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Chapter 18: Problem 17

Find the required ratios. The capacitance \(C\) of a capacitor is defined as the ratio of its charge \(q\) (in \(\mathrm{C}\) ) to the voltage \(V\). Find \(C\) (in \(\mathrm{F}\) ) for which \(q=5.00 \mu \mathrm{C}\) and \(V=200 \mathrm{V} \cdot(1 \mathrm{F}=1 \mathrm{C} / 1 \mathrm{V} .)\)

Short Answer

Expert verified
The capacitance \( C \) is \( 2.50 \times 10^{-8} \ \mathrm{F} \).

Step by step solution

01

Identify Given Values and Formula

We are given that the charge \( q \) on the capacitor is \( 5.00 \ \mu\mathrm{C} \) and the voltage \( V \) is \( 200 \ \mathrm{V} \). The formula for capacitance \( C \) is \( C = \frac{q}{V} \).
02

Convert Microcoulombs to Coulombs

Since \( 1 \ \mu\mathrm{C} = 1 \times 10^{-6} \ \mathrm{C} \), the charge \( q = 5.00 \ \mu\mathrm{C} \) converts to \( 5.00 \times 10^{-6} \ \mathrm{C} \).
03

Substitute Values into the Formula

Now substitute \( q = 5.00 \times 10^{-6} \ \mathrm{C} \) and \( V = 200 \ \mathrm{V} \) into the capacitance formula: \( C = \frac{5.00 \times 10^{-6}}{200} \).
04

Calculate the Capacitance

Divide the charge by the voltage to find the capacitance: \[ C = \frac{5.00 \times 10^{-6}}{200} = 2.50 \times 10^{-8} \ \mathrm{F} \].

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Key Concepts

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

Electrical Charge
Electrical charge is a fundamental property of matter, leading to interactions between subatomic particles like electrons and protons. It is measured in coulombs (C). Like charges repel, and opposite charges attract. In the context of capacitors, the charge is the quantity of electricity stored, which allows capacitors to hold electric potential energy.
  • A capacitor can be thought of as a device that stores electrical charge.
  • Charges can be positive (protons) or negative (electrons).
  • Free electrons in wires create a flow of electricity, or current.
In our example, the charge given is 5.00 microcoulombs (渭C), where microcoulombs is a unit of charge often used when dealing with capacitors, because their storage capacities involve minute amounts of charge.
Voltage
Voltage, also known as electric potential difference, is the energy per unit charge. It's what drives the movement of charge through a conductor and is measured in volts (V). Essentially, it's the push that makes electrical charges move in a circuit. In simple terms, you can think of it as the pressure driving the electric current through the circuit. Voltage is what powers devices.
  • High voltage implies a strong push for electrical charge flow but not necessarily a large current.
  • Voltage is fundamental for any electrical circuit to function, as it provides the necessary force to move charges.
In the given problem, we have a voltage of 200 volts applied across a capacitor. This voltage, combined with the electrical charge, determines the capacitance capability of the capacitor.
Unit Conversion
Converting units is crucial in physics to ensure that calculations are accurate and adhere to standard measurements. In our calculation, it's necessary to convert microcoulombs (渭C) to coulombs (C) because the standard unit of charge in the formula for capacitance is coulombs.
  • 1 渭C = 1 脳 10-6 coulombs (C).
  • Capacitance is calculated in farads (F), which is coulombs per volt (C/V).
Using unit conversion ensures we're using consistent units for charge and voltage, allowing us to correctly compute the capacitance. In our example, the charge is initially given in microcoulombs but needs to be converted to coulombs: 5.00 渭C becomes 5.00 脳 10-6 C, making the calculation straightforward.
Physics Education
Physics education aims to provide students with a comprehensive understanding of the natural laws governing the universe. This includes instilling a grasp of key concepts such as electrical charge, voltage, and how to perform vital calculations including unit conversions and determining capacitance. Capacitance plays a critical role in numerous modern electronic devices, including cameras, power supplies, and tuning circuits.
  • Find practical applications of physics concepts to make learning engaging.
  • Use examples such as this capacitance calculation to illustrate physics in everyday life.
  • Encourage hands-on experiments to solidify understanding of theory.
By understanding these principles, students not only excel in problem-solving exercises but also in real-world applications, preparing for careers that rely on physics principles.

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