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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?

Short Answer

Expert verified
(a) The inner radius of the hollow tube is approximately 0.455 cm. (b) The charge per unit length \( \lambda \) is 38.4 nC/m.

Step by step solution

01

Understanding the Formula

The capacitance per unit length of a cylindrical capacitor is given by the formula: \( C' = \frac{2 \pi \varepsilon_0}{\ln(b/a)} \), where \( a \) is the radius of the inner core, \( b \) is the inner radius of the hollow tube, and \( \varepsilon_0 \) is the permittivity of free space \( (\varepsilon_0 = 8.85 \times 10^{-12} \ \mathrm{F/m}) \). We need to rearrange this formula to find the value of \( b \).
02

Using Given Data to Find \( b \)

Using the total capacitance \( C = 36.7 \ \text{pF} = 36.7 \times 10^{-12} \ \mathrm{F} \) and knowing the total length \( L = 12.0 \ \mathrm{cm} = 0.12 \ \mathrm{m} \), we first find \( C' = \frac{C}{L} = \frac{36.7 \times 10^{-12}}{0.12} \). Next, substitute \( C' \) into the formula \( C' = \frac{2 \pi \varepsilon_0}{\ln(b/a)} \) and solve for \( \ln(b/a) \).
03

Calculate \( b \) From \( \ln(b/a) \)

With \( a = 0.250 \ \mathrm{cm} = 0.0025 \ \mathrm{m} \), substitute into \( \ln(b/a) \) to solve for \( b \). Rearrange the equation to get \( b = a \cdot e^{\ln(b/a)} \).
04

Verify the Calculation

Ensure the calculation of \( b \) is consistent with the physical design, i.e., \( b > a \). Calculate \( b \) to verify it makes sense in the context of this cylindrical capacitor.
05

Finding Charge per Unit Length \( \lambda \)

With the capacitance \( C \) known, given the potential \( V = 125 \ \mathrm{V} \), the charge \( Q = C \times V \), and the charge per unit length \( \lambda = \frac{Q}{L} \). Calculate \( \lambda \).

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Key Concepts

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

Capacitance Calculation
A cylindrical capacitor is a great way to visualize capacitance and how it varies with geometry. When calculating the capacitance for a cylindrical capacitor, the formula is quite specific. The capacitance per unit length \( C' \) is determined using the formula:
  • \( C' = \frac{2 \pi \varepsilon_0}{\ln(b/a)} \)
Here, \( a \) represents the radius of the inner conductive core, while \( b \) represents the inner radius of the outer hollow tube. The symbol \( \varepsilon_0 \) stands for the permittivity of free space, a fundamental constant in physics.

To find the capacitance for the entire length of the cylinder, multiply \( C' \) by the cylinder's length \( L \). In the given problem, the specific capacitance \( C \) for the entire capacitor is given, and using the above formula, you can rearrange to solve for other unknowns. This mathematical relationship is crucial for understanding how capacitance can change based on physical dimensions in cylindrical capacitors.
Inner Radius of Hollow Tube
The inner radius of the hollow tube, denoted as \( b \), is a critical dimension in the design of a cylindrical capacitor. Knowing the inner radius allows us to understand what space the capacitor has between its inner core and outer shell. To find \( b \), we must rearrange the capacitance formula for a cylindrical capacitor:
  • Start with \( C' = \frac{2 \pi \varepsilon_0}{\ln(b/a)} \).
  • Substitute \( C' = \frac{C}{L} \), where \( C \) is the total capacitance and \( L \) is the total length of the capacitor.
  • By solving for \( \ln(b/a) \), you can find \( b \).
Plugging in the known values, including that of the inner core radius \( a \), simplifies to solving for \( b \) using logarithmic and exponential functions. Always ensure your final value of \( b \) is greater than \( a \), as this physical constraint needs validation in practical designs.
Charge Per Unit Length
The concept of charge per unit length, symbolized as \( \lambda \), is meaningful because it relates the total charge stored on the capacitor to its length. For a cylindrical capacitor, calculate charge per unit length using:
  • Firstly, find the total charge \( Q \) using \( Q = C \times V \), where \( V \) is the voltage applied across the capacitor.
  • Once you have \( Q \), the charge per unit length \( \lambda \) is \( \lambda = \frac{Q}{L} \).
This calculation tells us how the charge is distributed along the capacitor, helping in understanding the field strength and potential distribution along the length of the cylinder. In practice, knowing \( \lambda \) can aid in the design and operation of electronic circuits involving cylindrical capacitors.
Permittivity of Free Space
The permittivity of free space, denoted \( \varepsilon_0 \), is a fundamental constant in electromagnetism. Its value is approximately \( 8.85 \times 10^{-12} \ \mathrm{F/m} \).

This constant appears in the formula for capacitance and expresses how an electric field affects, and is affected by, a medium.
  • In a vacuum, \( \varepsilon_0 \) describes the ability of electric field lines to travel through space.
  • It impacts how much energy can be stored as electric field configuration.
Within cylindrical capacitors, \( \varepsilon_0 \) plays a part in determining their capacitance by directly affecting the equation used to calculate \( C' \). Understanding \( \varepsilon_0 \) helps engineers predict how capacitors will perform in different configurations and under various electric potentials.

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Most popular questions from this chapter

When a 360 -nF air capacitor \(\left(1 \mathrm{nF}=10^{-9} \mathrm{F}\right)\) is connected to a power supply, the energy stored in the capacitor is \(1.85 \times 10^{-5} \mathrm{J}\) . While the capacitor is kept connected to the power supply, a slab of dielectric is inserted that completely fills the space between the plates. This increases the stored energy by \(2.32 \times 10^{-5} \mathrm{J}\) . (a) What is the potential difference between the capacitor plates? (b) What is the dielectric constant of the slab?

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.

Capacitance of a Thundercloud. The charge center of a thundercloud, drifting 3.0 \(\mathrm{km}\) above the earth's surface, contains 20 \(\mathrm{C}\) of negative charge. Assuming the charge center has a radius of \(1.0 \mathrm{km},\) and modeling the charge center and the earth's surface as parallel plates, calculate: (a) the capacitance of the system; (b) the potential difference between charge center and ground; (c) the average strength of the electric field between cloud and ground; (d) the electrical energy stored in the system.

The dielectric to be used in a parallel-plate capacitor has a dielectric constant of 3.60 and a dielectric strength of \(1.60 \times\) \(10^{7} \mathrm{V} / \mathrm{m}\) . The capacitor is to have a capacitance of \(1.25 \times 10^{-9} \mathrm{F}\) and must be able to withstand a maximum potential difference of 5500 \(\mathrm{V} .\) What is the minimum area the plates of the capacitor may have?

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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