/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 52 A capacitor is constructed of tw... [FREE SOLUTION] | 91Ó°ÊÓ

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A capacitor is constructed of two concentric conducting cylindrical shells. The radius of the inner cylindrical shell is \(2.35 \times 10^{-3} \mathrm{m}\) and that of the outer shell is \(2.50 \times 10^{-3} \mathrm{m} .\) When the cylinders carry equal and opposite charges of magnitude \(1.7 \times 10^{-10} \mathrm{C},\) the electric field between the plates has an average magnitude of \(4.2 \times 10^{4} \mathrm{V} / \mathrm{m}\) and is directed radially outward from the inner shell to the outer shell. Determine (a) the magnitude of the potential difference between the cylindrical shells and (b) the capacitance of this capacitor.

Short Answer

Expert verified
The potential difference is 6.3 V, and the capacitance is approximately 27 pF.

Step by step solution

01

Understanding the Setup

We have two cylindrical shells: an inner shell with radius \( r_1 = 2.35 \times 10^{-3} \text{ m} \) and an outer shell with radius \( r_2 = 2.50 \times 10^{-3} \text{ m} \). The charge magnitude on each shell is \( |Q| = 1.7 \times 10^{-10} \text{ C} \), and the electric field between them is given as \( E = 4.2 \times 10^4 \text{ V/m} \). We need to calculate the potential difference (part a) and the capacitance (part b).
02

Calculate the Potential Difference

The potential difference \( V \) between the shells can be found using the relation for a cylindrical capacitor: \( V = E \times (r_2 - r_1) \). Substitute \( E = 4.2 \times 10^4 \text{ V/m} \), \( r_2 = 2.50 \times 10^{-3} \text{ m} \), and \( r_1 = 2.35 \times 10^{-3} \text{ m} \) into the equation: \[V = 4.2 \times 10^4 \text{ V/m} \times (2.50 \times 10^{-3} \text{ m} - 2.35 \times 10^{-3} \text{ m})\]After performing the calculation, we find \( V = 6.3 \text{ V} \).
03

Calculate the Capacitance

Capacitance \( C \) is given by the formula \( C = \frac{Q}{V} \), where \( Q = 1.7 \times 10^{-10} \text{ C} \) and \( V = 6.3 \text{ V} \). Plug in these values:\[C = \frac{1.7 \times 10^{-10} \text{ C}}{6.3 \text{ V}}\]Solving this gives: \( C \approx 2.7 \times 10^{-11} \text{ F} \). In units more commonly used for capacitors: \( C \approx 27 \text{ pF} \).

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Key Concepts

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

Electric Field
The electric field is a vector field that represents the force exerted on a charge placed in the field. It is especially important in the context of capacitors, like our cylindrical shells, as it determines how charge moves between the conductors. In the case of a cylindrical capacitor, the electric field is directed radially outward from the inner shell to the outer shell. The strength or magnitude of the electric field is directly linked to the charges on the shells and the distance between them. Typically expressed in volts per meter (V/m), the electric field between the two cylindrical shells is given as 4.2 x 10^4 V/m in our exercise. This value represents the force per unit charge acting along the line between the cylinders, crucial for determining potential difference and capacitance.
Potential Difference
The potential difference between two points in an electric field is the work needed to move a unit charge from one point to the other. In our cylindrical capacitor setup, we are interested in the potential difference between the inner and outer shells. This can be calculated using the formula:
  • \( V = E \times (r_2 - r_1) \)
Here, \(E\) is the electric field between the shells, \(r_2\) is the radius of the outer shell, and \(r_1\) is the radius of the inner shell. By substituting our given values into the equation, we found the potential difference to be 6.3 volts. This potential difference is what creates an energy barrier for charges, meaning energy is required to move charges from one shell to the other.
Capacitance
Capacitance is a measure of a capacitor's ability to store charge per unit voltage. In our scenario with two concentric cylindrical shells, capacitance can be determined by using the formula:
  • \( C = \frac{Q}{V} \)
Where \(Q\) is the charge on each shell, and \(V\) is the potential difference we calculated earlier. Using this formula, the capacitance for our cylindrical capacitor was found to be approximately 27 picofarads (pF), or 2.7 x 10^-11 farads (F). Capacitors with high capacitance values can store more charge at a given potential difference than those with low capacitance, making them crucial for various electronic applications.
Concentric Cylindrical Shells
Concentric cylindrical shells refer to two or more cylinders that share a common center or axis, one placed inside the other. These geometric configurations are often used in capacitors where the inner shell can carry a charge opposite to that of the outer shell. This setup creates a uniform electric field between them, allowing for precise calculations of their electrical properties like potential difference and capacitance.
In the cylindrical capacitor discussed here, the inner cylinder has a radius of 2.35 x 10^-3 m, while the outer one has a slightly larger radius of 2.50 x 10^-3 m. By enclosing one cylinder within another, it efficiently utilizes space while maximizing the electric field strength and storage capabilities. Concentric cylindrical configurations are thus more effective for specific applications that require energy storage using minimal space.

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Most popular questions from this chapter

The electronic flash attachment for a camera contains a capacitor for storing the energy used to produce the flash. In one such unit, the potential difference between the plates of an \(850-\mu \mathrm{F}\) capacitor is \(280 \mathrm{V}\). (a) Determine the energy that is used to produce the flash in this unit. (b) Assuming that the flash lasts for \(3.9 \times 10^{-3}\) s, find the effective power or "wattage" of the flash.

Particle 1 has a mass of \(m_{1}=3.6 \times 10^{-6} \mathrm{kg},\) while particle 2 has a mass of \(m_{2}=6.2 \times 10^{-6} \mathrm{kg} .\) Each has the same electric charge. These particles are initially held at rest, and the two- particle system has an initial electric potential energy of 0.150 J. Suddenly, the particles are released and fly apart because of the repulsive electric force that acts (a) Two particles have different masses, but the same electrical charge \(q\) They are initially at rest. (b) At the instant following the release of the particles, they are flying apart due to the mutual force of electric repulsion. on each one (see the figure). The effects of the gravitational force are negligible, and no other forces act on the particles. Concepts: (i) What types of energy does the two-particle system have initially? (ii) What types of energy does the two-particle system have at the instant illustrated in part \(b\) of the drawing? (iii) Does the principle of conservation of energy apply to this problem? Explain. (iv) Does the conservation of linear momentum apply to the two particles as they fly apart? Explain. Calculations: At one instant following the release, the speed of particle 1 is measured to be \(v_{1}=170 \mathrm{m} / \mathrm{s} .\) What is the electric potential energy at this instant?

During a particular thunderstorm, the electric potential difference between a cloud and the ground is \(V_{\text {cloud }}-V_{\text {ground }}=1.3 \times 10^{8} \mathrm{V},\) with the cloud being at the higher potential. What is the change in an electron's electric potential energy when the electron moves from the ground to the cloud?

A particle is uncharged and is thrown vertically upward from ground level with a speed of \(25.0 \mathrm{m} / \mathrm{s}\). As a result, it attains a maximum height \(h\). The particle is then given a positive charge \(+q\) and reaches the same maximum height \(h\) when thrown vertically upward with a speed of \(30.0 \mathrm{m} / \mathrm{s}\). The electric potential at the height \(h\) exceeds the electric potential at ground level. Finally, the particle is given a negative charge \(-q .\) Ignoring air resistance, determine the speed with which the negatively charged particle must be thrown vertically upward, so that it attains exactly the maximum height h. In all three situations, be sure to include the effect of gravity.

An empty parallel plate capacitor is connected between the terminals of a \(9.0-\mathrm{V}\) battery and charged up. The capacitor is then disconnected from the battery, and the spacing between the capacitor plates is doubled. As a result of this change, what is the new voltage between the plates of the capacitor?

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