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Ch 24: Capacitance and Dielectrics
Young & Freedman Calc - University Physics 14th Edition
Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Not the one you use?Change textbook
Chapter 24, Problem 9

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 10.010.0 pC. The inner cylinder has radius 0.500.50 mm, the outer one has radius 5.005.00 mm, and the length of each cylinder is 18.018.0 cm.
(a) What is the capacitance?
(b) What applied potential difference is necessary to produce these charges on the cylinders?

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1
Step 1: Understand the geometry of the capacitor. We have two coaxial cylinders, one inside the other, forming a cylindrical capacitor. The inner cylinder has a radius of 0.50 mm, and the outer cylinder has a radius of 5.00 mm. Both cylinders have a length of 18.0 cm.
Step 2: Use the formula for the capacitance of a cylindrical capacitor. The capacitance C is given by the formula: C=2πεlln(ba), where ε is the permittivity of free space, l is the length of the cylinders, a is the radius of the inner cylinder, and b is the radius of the outer cylinder.
Step 3: Convert all units to meters for consistency. The inner radius is 0.50 mm = 0.0005 m, the outer radius is 5.00 mm = 0.005 m, and the length is 18.0 cm = 0.18 m.
Step 4: Calculate the capacitance using the formula from Step 2. Substitute the values: a = 0.0005 m, b = 0.005 m, l = 0.18 m, and ε = 8.85 x 10-12 F/m.
Step 5: To find the applied potential difference, use the formula V=QC, where Q is the charge on the cylinders (10.0 pC = 10.0 x 10-12 C) and C is the capacitance calculated in Step 4.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Capacitance of a Coaxial Capacitor

Capacitance is a measure of a capacitor's ability to store charge per unit voltage. For coaxial cylinders, the capacitance can be calculated using the formula C = (2πε₀L) / ln(b/a), where ε₀ is the permittivity of free space, L is the length of the cylinders, and a and b are the radii of the inner and outer cylinders, respectively. This formula accounts for the cylindrical geometry and the electric field distribution between the cylinders.
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Electric Potential Difference

The electric potential difference, or voltage, between two points is the work done per unit charge to move a charge between those points. For a coaxial capacitor, the potential difference V can be found using the relationship V = Q/C, where Q is the charge on the cylinders and C is the capacitance. This concept is crucial for determining the voltage required to achieve a specific charge distribution on the capacitor.
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Permittivity of Free Space

Permittivity of free space, denoted as ε₀, is a fundamental physical constant that characterizes the ability of a vacuum to permit electric field lines. It appears in the equations for capacitance and electric fields, influencing how electric fields interact with materials. In the context of a coaxial capacitor, ε₀ is used to calculate the capacitance, reflecting how the electric field is distributed between the charged cylinders.
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Related Practice
Textbook Question

A 5.005.00-μ\(\mu\)F parallel-plate capacitor is connected to a 12.012.0 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?

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Textbook Question

A parallel-plate air capacitor is to store charge of magnitude 240.0240.0 pC on each plate when the potential difference between the plates is 42.042.0 V.

(a) If the area of each plate is 6.806.80 cm2, what is the separation between the plates?

(b) If the separation between the two plates is double the value calculated in part (a), what potential difference is required for the capacitor to store charge of magnitude 240.0240.0 pC on each plate?

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Textbook Question

Figure E24.1424.14 shows a system of four capacitors, where the potential difference across ab is 50.050.0 V. How much charge is stored in each of the 10.010.0-μ\(\mu\)F and the 9.09.0-μ\(\mu\)F capacitors?

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Textbook Question

Figure E24.1424.14 shows a system of four capacitors, where the potential difference across ab is 50.050.0 V. How much charge is stored by this combination of capacitors?

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Textbook Question

A parallel-plate air capacitor of capacitance 245245 pF has a charge of magnitude 0.148 0.148 μ\(\mu\)C on each plate. The plates are 0.3280.328 mm apart.

(a) What is the potential difference between the plates?

(b) What is the area of each plate?

(c) What is the electric field magnitude between the plates?

(d) What is the surface charge density on each plate?

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Textbook Question

A spherical capacitor contains a charge of 3.303.30 nC when connected to a potential difference of 220220 V. If its plates are separated by vacuum and the inner radius of the outer shell is 4.004.00 cm, calculate: (a) the capacitance; (b) the radius of the inner sphere; (c) the electric field just outside the surface of the inner sphere.

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