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

Short Answer

Expert verified
The inner radius of the hollow tube is approximately 0.285 cm, and the charge per unit length \( \lambda \) is approximately 3.82 nC/m.

Step by step solution

01

Understanding the Capacitance Formula for a Cylindrical Capacitor

The capacitance formula for a cylindrical capacitor is given by \( C = \frac{2\pi\varepsilon_0 L}{\ln\left(\frac{b}{a}\right)} \), where \( C \) is the capacitance, \( a \) is the radius of the inner conductor, \( b \) is the radius of the outer conductor, and \( L \) is the length of the cylinder. We need to solve this equation for \( b \).
02

Substitute Known Values and Solve for b

We know \( C = 36.7 \text{ pF} = 36.7 \times 10^{-12} \text{ F} \), \( a = 0.250 \text{ cm} = 0.0025 \text{ m} \), \( \varepsilon_0 = 8.85 \times 10^{-12} \text{ F/m} \), and \( L = 0.12 \text{ m} \). Substitute these into the formula: \( 36.7 \times 10^{-12} = \frac{2\pi(8.85 \times 10^{-12})(0.12)}{\ln\left(\frac{b}{0.0025}\right)} \). Solve this equation for \( b \).
03

Calculate the Natural Logarithm and Solve for b

Re-arrange the equation:\( \ln\left(\frac{b}{0.0025}\right) = \frac{2\pi(8.85 \times 10^{-12})(0.12)}{36.7 \times 10^{-12}} \). Calculate the right-hand side to find the value of the natural logarithm. Solve for \( b \) by exponentiating both sides and multiplying by \( a = 0.0025 \).
04

Calculate the Charge per Unit Length \( \lambda \)

The charge per unit length on the capacitor is given by \( \lambda = C \times V / L \), where \( V = 125 \text{ V} \) is the voltage. Substitute the previously found capacitance and values into the formula to calculate \( \lambda \).
05

Substitute Values into \( \lambda \) Formula

\( \lambda = \frac{36.7 \times 10^{-12} \times 125}{0.12} \). Calculate this expression to find the charge per unit length \( \lambda \).

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Key Concepts

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

Capacitance Formula
When dealing with a cylindrical capacitor, understanding the capacitance formula is crucial. This formula determines how much charge a capacitor can store for a given electric potential. For a cylindrical capacitor, the formula is: \[ C = \frac{2\pi\varepsilon_0 L}{\ln\left(\frac{b}{a}\right)} \]Here,
  • \( C \) is the capacitance in farads (F).
  • \( \varepsilon_0 \) is the permittivity of free space, which is approximately \( 8.85 \times 10^{-12} \text{ F/m} \).
  • \( L \) is the length of the capacitor in meters.
  • \( a \) is the radius of the inner conductor.
  • \( b \) is the radius of the outer conductor.
This formula accounts for how the geometry of the capacitor affects its ability to store charge. In this case, the natural logarithm function, \( \ln\left(\frac{b}{a}\right) \), tells us how the radii of the cylinders are related to the capacitance. By manipulating this equation, you can solve for any unknown variable if the other values are provided.
Charge per Unit Length
The concept of charge per unit length, represented by \( \lambda \), is important when analyzing cylindrical capacitors. This quantity describes how charge is distributed along the length of the capacitor.The formula for calculating the charge per unit length is: \[ \lambda = \frac{C \times V}{L} \]Where,
  • \( \lambda \) is the charge per unit length in coulombs per meter (C/m).
  • \( C \) is the capacitance in farads.
  • \( V \) is the voltage applied across the capacitor in volts.
  • \( L \) is the length of the cylinder in meters.
This formula shows that the charge per unit length is directly proportional to both the capacitance and the voltage, and inversely proportional to the length of the cylinder. When voltage is applied to the capacitor, charge begins to distribute uniformly along its length. Knowing \( \lambda \) helps in analyzing the electric field distribution around the capacitor.
Electric Field
The electric field in the region between the two conductors of a cylindrical capacitor is an essential aspect of its operation. This field results from the separation of charges on the conductor surfaces and plays a key role in how capacitors store and release electrical energy.The electric field \( E \) at a distance \( r \) from the axis of a cylindrical capacitor with an inner radius \( a \) and an outer radius \( b \) is given by:\[ E(r) = \frac{\lambda}{2\pi\varepsilon_0 r} \]Here,
  • \( E(r) \) is the magnitude of the electric field at distance \( r \).
  • \( \lambda \) is the charge per unit length.
  • \( r \) is the radial distance from the axis of the cylinder.
  • \( \varepsilon_0 \) is the permittivity of free space.
This formula shows that the electric field decreases with an increase in the distance \( r \) from the axis. The electric field inside the capacitor essentially creates forces on the charges, compelling them towards the opposite plates, which is the property that makes capacitors useful for storing energy. Understanding the behavior of the electric field within the capacitor is key to analyzing its charging and discharging processes.

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Most popular questions from this chapter

When a 360-nF air capacitor (11 nF = 10\(^{-9}\) F) is connected to a power supply, the energy stored in the capacitor is 1.85 \(\times\) 10\(^{-5}\) 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\(^{-5}\) J. (a) What is the potential difference between the capacitor plates? (b) What is the dielectric constant of the slab?

The plates of a parallel-plate capacitor are 3.28 mm apart, and each has an area of 9.82 cm\(^2\). Each plate carries a charge of magnitude 4.35 \(\times\) 10\(^{-8}\) C. The plates are in vacuum. What is (a) the capacitance; (b) the potential difference between the plates; (c) the magnitude of the electric field between the plates?

A 0.350-m-long cylindrical capacitor consists of a solid conducting core with a radius of 1.20 mm and an outer hollow conducting tube with an inner radius of 2.00 mm. The two conductors are separated by air and charged to a potential difference of 6.00 V. Calculate (a) the charge per length for the capacitor; (b) the total charge on the capacitor; (c) the capacitance; (d) the energy stored in the capacitor when fully charged.

A parallel-plate air capacitor has a capacitance of 920 pF. The charge on each plate is 3.90 \(\mu\)C. (a) What is the potential difference between the plates? (b) If the charge is kept constant, what will be the potential difference if the plate separation is doubled? (c) How much work is required to double the separation?

A 5.80-\(\mu\)F, parallel-plate, air capacitor has a plate separation of 5.00 mm and is charged to a potential difference of 400 V. Calculate the energy density in the region between the plates, in units of J/m\(^3\).

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