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Chapter 26: Problem 2

2\. Charged Belt A charged belt, \(50 \mathrm{~cm}\) wide, travels at \(30 \mathrm{~m} / \mathrm{s}\) between a source of charge and a sphere. The belt carries charge into the sphere at a rate corresponding to \(100 \mu \mathrm{A}\). Compute the surface charge density on the belt.

Short Answer

Expert verified
The surface charge density on the belt is \(6.\bar{6} \times 10^{-6} \text{ C/m}^2\).

Step by step solution

01

- Convert Units

Convert all given units to SI units for consistency. Given: width of belt, \(50 \text{ cm} = 0.5 \text{ m}\), speed of belt, \(30 \text{ m/s}\), and the current, \(100 \text{ }\text{渭}\text{A} = 100 \times 10^{-6} \text{ A} = 1 \times 10^{-4} \text{ A}\).
02

- Calculate Charge per Time

The current (\text{I}) represents the charge per unit of time. Thus, the charge per second (Q) is \(Q = 1 \times 10^{-4} \text{ C/s}\). This is the charge that the belt delivers to the sphere per second.
03

- Determine Charge per Length

Calculate the amount of charge on a unit length of the belt. Since the belt travels at \(30 \text{ m/s}\), in one second the belt travels \(30 \text{ m}\). Thus, the charge per meter length of the belt is \(\frac{Q}{30 \text{ m}} = \frac{1 \times 10^{-4}}{30} \text{ C/m} = 3.\bar{3} \times 10^{-6} \text{ C/m}\).
04

- Calculate Surface Charge Density

The surface charge density (\text{蟽}) is the charge per unit area. The width of the belt is \(0.5 \text{ m}\). Thus, the area per meter length of the belt is \(0.5 \text{ m}^2\). Surface charge density is then \(蟽 = \frac{3.\bar{3} \times 10^{-6} \text{ C/m}}{0.5 \text{ m}^2} = 6.\bar{6} \times 10^{-6} \text{ C/m}^2\).

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Key Concepts

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

Unit Conversion
When solving physics problems, consistency in units is critical. Suppose a belt's width is given as 50 cm (centimeters). To use it in calculations, convert to SI units: 50 cm = 0.5 m (meters). Similarly, if the electric current is given in microamperes (渭A), convert it to amperes (A). Here, 100 渭A is equivalent to 100 脳 10-6 A or 1 脳 10-4 A. Using consistent units ensures that the formulas and calculations yield correct results.
Electric Current
Electric current, denoted by I, is the flow rate of electric charge. In this context, 100 渭A (1 脳 10-4 A) flows through the charged belt to a sphere. The current measures the charge's movement over time. Therefore, charge (Q) over a given time (t) can be found using:

Q = I 脳 t

For one second, Q = 1 脳 10-4 C (coulombs per second). This indicates how much charge the belt carries to the sphere each second.
Charge per Unit Length
Understanding the charge distributed along the belt's length is crucial. The belt's speed is 30 m/s, meaning it moves 30 meters in one second. The charge per meter length (位) shows how this charge distributes along the belt. Calculate 位 using:

位 = Q / length

For a belt traveling 30 meters in one second with a charge of 1 脳 10-4 C, 位 becomes:

位 = (1 脳 10-4 C) / (30 m) = 3.33 脳 10-6 C/m

This indicates how much charge is present per meter along the moving belt.
Surface Charge Density Calculation
Finally, surface charge density (蟽) measures the amount of charge per unit area. For a belt 0.5 meters wide, calculate the area per meter length:

Area = length 脳 width = 1 m 脳 0.5 m = 0.5 m虏

With charge per meter length (位) as 3.33 脳 10-6 C/m, use:

蟽 = 位 / area

Therefore, 蟽 = (3.33 脳 10-6 C/m) / (0.5 m虏) = 6.66 脳 10-6 颁/尘虏

This surface charge density describes the charge distribution across the area of the belt.

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Most popular questions from this chapter

13\. Two Conductors Two conductors are made of the same material and have the same length. Conductor \(A\) is a solid wire of diameter \(1.0 \mathrm{~mm}\). Conductor \(B\) is a hollow tube of outside diameter \(2.0\) \(\mathrm{mm}\) and inside diameter \(1.0 \mathrm{~mm}\). What is the resistance ratio \(R_{A} / R_{B}\), measured between their ends?

45\. Building a Water Heater The nickel-chromium alloy Nichrome has a resistivity of about \(10^{-6} \Omega-\mathrm{m} .\) Suppose you want to build a small heater out of a coil of Nichrome wire and a \(6 \mathrm{~V}\) battery in order to heat \(30 \mathrm{ml}\) of water from a temperature of \(20 \mathrm{C}\) to \(40 \mathrm{C}\) in \(1 \mathrm{~min}\). Assume the battery has negligible internal resistance. (a) How much heat energy (in joules) do you need to do this? (b) How much power (in watts) do you need to do it in the time indicated? (c) What resistance should your Nichrome coil have in order to produce this much power in heat? (d) Can you create a coil having these properties? (Hint: Can you find a plausible length and cross-sectional area for your wire that will give you the resistance you need?) (e) If the internal resistance of the battery were \(1 / 3 \Omega\), how would it affect your calculation? (Only explain what you would have to do; don't recalculate the size of your coil.)

12\. A Certain Wire A certain wire has a resistance \(R\). What is the resistance of a second wire, made of the same material, that is half as long and has half the diameter?

7\. Conducting Wire A conducting wire has a \(1.0 \mathrm{~mm}\) diameter, a \(2.0 \mathrm{~m}\) length, and a \(50 \mathrm{~m} \Omega\) resistance. What is the resistivity of the material?

38\. Nichrome A wire of Nichrome (a nickel-chromium-iron alloy commonly used in heating elements) is \(1.0 \mathrm{~m}\) long and \(1.0 \mathrm{~mm}^{2}\) in cross-sectional area. It carries a current of \(4.0 \mathrm{~A}\) when a \(2.0 \mathrm{~V}\) potential difference is applied between its ends. Calculate the conductivity \(\sigma\) of Nichrome.
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