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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.

Short Answer

Expert verified
(a) Charge per length λ2.68×1011C/m. (b) Total charge Q9.38×1012C. (c) Capacitance C1.56×1012F. (d) Energy stored U2.81×1011J.

Step by step solution

01

Identify the formula for voltage difference (V) in a cylindrical capacitor

In a cylindrical capacitor, the electric potential difference V between the core and outer tube is given by the formula V=λ2πε0ln(r2r1) where λ is the charge per unit length, ε0 is the permittivity of free space (8.85×1012C2/Nm2), r1 is the radius of the inner core, and r2 is the radius of the outer hollow conductor.
02

Calculate charge per length (λ)

Rearrange the formula from Step 1 to solve for charge per unit length λ: λ=2πε0Vln(r2/r1) Substitute V=6.00V, r1=1.20×103m, r2=2.00×103m and ε0=8.85×1012C2/Nm2:λ=2π(8.85×1012)(6.00)ln(2.00/1.20)2.68×1011C/m.
03

Calculate the total charge (Q) on the capacitor

The total charge Q is the charge per unit length λ times the length of the capacitor L:Q=λ×L Given L=0.350m, and using the λ calculated:Q=(2.68×1011C/m)(0.350m)9.38×1012C.
04

Determine the capacitance (C)

The capacitance of a cylindrical capacitor is given by:C=2πε0Lln(r2/r1)Substitute the given values L=0.350m, r1=1.20×103m, r2=2.00×103m, and ε0=8.85×1012C2/Nm2:C=2π(8.85×1012)(0.350)ln(2.00/1.20)1.56×1012F.
05

Calculate the energy stored (U)

The energy stored in the capacitor when fully charged can be calculated using:U=12CV2Substitute C=1.56×1012F and V=6.00V:U=12(1.56×1012)(6.002)2.81×1011J.

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Key Concepts

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

Charge per Length in a Cylindrical Capacitor
In the context of capacitors, especially cylindrical ones, it is important to understand the term "charge per length." This refers to how much electric charge is distributed along a given length of the cylindrical surface. The symbol λ is commonly used to represent this quantity.

The cylindrical capacitor features a solid core and an outer hollow tube. To find the charge per length λ, we use the formula:
  • λ=2πε0Vln(r2/r1)
In this formula:
  • ε0 signifies the permittivity of free space, approximately 8.85×1012C2/Nm2.
  • V is the potential difference between the core and tube.
  • r1 and r2 are the radii of the inner core and outer tube, respectively.
To solve an example from a given problem, you might substitute specific values such as V=6.00V, r1=1.20×103m, and r2=2.00×103m. This gives a charge per length λ2.68×1011C/m.

Understanding charge per length lays a foundation for calculating total charge and other properties of capacitors.
Capacitance of a Cylindrical Capacitor
Capacitance is a measure of the ability of a capacitor to store an electrical charge. In cylindrical capacitors, it depends on the geometry and material properties. For a cylindrical capacitor, the formula for capacitance C is:
  • C=2πε0Lln(r2/r1)
Here:
  • L is the length of the cylindrical capacitor.
  • Again, r1 and r2 are the radii of the inner and outer parts, respectively.
  • ε0 is the permittivity of free space, as before.
Using our example values, such as L=0.350m, you calculate the capacitance to be approximately C1.56×1012F.

The capacitance indicates how much charge the capacitor can hold when a particular voltage is applied. This makes capacitance a central quantity in the analysis of electrical circuits.
Energy Stored in a Capacitor
The energy stored in a capacitor is crucial for understanding how capacitors function in circuits. This stored energy can be calculated using the formula:
  • U=12CV2
Where:
  • C is the capacitance found earlier, 1.56×1012F in our drumroll example.
  • V is the potential difference, given as 6.00V.
Substituting these values into the formula, the energy stored U is 2.81×1011J.

This energy is essentially the potential energy that can be released into a circuit to do work, such as lighting a bulb or powering a device. It highlights the utility of capacitors in storing and releasing energy quickly when needed.

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Most popular questions from this chapter

A parallel-plate air capacitor has a capacitance of 920 pF. The charge on each plate is 3.90 μC. (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?

An air capacitor is made from two flat parallel plates 1.50 mm apart. The magnitude of charge on each plate is 0.0180 μC when the potential difference is 200 V. (a) What is the capacitance? (b) What is the area of each plate? (c) What maximum voltage can be applied without dielectric breakdown? (Dielectric breakdown for air occurs at an electric-field strength of 3.0 × 106 V/m.) (d) When the charge is 0.0180 μC, what total energy is stored?

A spherical capacitor is formed from two concentric, spherical, conducting shells separated by vacuum. The inner sphere has radius 15.0 cm and the capacitance is 116 pF. (a) What is the radius of the outer sphere? (b) If the potential difference between the two spheres is 220 V, what is the magnitude of charge on each sphere?

A 5.00μF parallel-plate capacitor is connected to a 12.0-V battery. After the capacitor is fully charged, the battery is disconnected without loss of any of the charge on the plates. (a) A voltmeter is connected across the two plates without discharging them. What does it read? (b) What would the voltmeter read if (i) the plate separation were doubled; (ii) the radius of each plate were doubled but their separation was unchanged?

An air capacitor is made by using two flat plates, each with area A, separated by a distance d. Then a metal slab having thickness a (less than d) and the same shape and size as the plates is inserted between them, parallel to the plates and not touching either plate (Fig. P24.62). (a) What is the capacitance of this arrangement? (b) Express the capacitance as a multiple of the capacitance C0 when the metal slab is not present. (c) Discuss what happens to the capacitance in the limits a 0 and a d.

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