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A capacitor is made from two hollow, coaxial, iron cylinders, one inside the other. The inner cylinder is negatively charged and the outer is positively charged; the magnitude of the charge on each is 10.0 \(\mathrm{pC}\) . The inner cylinder has radius \(0.50 \mathrm{mm},\) the outer one has radius \(5.00 \mathrm{mm},\) and the length of each cylinder is 18.0 \(\mathrm{cm} .\) (a) What is the capacitance? (b) What applied potential difference is necessary to produce these charges on the cylinders?

Short Answer

Expert verified
(a) Capacitance is 4.376 pF. (b) Potential difference is 2.285 V.

Step by step solution

01

Understand the Parameters

Given parameters include the radius of the inner cylinder \( r_1 = 0.50 \) mm = 0.0005 m, the radius of the outer cylinder \( r_2 = 5.00 \) mm = 0.005 m, the length of both cylinders \( L = 18.0 \) cm = 0.18 m, and the charge magnitude on each cylinder \( Q = 10.0 \) pC = 10.0 × 10^{-12} C.
02

Formula for Capacitance of Coaxial Cylinders

The capacitance \( C \) of a pair of coaxial cylinders is calculated using the formula \[ C = \frac{2 \pi \varepsilon_0 L}{\ln(\frac{r_2}{r_1})} \], where \( \varepsilon_0 = 8.854 \times 10^{-12} \mathrm{F/m} \) is the permittivity of free space.
03

Calculate ln(r2/r1)

Compute the natural logarithm part of the formula: \( \ln(\frac{r_2}{r_1}) = \ln(\frac{0.005}{0.0005}) = \ln(10) \approx 2.302 \).
04

Calculate Capacitance C

Plug the values into the formula: \[ C = \frac{2 \pi \times 8.854 \times 10^{-12} \times 0.18}{2.302} \]. Compute to get \[ C \approx 4.376 \times 10^{-12} \text{F} \text{ (or 4.376 pF)} \].
05

Potential Difference Using Capacitance

Using the formula \( V = \frac{Q}{C} \), where \( Q = 10.0 \times 10^{-12} \) C, calculate the potential difference: \[ V = \frac{10.0 \times 10^{-12}}{4.376 \times 10^{-12}} \].
06

Calculate Potential Difference V

Compute \( V = \frac{10.0}{4.376} \approx 2.285 \text{ V} \).

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Key Concepts

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

Coaxial Cylinders
Coaxial cylinders are an essential concept when dealing with the structure of a specific type of capacitor. Imagine two hollow, cylindrical tubes placed one inside the other, much like a telescope extending outwards. The crucial distinction here is that these cylinders have different radii. The inner cylinder is completely enclosed by the outer one. Each of these cylinders in the setup carries an equal but opposite charge.
The inner cylinder is negatively charged, while the outer one carries a positive charge. This charge setup is typical in capacitors to store energy by maintaining a potential difference between the conductors. Such a cylindrical configuration is highly efficient for certain applications due to its ability to offer high capacitance with relatively large surface areas while occupying a small volume.
In practical uses, this design helps in ensuring minimal losses of energy as well as a uniform electric field between the coaxial cylinders. By manipulating the sizes of these cylinders and using materials with varying dielectric properties, coaxial cylinder capacitors can be tailored for specific functions in electrical circuits.
Capacitance Calculation
The calculation of capacitance for a pair of coaxial cylinders involves understanding how this property is determined by their geometric setup and material properties. Capacitance, symbolized as\( C \), measures a system's ability to store charge per unit voltage.
For coaxial cylinders, the formula to determine capacitance is:\[ C = \frac{2 \pi \varepsilon_0 L}{\ln(\frac{r_2}{r_1})} \]Here, \( \varepsilon_0 \) is the permittivity of free space, \( L \) is the length of the cylinders, and \( r_1 \) and \( r_2 \) are the radii of the inner and outer cylinders, respectively. This formula helps in calculating \( C \) by accounting for both the physical dimensions of the cylinders and the material between them.
One of the critical parts of this formula is the natural logarithm of the ratio of the outer to the inner radius, \( \ln(\frac{r_2}{r_1}) \). This term shows how the spatial distance between the two cylindrical surfaces influences the system's capacitance. Once the values of the dimensions and permittivity are known, the capacitance can be easily calculated, providing valuable insights into how effectively a capacitor can store charge.
Potential Difference
The potential difference in a capacitor context refers to the voltage across the two conductors, or plates, of the capacitor. It is a vital part of understanding how capacitors function, as it is directly related to the energy storage capability of the capacitor.
For coaxial cylinders, to find the potential difference \( V \) between the inner and outer cylinders, you can use the formula:\[ V = \frac{Q}{C} \]Where \( Q \) is the charge on the cylinders and \( C \) is the capacitance calculated earlier. This equation derives from the basic relationship in a capacitor, where the potential difference is simply the charge divided by the capacitance.
A higher potential difference between the inner and outer cylinder indicates a stronger electric field and more energy stored within the capacitor. Understanding this concept is crucial, as it can help in designing circuits where specific voltage ratings must be adhered to in order to ensure desired performance and safety standards. By controlling the potential difference, you can adjust how the capacitor interacts within an electrical network.

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Most popular questions from this chapter

A \(5.00-\) pF, parallel-plate, air-filled capacitor with circular plates is to be used in a circuit in which it will be subjected to potentials of up to \(1.00 \times 10^{2} \mathrm{V}\) . The electric field between the plates is to be no greater than \(1.00 \times 10^{4} \mathrm{N} / \mathrm{C}\) . As a budding electrical engineer for Live-Wire Electronics, your tasks are to (a) design the capacitor by finding what its physical dimensions and separation must be; ( b) find the maximum charge these plates can hold.

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 capacitor is made from two hollow, coaxial copper cylinders, one inside the other. There is air in the space between the cylinders. The inner cylinder has net positive charge and the outer cylinder has net negative charge. The inner cylinder has radius \(2.50 \mathrm{mm},\) the outer cylinder has radius \(3.10 \mathrm{mm},\) and the length of each cylinder is 36.0 \(\mathrm{cm} .\) If the potential difference between the surfaces of the two cylinders is \(80.0 \mathrm{V},\) what is the magnitude of the electric field at a point between the two cylinders that is a distance of 2.80 \(\mathrm{mm}\) from their common axis and midway between the ends of the cylinders?

Capacitance of an Oscilloscope. Oscilloscopes have parallel metal plates inside them to deflect the electron beam. These plates are called the deffecting plates. Typically, they are squares 3.0 \(\mathrm{cm}\) on a side and separated by \(5.0 \mathrm{mm},\) with vacuum in between. What is the capacitance of these deflecting plates and hence of the oscilloscope? (Note: This capacitance can sometimes have an effect on the circuit you are trying to study and must be taken into consideration in your calculations.)

A 12.5 -\muF capacitor is connected to a power supply that keeps a constant potential difference of 24.0 \(\mathrm{V}\) across the plates. A piece of material having a dielectric constant of 3.75 is placed between the plates, completely filling the space between them. (a) How much energy is stored in the capacitor before and after the dielectric is inserted? (b) By how much did the energy change during the insertion? Did it increase or decrease?

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