/*! 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 13 You measure the capacitance \(C_... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 24: Problem 13

You measure the capacitance \(C_{1}\) of a capacitor by doing the following: First connect capacitors \(C_{1}\) and \(C_{2}\) in series to a power supply that provides a voltage \(V\) that can be varied. The capacitance of \(C_{2}\) is known to be \(3.00 \mu \mathrm{F}\). Then vary the applied voltage \(V,\) and for each value of \(V\) measure the voltage \(V_{2}\) across \(C_{2}\). After plotting your data as \(V_{2}\) versus \(V,\) you find that the data fall close to a straight line that has slope \(0.650 .\) What is the capacitance \(C_{1} ?\)

Short Answer

Expert verified
The capacitance \(C_{1}\) equals to 1.95 \mu \mathrm{F}.

Step by step solution

01

Define the series capacitance relation

When two capacitors are in series, the total capacitance of the system is given by the reciprocal of the sum of the reciprocals of the individual capacitances, \(1/C_{total} = 1/C_{1} + 1/C_{2}\). This relation can be re-arranged to find \(C_{1}\): \(1/C_{1} = 1/C_{total} - 1/C_{2}\)
02

Extract information from the measurements

The data collected and plotted show a linear relation between \(V_{2}\) and \(V\) with slope equal to 0.650. It can be interpreted by examining the voltage distribution relation for capacitors in series, \(V_{1}/V_{2} = C_{2}/C_{1}\), which implies \(V_{2}/V = C_{1}/C_{2}\). The slope of the graph effectively gives this ratio. Thus, \(C_{1}/C_{2} = 0.650\).
03

Calculate the value of \(C_{1}\)

By using the extracted ratio from step 2, we can express \(C_{1}\) in terms of \(C_{2}\): \(C_{1} = 0.650 * C_{2}\). Therefore, by substituting \(C_{2} = 3.00 \mu \mathrm{F}\) into this equation, we have \(C_{1} = 0.650 * 3.00 \mu \mathrm{F} = 1.95 \mu \mathrm{F}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Series Capacitors
When capacitors are placed in series, their total capacitance is different from simply adding them together. In a series configuration, the overall capacitance (C_{total}) is determined by the reciprocal formula:
  • \( \frac{1}{C_{total}} = \frac{1}{C_{1}} + \frac{1}{C_{2}} + \ldots \)
This formula is important because it influences how capacitors store energy when connected in a circuit. In practical terms, adding capacitors in series actually reduces the total capacitance.
Consider this: imagine capacitors as water tanks. Connecting them in series is like stacking them vertically 鈥� the overall water capacity doesn't simply add up, but depends on the inverse of the individual tank capacities.
Voltage Distribution
The voltage across capacitors in series shares a specific relationship contingent on their capacitance values. Each capacitor in series will carry the same charge, but the voltage across them varies based on their capacitance:
  • \( V_{1}/V_{2} = C_{2}/C_{1} \)
So, for two capacitors, the voltage across each capacitor is inversely proportional to its capacitance.
This means the larger the capacitance, the smaller the voltage across it. For example, if you have one large and one small capacitor, the smaller one will have a higher voltage across it than the larger one. This voltage distribution helps in understanding and predicting how circuits will behave when capacitors are in series.
Linear Graph Analysis
Graphs can provide valuable information about the relationship between voltage and capacitance. In this exercise, the plotted data between voltage \(V_{2}\) across capacitor \(C_{2}\) and the total applied voltage \(V\) results in a straight line.
This linearity showcases a proportional relationship, characterized by the graph's slope:
  • The line slope tells us how one capacitor's voltage relates to the other's.
In formulas, this means the slope \(m = 0.650\) provides an insight into the ratio \(C_{1}/C_{2}\). The straight-line graph simplifies finding this ratio, making it easier to perform further calculations, like determining unknown capacitances.
Capacitor Ratio Calculation
Finding the capacitance of a capacitor when another is known requires understanding voltage distribution and linear relationships. In this situation, the graph's slope aids in determining the ratio of two capacitors:
  • The formula \( V_{2}/V = C_{1}/C_{2} \) becomes crucial.
By knowing the slope \(0.650\) and the capacitance \(C_{2} = 3.00 \, \mu F\), we can find \(C_{1}\) easily. Here's how:
  • \( C_{1} = 0.650 \times 3.00 \, \mu F = 1.95 \, \mu F \)
This process highlights the significance of measuring slopes on graphs when deriving unknown values. It's a nice combination of visual data analysis and mathematical calculation, leading to effective problem-solving.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Two identical air-filled parallel-plate capacitors \(C_{1}\) and \(C_{2}\), each with capacitance \(C,\) are connected in series to a battery that has voltage \(V\). While the two capacitors remain connected to the battery, a dielectric with dielectric constant \(K>1\) is inserted between the plates of one of the capacitors, completely filling the space between them. Let \(U_{0}\) be the total energy stored in the two capacitors without the dielectric and \(U\) be the total energy stored after the dielectric is inserted. In terms of \(K,\) what is the ratio \(U / U_{0} ?\) Does the total stored energy increase, decrease, or stay the same after the dielectric is inserted?

Two air-filled parallel-plate capacitors with capacitances \(C_{1}\) and \(C_{2}\) are connected in parallel to a battery that has a voltage of \(36.0 \mathrm{~V}\) \(C_{1}=4.00 \mu \mathrm{F}\) and \(C_{2}=6.00 \mu \mathrm{F}\). (a) What is the total positive charge stored in the two capacitors? (b) While the capacitors remain connected to the battery, a dielectric with dielectric constant 5.00 is inserted between the plates of capacitor \(C_{1}\), completely filling the space between them. Then what is the total positive charge stored on the two capacitors? Does the insertion of the dielectric cause total charge stored to increase or decrease?

A fuel gauge uses a capacitor to determine the height of the fuel in a tank. The effective dielectric con- stant \(K_{\text {eff }}\) changes from a value of 1 when the tank is empty to a value of \(K\), the dielectric constant of the fuel, when the tank is full. The appropriate electronic circuitry can determine the effective dielectric constant of the combined air and fuel between the capacitor plates. Each of the two rectangular plates has a width \(w\) and a length \(L\) (Fig. \(\mathbf{P 2 4 . 6 6}\) ). The height of the fuel between the plates is \(h\). You can ignore any fringing effects. (a) Derive an expression for \(K_{\text {eff }}\) as a function of \(h\). (b) What is the effective dielectric constant for a tank \(\frac{1}{4}\) full, \(\frac{1}{2}\) full, and \(\frac{3}{4}\) full if the fuel is gasoline \((K=1.95) ?\) (c) Repeat part (b) for methanol \((K=33.0)\). (d) For which fuel is this fuel gauge more practical?

After combing your hair on a dry day, some of your hair stands up, forced away from your head by electrostatic repulsion. (a) Estimate the length \(L\) of your hair. (b) Using the average linear mass density of hair, which is \(65 \mu \mathrm{g} / \mathrm{cm},\) estimate the mass \(m\) of one of your hairs. (c) Estimate the number \(N\) of hairs that stand after combing. (d) Assume that the comb has taken away a charge \(-2 Q,\) and that your hair has therefore gained an amount of charge \(2 Q .\) Assume further that half of this charge resides next to your head and the other half is distributed at the ends of the \(N\) strands that stand up. Assume the electrostatic force that lifted a hair was twice its weight. Show that this leads to \(2 m g=\frac{1}{4 \pi \epsilon_{0}} \frac{Q^{2} / N}{L^{2}}\) Use this equation to estimate the charge \(Q\) that resides on your head. (e) If your head were a sphere with radius \(R\), it would have a capacitance of \(4 \pi \epsilon_{0} R .\) Estimate the radius of your head; then use your estimate to determine your head's capacitance. (f) Use your result to estimate the potential attained due to combing. (The surprising result illustrates an interesting feature of static electricity.)

Two parallel plates have equal and opposite charges. When the space between the plates is evacuated, the electric field is \(E=3.20 \times 10^{5} \mathrm{~V} / \mathrm{m} .\) When the space is filled with dielectric, the electric field is \(E=2.50 \times 10^{5} \mathrm{~V} / \mathrm{m} .\) (a) What is the charge density on each surface of the dielectric? (b) What is the dielectric constant?
See all solutions

Recommended explanations on Physics Textbooks

Magnetism

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Geometrical and Physical Optics

Read Explanation

Mechanics and Materials

Read Explanation

Fluids

Read Explanation

Scientific Method Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.