 10^{-12} \times 45 = 2.25 \times 10^{-10} \mathrm{C} \).
05

Calculate Modified Capacitance with Dielectric

The capacitance with the dielectric \( C = K \times C_0 \). Substitute \( K = 2.70 \) and \( C_0 = 5.00 \mathrm{pF} \) to get \( C = 2.70 \times 5.00 \mathrm{pF} = 13.50 \mathrm{pF} \).
06

Calculate Maximum Charge with Dielectric

Repeat the charge calculation with the dielectric: \( Q = C \times V \). Substitute \( C = 13.50 \times 10^{-12} \mathrm{F} \) and the same \( V = 45 \mathrm{V} \) to get \( Q = 13.50 \times 10^{-12} \times 45 = 6.075 \times 10^{-10} \mathrm{C} \).

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Key Concepts

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

Dielectric Constant
Let's delve into the dielectric constant, often represented as \( K \). It is a dimensionless number that describes how much electric charge the material can store compared to a vacuum. When a dielectric material is inserted between the plates of a capacitor, it affects the capacitor's ability to hold charge. This is because the dielectric reduces the electric field within the capacitor.
  • This reduction happens because the dielectric becomes polarized in the presence of an electric field, effectively opposing the field.
  • As a result, the capacitor can store more charge for the same applied voltage, which increases the capacitance by a factor of \( K \).
  • The higher the dielectric constant, the better the material is at increasing the capacitance of the capacitor.
Understanding how the dielectric constant works is crucial in designing capacitors for different applications, where adjusting capacitance without changing physical dimensions is desirable.
Electric Field
An electric field in a capacitor is created by the voltage difference between its plates. It is an important factor because it determines how much voltage the capacitor can handle.
  • The electric field \( E \) is defined as the voltage \( V \) across the plates divided by the distance \( d \) between them: \( E = \frac{V}{d} \).
  • This relationship means that for a given plate separation, increasing the electric field requires a higher potential difference.
  • The maximum electric field is limited by the dielectric strength of the material between the capacitor plates.
A strong electric field can cause the dielectric to become conductive, leading to a breakdown. That's why it's crucial to keep the electric field within safe limits to ensure that the capacitor functions correctly. By understanding electric fields, one can design and utilize capacitors efficiently.
Capacitance
Capacitance, denoted by \( C \), is the ability of a capacitor to store an electric charge. It's a fundamental property that defines the relationship between the stored charge \( Q \) and the voltage \( V \) across the plates: \( Q = C \times V \).
  • Capacitance depends on the surface area of the plates, the distance between them, and the type of material (dielectric) between the plates.
  • If a dielectric is present, the capacitance increases by a factor called the dielectric constant \( K \).
  • This increase allows the capacitor to store more charge for the same voltage.
In practical applications, capacitance is a critical parameter. It influences how much energy a capacitor can store and how it discharges. For example, in electronic circuits, capacitors with specific capacitance values are chosen to achieve desired timing and filtering effects.

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Most popular questions from this chapter

B10 Potential in Human Cells. Some cell walls in the human body have a layer of negative charge on the inside surface and a layer of positive charge of equal magnitude on the outside surface. Suppose that the charge density on either surface is \(\pm 0.50 \times 10^{-3} \mathrm{C} / \mathrm{m}^{2},\) the cell wall is 5.0 \(\mathrm{nm}\) thick, and the cell-wall material is air. (a) Find the magnitude of \(\vec{\boldsymbol{E}}\) in the wall-between the two layers of charge. (b) Find the potential difference between the inside and the outside of the cell. Which is at the higher potential? (c) Atypical cell in the human body has a volume of\(10^{-16} \mathrm{m}^{3} .\) Estimate the total electric-field energy stored in the wall of a cell of this size. (Hint: Assume that the cell is spherical, and calculate the volume of the cell wall.) (d) In reality, the cell wall is made up, not of air, but of tissue with a dielectric constant of \(5.4 .\) Repeat parts (a) and (b) in this case.

A cylindrical capacitor consists of a solid inner conducting core with radius \(0.250 \mathrm{cm},\) surrounded by an outer hollow conducting tube. The two conductors are separated by air, and the length of the cylinder is 12.0 \(\mathrm{cm}\) . The capacitance is 36.7 pF. (a) Calculate the inner radius of the hollow tube. (b) When the capacitor is charged to \(125 \mathrm{V},\) what is the charge per unit length \(\lambda\) on the capacitor?

A \(10.0-\mu \mathrm{F}\) parallel-plate capacitor with circular plates is connected to a 12.0 -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?

Electronic flash units for cameras contain a capacitor for storing the energy used to produce the flash. In one such unit, the flash lasts for \(\frac{1}{677}\) s with an average light power output of \(2.70 \times 10^{5} \mathrm{W}\) (a) If the conversion of electrical energy to light is 95\(\%\) efficient (the rest of the energy goes to thermal energy), how much energy must be stored in the capacitor for one flash? (b) The capacitor has a potential difference between its plates of 125 \(\mathrm{V}\) when the stored energy equals the value calculated in part (a). What is the capacitance?

Capacitance of the Earth. Consider a spherical capacitor with one conductor being a solid conducting sphere of radius \(R\) and the other conductor being at infinity. (a) Use Eq. (24.1) and what you know about the potential at the surface of a conducting sphere with charge \(Q\) to derive an expression for the capacitance of the charged sphere. (b) Use your result in part (a) to calculate the capacitance of the earth. The earth is a good conductor and has a radius of 6380 \(\mathrm{km} .\) Compare your results to the capacitance of typical capacitors used in electronic circuits, which ranges from 10 \(\mathrm{pF}\) to 100 \(\mathrm{pF} .\)
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