/*! 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 3 An isolated charged conducting s... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

An isolated charged conducting sphere of radius \(12.0 \mathrm{cm}\) creates an electric field of \(4.90 \times 10^{4} \mathrm{N} / \mathrm{C}\) at a distance \(21.0 \mathrm{cm}\) from its center. (a) What is its surface charge density? (b) What is its capacitance?

Short Answer

Expert verified
The surface charge density of the conducting sphere is around 1.15 x 10^-6 C/m^2 and its capacitance is approximately 33.9 pF.

Step by step solution

01

Calculate the charge on the sphere

Since the electric field at a distance r from the center of a spherical charge \(Q\) is given by: \(E = \frac{Q}{4πε_0r^2}\). The charge on the sphere can be calculated by rearranging the formula to: \(Q = 4πε_0xEr^2\). Given \(E=4.90 \times 10^{4} N/C, r=21.0cm=0.21m, ε_0=8.85 \times 10^{-12} C^2/Nm^2\), we can substitute these values into the equation to obtain the charge.
02

Calculate the surface charge density

The surface charge density \(σ\) of the sphere is given by \(σ= \frac{Q}{A}\), where \(A = 4πr^2\) is the surface area of the sphere. We can substitute the charge obtained in step 1 and the given radius \(a = 12.0cm = 0.12m\) into this formula to calculate the surface charge density.
03

Compute the capacitance of the sphere

The capacitance \(C\) of a single spherical conductor is given by the formula \(C = \frac{4πε_0a}{1}\), where \(a\) is the radius of the sphere. By substituting the given radius and the constant \(ε_0\) into the formula, we can calculate the capacitance.

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 91Ó°ÊÓ!

Key Concepts

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

Surface Charge Density
Surface charge density is a measure of how much electric charge is accumulated per unit area on the surface of a conductor. This concept is crucial in understanding how charges distribute themselves on surfaces and their effect on electric fields.To calculate surface charge density, represented by \( \sigma \), you need to know the total charge \( Q \) on the surface and the area \( A \) over which it is spread. The formula is:\[\sigma = \frac{Q}{A}\]The area \( A \) for a sphere is given by \( A = 4 \pi r^2 \), where \( r \) is the radius.In our exercise, after determining the charge on the sphere using the electric field information given, we use this formula to find out how much charge is spread per square meter of the sphere’s surface. This is essential to understand the amount of charge concentrated in a specific area.
Electric Field
An electric field is a region around a charged object where forces are exerted on other charges. The strength and direction of this field depend on the magnitude of the charge and the distance from the charge.The electric field \( E \) at a distance \( r \) from the center of a uniformly charged sphere is given by the formula:\[ E = \frac{Q}{4\pi\varepsilon_0 r^2} \]where:
  • \( Q \) is the charge enclosed by the Gaussian surface,
  • \( \varepsilon_0 \) is the permittivity of free space, approximately \(8.85 \times 10^{-12} \text{ C}^2/\text{Nm}^2\),
  • \( r \) is the radial distance from the sphere's center.
Using this equation, the charge causing the given electric field strength can be determined. Knowing the electric field helps us understand the influence of the sphere's charge over a certain distance.
Capacitance
Capacitance refers to the ability of a system to store electric charge. It is a key concept in electrostatics, particularly when discussing conductors and insulators.The capacitance \( C \) of an isolated spherical conductor is given by:\[ C = 4 \pi \varepsilon_0 a \]where:
  • \( a \) is the radius of the sphere
  • \( \varepsilon_0 \) is the permittivity of free space.
This formula shows that the capacitance of a sphere depends only on its physical size and the properties of the surrounding space and not on the charge or potential.Finding the capacitance is crucial because it tells us the sphere’s potential to store charge. It plays a significant role in applications where charge storage and management are needed. In our exercise, we calculate the sphere's capacitance using the given radius.

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

A 1 -megabit computer memory chip contains many \(60.0-\mathrm{fF}\) capacitors. Fach capacitor has a plate area of \(21.0 \times 10^{-12} \mathrm{m}^{2} .\) Determine the plate separation of such a capacitor (assume a parallel-plate configuration). The order of magnitude of the diameter of an atom is \(10^{-10} \mathrm{m}=0.1 \mathrm{nm} .\) Express the plate separation in nanometers.

The immediate cause of many deaths is ventricular fibrillation, uncoordinated quivering of the heart as opposed to proper beating. An electric shock to the chest can cause momentary paralysis of the heart muscle, after which the heart will sometimes start organized beating again. A defibrillator (Fig. 26.14 ) is a device that applies a strong electric shock to the chest over a time interval of a few milliseconds. The device contains a capacitor of several microfarads, charged to several thousand volts. Electrodes called paddles, about \(8 \mathrm{cm}\) across and coated with conducting paste, are held against the chest on both sides of the heart. Their handles are insulated to prevent injury to the operator, who calls, "Clear!" and pushes a button on one paddle to discharge the capacitor through the patient's chest. Assume that an energy of \(300 \mathrm{J}\) is to be delivered from a \(30.0-\mu \mathrm{F}\) capacitor. To what potential difference must it be charged?

The inner conductor of a coaxial cable has a radius of \(0.800 \mathrm{mm},\) and the outer conductor's inside radius is \(3.00 \mathrm{mm} .\) The space between the conductors is filled with polyethylene, which has a diclectric constant of 2.30 and a dielectric strength of \(18.0 \times 10^{6} \mathrm{V} / \mathrm{m} .\) What is the maximum potential difference that this cable can withstand?

Find the capacitance of the Earth. (Suggestion: The outer conductor of the "spherical capacitor" may be considered as a conducting sphere at infinity where \(V\) approaches zero.)

A \(20.0-\mu \mathrm{F}\) spherical capacitor is composed of two concentric metal spheres, one having a radius twice as large as the other. The region between the spheres is a vacuum. Determine the volume of this region.

See all solutions

Recommended explanations on Physics Textbooks

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.