/*! 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 21 A metal sphere mounted on an ins... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 25: Problem 21

A metal sphere mounted on an insulating rod carries a charge of \(6.0 \mathrm{nC}\) when its potential is \(200 \mathrm{~V}\) higher than its surroundings. What is the capacitance of the capacitor formed by the sphere and its surroundings? $$ C=\frac{q}{V}=\frac{6.0 \times 10^{-9} \mathrm{C}}{200 \mathrm{~V}}=30 \mathrm{pF} $$

Short Answer

Expert verified
The capacitance is 30 pF.

Step by step solution

01

Understanding the Formula

The capacitance of a system can be calculated using the formula: \( C = \frac{q}{V} \) where \( C \) is the capacitance, \( q \) is the charge, and \( V \) is the potential difference.
02

Identifying Given Values

Here, the charge \( q \) is given as \( 6.0 \times 10^{-9} \mathrm{C} \) and the potential difference \( V \) is \( 200 \mathrm{~V} \). We need these values to substitute into the formula.
03

Substituting Values into the Formula

Substitute the given values into the formula: \( C = \frac{6.0 \times 10^{-9} \mathrm{C}}{200 \mathrm{~V}} \).
04

Calculating Capacitance

Perform the division: \( \frac{6.0 \times 10^{-9}}{200} = 3.0 \times 10^{-11} \mathrm{F} \).
05

Converting Units

The result \( 3.0 \times 10^{-11} \mathrm{F} \) can be expressed in picofarads (pF) since \( 1 \mathrm{F} = 10^{12} \mathrm{pF} \). Thus, \( 3.0 \times 10^{-11} \mathrm{F} = 30 \mathrm{pF} \).

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.

Electric Charge
Electric charge is a fundamental property of certain subatomic particles, such as electrons and protons, which gives rise to the electromagnetic force. This is one of the four fundamental forces in nature.
The unit of electric charge is the coulomb (C). Sometimes, charges are expressed in smaller units like nanocoulombs (nC) or picocoulombs (pC), which are more convenient for everyday situations due to the large magnitude of a coulomb.
There are two types of electric charges: positive and negative. Like charges repel each other while opposite charges attract.
  • The charge on the sphere given in the exercise was 6.0 nC, where nano means one-billionth of a coulomb.
  • Charges can be transferred from one object to another resulting in objects being charged by various methods like friction or induction.
This property is crucial for understanding how systems like capacitors store and transfer energy. Understanding electric charge allows us to solve problems involving capacitance, as it reflects the amount of electric field present in the system.
Potential Difference
Potential difference, also known as voltage (V), is the difference in electrical potential between two points. It represents the work needed to move a charge between those points.
This is measured in volts (V) and is a critical factor in the study of electrical circuits and systems.
  • In our exercise, a potential difference of 200 V was given, meaning that the sphere's potential was 200 volts higher than the surroundings.
  • The potential difference can drive current from higher to lower potential areas, similar to water flowing downhill.

A potential difference indicates the capacity of a system to perform work on charges. It is an essential part of calculating capacitance since it contributes to how much charge can be stored for a given energy level.
Capacitors
Capacitors are devices used to store electric charge and energy in an electric field. They consist typically of two conductive plates separated by an insulating material called a dielectric.
Capacitors are found in various applications, from electronic circuits to power systems, where they may be used to filter signals, store energy temporarily, or manage power supply changes.
  • In the exercise, the problem was essentially describing a simple capacitor where the metal sphere acts as one plate, and its surrounding acts as another.
  • The formula for capacitance is \( C = \frac{q}{V} \), which describes the ratio of the charge stored to the potential difference applied.

The unit of capacitance is the farad (F), with common subunits being microfarads (碌F) and picofarads (pF) because farads are typically a large unit for practical use.
Understanding capacitors and their roles in circuits can help predict how much energy they can store and supply in transient states or steady conditions, making them versatile and essential in modern electronics.

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

An electron is moving in the \(+x\) -direction with a speed of \(5.0 \times 10^{6} \mathrm{~m} / \mathrm{s}\). There is an electric field of \(3.0 \mathrm{kV} / \mathrm{m}\) in the \(+x\) -direction. What will be the electron's speed after it has moved \(1.00 \mathrm{~cm}\) along the field?

Figure \(25-4\) depicts two large, closely spaced metal plates (perpendicular to the page) connected to a 120-V battery. Assume the plates to be in vacuum and to be much larger than shown. Find (a) \(E\) between the plates, \((b)\) the force experienced by an electron between the plates, \((c)\) the \(\mathrm{PE}_{E}\) lost by an electron as it moves from plate- \(B\) to plate- \(A\), and \((d)\) the speed of the electron released from plate- \(B\) just before striking plate- \(A\). (a) \(E\) is directed from the positive plate- \(A\) to the negative plate- \(B .\) It is uniform between large parallel plates and is given by $$ E=\frac{V}{d}=\frac{120 \mathrm{~V}}{0.020 \mathrm{~m}}=6000 \mathrm{~V} / \mathrm{m}=6.0 \mathrm{kV} / \mathrm{m} $$ directed from left to right. (b) \(\quad F_{E}=q E=\left(-1.6 \times 10^{-19} \mathrm{C}\right)(6000 \mathrm{~V} / \mathrm{m})=-9.6 \times 10^{-16} \mathrm{~N}\) The minus sign tells us that \(\overrightarrow{\mathbf{F}}_{E}\) is directed oppositely to \(\overrightarrow{\mathbf{E}}\). Since plate- \(A\) is positive, the electron is attracted by it. The force on the electron is toward the left. (c) Change in \(\mathrm{PE}_{E}=V q=(120 \mathrm{~V})\left(-1.6 \times 10^{-19} \mathrm{C}\right)=-1.92 \times 10^{-17} \mathrm{~J}=1.9 \times 10^{-17} \mathrm{~J}\) Notice that \(V\) is a potential rise from \(B\) to \(A\) \((d)\) \(\mathrm{PE}_{E}\) lost \(=\mathrm{KE}\) gained \(1.92 \times 10^{-17} \mathrm{~J}=\frac{1}{2} m v_{f}^{2}-\frac{1}{2} m v_{i}^{2}\) \(1.92 \times 10^{-17} \mathrm{~J}=\frac{1}{2}\left(9.1 \times 10^{-31} \mathrm{~kg}\right) v_{f}^{2}-0\) from which \(v_{f}=6.5 \times 10^{6} \mathrm{~m} / \mathrm{s}\).

Find the electrical potential energy of three point charges placed in vacuum as follows on the \(x\) -axis: \(+2.0 \mu \mathrm{C}\) at \(x=0,+3.0 \mu \mathrm{C}\) at \(x=20 \mathrm{~cm}\), and \(+6.0 \mu \mathrm{C}\) at \(x=50 \mathrm{~cm}\). Take the \(\mathrm{PE}_{E}\) to be zero when the charges are separated far apart. $$ V_{x=0.2}=k_{0} \frac{2.0 \mu \mathrm{C}}{0.20 \mathrm{~m}}=\left(9.0 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}\right)\left(\frac{2 \times 10^{-6} \mathrm{C}}{0.20 \mathrm{~m}}\right)=9.0 \times 10^{4} \mathrm{~V} $$ Therefore the work required to bring in the \(3 \mu \mathrm{C}\) charge is $$ W_{3 \mu \mathrm{C}}=q V_{x=0.2}=\left(3.0 \times 10^{-6} \mathrm{C}\right)\left(9.0 \times 10^{4} \mathrm{~V}\right)=0.270 \mathrm{~J} $$ Finally bring the \(6.0 \mu \mathrm{C}\) charge in to \(x=050 \mathrm{~m}\). The potential there due to the two charges already present is $$ V_{x=0.5}=k_{0}\left(\frac{2.0 \times 10^{-6} \mathrm{C}}{0.50 \mathrm{~m}}+\frac{3.0 \times 10^{-6} \mathrm{C}}{0.30 \mathrm{~m}}\right)=12.6 \times 10^{4} \mathrm{~V} $$ Therefore the work required to bring in the \(6.0 \mu \mathrm{C}\) charge is $$ W_{6 \mu \mathrm{C}}=q V_{x=0.5}=\left(6.0 \times 10^{-6} \mathrm{C}\right)\left(12.6 \times 10^{4} \mathrm{~V}\right)=0.756 \mathrm{~J} $$ Adding the amounts of work required to assemble the charges gives the energy stored in the system: $$ \mathrm{PE}_{E}=0.270 \mathrm{~J}+0.756 \mathrm{~J}=1.0 \mathrm{~J} $$ Can you show that the order in which the charges are brought in from infinity does not affect this result?

The nucleus of a tin atom in vacuum has a charge of \(+50 e\). (a) Find the absolute potential \(V\) at a radial distance of \(1.0 \times 10^{-12} \mathrm{~m}\) from the nucleus. (b) If a proton is released from this point, how fast will it be moving when it is \(1.0 \mathrm{~m}\) from the nucleus? (a) \(V=k_{0} \frac{q}{r}=\left(9.0 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}\right) \frac{(50)\left(1.6 \times 10^{-19} \mathrm{C}\right)}{10^{-12} \mathrm{~m}}=72 \mathrm{kV}\) (b) The proton is repelled by the nucleus and flies out to infinity. The absolute potential at a point is the potential difference between the point in question and infinity. Hence, there is a potential drop of \(72 \mathrm{kV}\) as the proton flies to infinity. Usually we would simply assume that \(1.0 \mathrm{~m}\) is far enough from the nucleus to consider it to be at infinity. But, as a check, compute \(V\) at \(r=1.0 \mathrm{~m}\) : $$ V_{1 \mathrm{~m}}=k_{0} \frac{q}{r}=\left(9.0 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}\right) \frac{(50)\left(1.6 \times 10^{-19} \mathrm{C}\right)}{1.0 \mathrm{~m}}=7.2 \times 10^{-8} \mathrm{~V} $$ which is essentially zero in comparison with \(72 \mathrm{kV}\). As the proton falls through \(72 \mathrm{kV}\), $$ \begin{aligned} \text { KE gained } &=\mathrm{PE}_{E} \text { lost } \\ \frac{1}{2} m v_{f}^{2}-\frac{1}{2} m v_{i}^{2} &=q V \\ \frac{1}{2}\left(1.67 \times 10^{-27} \mathrm{~kg}\right) v_{f}^{2}-0 &=\left(1.6 \times 10^{-19} \mathrm{C}\right)(72000 \mathrm{~V}) \end{aligned} $$ from which \(v_{f}=3.7 \times 10^{6} \mathrm{~m} / \mathrm{s}\)

An electron starts from rest and falls through a potential rise of \(80 \mathrm{~V}\). What is its final speed? Positive charges fall through potential drops; negative charges, such as electrons, fall through potential rises. $$ \text { Change in } \mathrm{PE}_{E}=V q=(80 \mathrm{~V})\left(-1.6 \times 10^{-19} \mathrm{C}\right)=-1.28 \times 10^{-17 \mathrm{~J}} $$ This lost \(\mathrm{PE}_{E}\) appears as \(\mathrm{KE}\) of the electron: $$ \begin{array}{l} \text { and }\\\ \begin{aligned} \mathrm{PE}_{E} \text { lost } &=\mathrm{KE} \text { gained } \\ 1.28 \times 10^{-17} \mathrm{~J} &=\frac{1}{2} m v_{f}^{2}-\frac{1}{2} m v_{i}^{2}=\frac{1}{2} m v_{f}^{2}-0 \\ v_{f} &=\sqrt{\frac{\left(1.28 \times 10^{-17} \mathrm{~J}\right)(2)}{9.1 \times 10^{-31} \mathrm{~kg}}}=5.3 \times 10^{6} \mathrm{~m} / \mathrm{s} \end{aligned} \end{array} $$
See all solutions

Recommended explanations on Physics Textbooks

Torque and Rotational Motion

Read Explanation

Thermodynamics

Read Explanation

Electromagnetism

Read Explanation

Magnetism

Read Explanation

Physical Quantities And Units

Read Explanation

Radiation

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.