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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 \(36.7 \mathrm{pF}\). (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
The inner radius of the hollow tube is calculated from the given dimensions and capacitance value. The charge per unit length is found from the relationship between charge, voltage, and capacitance using the given voltage and the calculated charge.

Step by step solution

01

Identifying Given Values

Firstly, identify the values provided: The radius of the inner conducting core \(r_1 = 0.250 cm\), the length \(l = 12.0 cm\), and the capacitance \(C = 36.7 pF\). If needed, convert these values to base units (meters, Farads), remembering that \(1pF = 1x10^{-12} F\).
02

Formula for Capacitance in Cylindrical Capacitors

Use the formula for capacitance in cylindrical capacitors: \(C = 2\pi \epsilon_0 \frac{l}{ln(r_2/r_1)}\), where \(\epsilon_0 = 8.85x10^{-12} F/m\). This formula can be rearranged to solve for the unknown \(r_2\), the radius of the outer conductor: \(r_2 = r_1 e^{\frac{2\pi \epsilon_0 l}{C}}\).
03

Calculating the Inner Radius of the Tube

Substitute the identified values into the rearranged formula and calculate \(r_2\).
04

Using Capacitance to Determine Charge

From physics, the formula \(Q = CV\) can be used to calculate the charge \(Q\) of the capacitor. Given that the voltage \(V = 125 V\), substitute these values into the formula to determine the charge of the capacitor.
05

Calculate Charge Per Unit Length

To determine the charge per unit length \(\lambda\), it is necessary to divide the total charge \(Q\) by the length of the cylinder: \(\lambda = Q / l\). Substitute the values and calculate \(\lambda\).

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Key Concepts

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

Capacitance Formula
Understanding the capacitance formula is critical when studying capacitors, and a cylindrical capacitor offers an excellent example of how geometry can affect capacitance. Capacitance, denoted by the symbol \( C \), is a measure of a capacitor's ability to store electric charge. In the context of a cylindrical capacitor, which is comprised of an inner conductor and an outer conducting shell, the capacitance can be described using the formula:

\[ C = 2\pi \epsilon_0 \frac{l}{\ln(r_2/r_1)} \]

Here, \( \epsilon_0 \) is the permittivity of free space, which is a constant \(8.85 \times 10^{-12} F/m\). The symbols \( r_1 \) and \( r_2 \) represent the radii of the inner and outer conductors, respectively, while \( l \) is the length of the cylindrical capacitors. This formula is derived from the integral of the electric field between the two conductors over the volume of the capacitor. It's also worth noting that when the outer conductor's radius \( r_2 \) is much larger than \( r_1 \) or the length \( l \) of the capacitor is very large, the end effects become negligible, which makes the formula a good approximation. When solving problems involving cylindrical capacitors, the values need to be inserted in standard units of Farads for capacitance and meters for length and radius.
Electric Charge
A fundamental concept linked to capacitance is electric charge. The electric charge, typically represented by \( Q \), is the amount of electricity held by an object. For capacitors, the relationship between voltage \( V \), capacitance \( C \) and the stored charge \( Q \) is given by the formula:

\[ Q = CV \]

This simple equation is frequently used to determine how much charge a capacitor can store when a certain voltage is applied across its terminals. In the context of the cylindrical capacitor from our exercise, once the capacitance and the voltage applied are known, you can calculate the total charge stored in the capacitor. It's important to distinguish between the total charge and the charge distribution, however. In the case of a cylindrical capacitor, this charge distribution is not uniform due to the cylindrical geometry, but \( Q \) gives us the aggregate charge on each conductor (not considering sign).
Charge Per Unit Length
When dealing with cylindrical capacitors, the 'charge per unit length' (\lambda) is a useful concept that describes how charge is spread out along the length of the cylindrical conductors. It is calculated by taking the total charge \( Q \) and dividing it by the length \( l \) of the capacitor:

\[ \lambda = \frac{Q}{l} \]

This concept is particularly important in the context of long, coaxial cables or cylindrical capacitors which are commonly used in electrical and electronics applications. Understanding \( \lambda \) helps in grasping how charge is distributed over a component's geometry and is critical in predicting the electric field within and outside the cylindrical structure. In practical scenarios, knowing \( \lambda \) helps in characterizing the capacitance properties of transmission lines and, by extension, their effects on signal transmission and energy loss.

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Most popular questions from this chapter

Two identical air-filled parallel-plate capacitors \(C_{1}\) and \(C_{2}\) are connected in series to a battery that has voltage \(V .\) The charge on each capacitor is \(Q_{0}\). While the two capacitors remain connected to the battery, a dielectric with dielectric constant \(K>1\) is inserted between the plates of capacitor \(C_{1},\) completely filling the space between them. In terms of \(K\) and \(Q_{0},\) what is the charge on capacitor \(C_{1}\) after the dielectric is inserted? Does the charge on \(C_{1}\) increase, decrease, or stay the same?

A parallel-plate air capacitor has a capacitance of \(920 \mathrm{pF}\). The charge on each plate is \(3.90 \mu \mathrm{C}\). (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?

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 \(\frac{1}{675} \mathrm{~s}\) 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?

DATA Your electronics company has several identical capacitors with capacitance \(C_{1}\) and several others with capacitance \(C_{2}\). You must determine the values of \(C_{1}\) and \(C_{2}\) but don't have access to \(C_{1}\) and \(C_{2}\) individually. Instead, you have a network with \(C_{1}\) and \(C_{2}\) connected in series and a network with \(C_{1}\) and \(C_{2}\) connected in parallel. You have a \(200.0 \mathrm{~V}\) battery and instrumentation that measures the total \(\mathrm{en}\) ergy supplied by the battery when it is connected to the network. When the parallel combination is connected to the battery, \(0.180 \mathrm{~J}\) of energyis stored in the network. When the series combination is connected. \(0.0400 \mathrm{~J}\) of energy is stored. You are told that \(C_{1}\) is greater than \(C_{2}\) (a) Calculate \(C_{1}\) and \(C_{2}\). (b) For the series combination, does \(C_{1}\) or \(C_{2}\) store more charge, or are the values equal? Does \(C_{1}\) or \(C_{2}\) store more energy, or are the values equal? (c) Repeat part (b) for the parallel combination.

Two parallel plates have equal and opposite charges. When the space between the plates is evacuated, the electric field is \(E=3.20 \times 10^{5} \mathrm{~V} / \mathrm{m} .\) When the space is filled with dielectric, the electric field is \(E=2.50 \times 10^{5} \mathrm{~V} / \mathrm{m} .\) (a) What is the charge density on each surface of the dielectric? (b) What is the dielectric constant?

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