/*! 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 Problem

We have a metal sphere charged to a potential of 200 V above its surroundings, carrying a charge of 6.0 nC. We need to determine the capacitance of the system formed by the sphere and its surroundings.
02

Equation for Capacitance

Capacitance \(C\) is defined as the ratio of the charge \(q\) to the potential difference \(V\), given by the formula \( C = \frac{q}{V} \). Here, \(q = 6.0 \times 10^{-9} \) C and \(V = 200\) V.
03

Substitution of Values

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

Calculation of Capacitance

Perform the division: \( 6.0 \times 10^{-9} \text{ C} \div 200 \text{ V} = 3.0 \times 10^{-11}\, \text{F}\), which corresponds to 30 pF (picoFarads, where 1 pF = \(10^{-12}\) F).
05

Verification of Units

Ensure the units are consistent: Charge in Coulombs and Potential in Volts yield Capacitance in Farads, confirming that the results make sense.

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 matter that causes it to experience a force when placed in an electromagnetic field. The unit of electric charge is the Coulomb (C).
If you have ever rubbed a balloon on your hair and watched it stick to a wall, you have witnessed electric charge in action! In the context of this exercise, the metal sphere carries a charge of 6.0 nanoCoulombs (nC).
A nanoCoulomb is a billionth of a Coulomb, which is often represented as:
  • 1 nC = 1 x 10-9 C
The sphere becomes charged, indicating that it either has an excess or a deficiency of electrons.
A positively charged object has a deficiency of electrons, whereas a negatively charged object has an excess of them. Electric charges can lead to the creation of electric fields, which is a topic that is heavily related to capacitance.
Potential difference
Potential difference, commonly known as voltage, is the difference in electrical potential energy between two points. It measures the potential energy required to move a charge between two points.
The formula for potential difference is:
  • V = W/q
where:
  • V is the potential difference,
  • W is the work done to move the charge, and
  • q is the charge.
In this exercise, the sphere's potential is 200 volts (V) higher than its surroundings.
This means that a charged particle moving between the sphere and its surroundings would experience an energy change equivalent to 200 Joules per Coulomb of charge. Potential difference is crucial as it drives the flow of electric charge, much like how a waterfall causes water to flow downward.
Metal sphere capacitance
The capacitance of a system is its ability to store electric charge per unit of potential difference. Capacitance is a measure of how much charge a body can hold at a given potential difference.
It is represented by the equation:
  • C = q/V
Where:
  • C is the capacitance,
  • q is the charge stored, and
  • V is the potential difference.
In this exercise, the capacitance of the metal sphere is calculated to be 30 picoFarads (pF), where:
  • 1 picoFarad = 10-12 Farads
To calculate, substitute the charge and potential difference into the formula:
  • C = 6.0 x 10-9 C / 200 V = 30 x 10-12 F
Thus, the sphere's capacity to store electric charge is directly related to its size and the surrounding environment.
Metal sphere capacitance plays a significant role in designing circuits and in the function of electronic devices.

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

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. Compute how much work must be done to bring the charges from infinity to their places on the axis. Bring in the \(2.0 \mu \mathrm{C}\) charge first; this requires no work because there are no other charges in the vicinity. Next bring in the \(3.0 \mu \mathrm{C}\) charge, which is repelled by the \(+2.0 \mu \mathrm{C}\) charge. The potential difference between infinity and the position to which we bring it is due to the \(+2.0 \mu \mathrm{C}\) charge and is $$ V_{x=0.2}=k_{0} \frac{2.0 \mu \mathrm{C}}{0.20 \mathrm{~m}}=\left(9.0 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}\right)\left(\frac{2 \times 10^{-6} \mathrm{C}}{0.20 \mathrm{~m}}\right)=9.0 \times 10^{4} \mathrm{v} $$ Therefore the work required to bring in the \(3 \mu \mathrm{C}\) charge is \(W_{3 \mu C}=q V_{x=0.2}=\left(3.0 \times 10^{-6} \mathrm{C}\right)\left(9.0 \times 10^{4} \mathrm{~V}\right)=0.270 \mathrm{~J}\) 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 \(W_{6 \mu C}=q V_{x=0.5}=\left(6.0 \times 10^{-6} \mathrm{C}\right)\left(12.6 \times 10^{4} \mathrm{~V}\right)=0.756 \mathrm{~J}\) 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?

Two point charges, \(+q\) and \(-q\), are separated by a distance \(d\) in air. Where, besides at infinity, is the absolute potential zero? At the point (or points) in question, \(0=k_{0} \frac{q}{r_{1}}+k_{0} \frac{-q}{r_{2}} \quad\) or \(\quad r_{1}=r_{2}\) This condition holds everywhere on a plane, which is the perpendicular bisector of the line joining the two charges. Therefore, the absolute potential is zero everywhere on that plane.

An electron has a speed of \(6.0 \times 10^{5} \mathrm{~m} / \mathrm{s}\) as it passes point- \(A\) on its way to point- \(B\). Its speed at \(B\) is \(12 \times 10^{5} \mathrm{~m} / \mathrm{s}\). What is the potential difference between \(A\) and \(B\), and which is at the higher potential?

(a) Calculate the capacitance of a capacitor consisting of two parallel plates separated by a layer of paraffin wax \(0.50 \mathrm{~cm}\) thick, the area of each plate being \(80 \mathrm{~cm}^{2}\). The dielectric constant for the wax is \(2.0 .(b)\) If the capacitor is connected to a \(100-V\) source, calculate the charge on the capacitor and the energy stored in the capacitor.

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

Recommended explanations on Physics Textbooks

Rotational Dynamics

Read Explanation

Radiation

Read Explanation

Physical Quantities And Units

Read Explanation

Linear Momentum

Read Explanation

Quantum Physics

Read Explanation

Electric Charge Field and Potential

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.