/*! 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 60 A solid sphere \(10 \mathrm{cm}\... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 21: Problem 60

A solid sphere \(10 \mathrm{cm}\) in radius carries a \(40-\mu \mathrm{C}\) charge distributed uniformly throughout its volume. It's surrounded by a concentric shell \(20 \mathrm{cm}\) in radius, also uniformly charged with \(40 \mu \mathrm{C}\). Find the electric field (a) \(5.0 \mathrm{cm},\) (b) \(15 \mathrm{cm},\) and (c) \(30 \mathrm{cm}\) from the center.

Short Answer

Expert verified
The electric field at 5.0cm is 9.0\(脳10^{5} NC^{-1}\), at 15cm is 4.0\(脳10^{5} NC^{-1}\) and at 30cm is 4.0\(脳10^{5} NC^{-1}\).

Step by step solution

01

The Principle of Superposition

To find the electric field at a point in space, you must, according to the principle of superposition, find the electric field created by each charge distribution and then add them up.
02

Field Inside Solid Sphere

At distance 5.0cm which is inside the solid sphere, the electric field is given by \(E = \frac{KQr}{R^{3}}\), where \(K = 9.0脳10^{9} Nm^{2}C^{鈭�2}\) (Coulomb's constant), \(Q = 40碌C\) (charge on the sphere), \(r = 5.0cm\) (distance from center), and \(R = 10cm\) (radius of the sphere). Calculate the electric field.
03

Field Between Solid Sphere and Shell

At distance 15cm which is between the sphere and the shell, the electric field will only be due to the sphere as the field inside of a shell is zero. Hence the electric field is given by \(E = \frac{KQ}{r^{2}}\), where \(r = 15cm\). Calculate the electric field.
04

Field Outside Both Sphere and Shell

At distance 30cm which is outside both the sphere and the shell, both the sphere and the shell will contribute to the total electric field. Since both the sphere and shell have the same charge, their contribution to the total field at this point will be the same. Hence the electric field is given by \(E = \frac{2KQ}{r^{2}}\), where \(r = 30cm\). Calculate the electric field.

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!

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

What surface charge density on an infinite sheet will produce a 1.4-kN/C electric field?

A long, thin wire carrying \(5.6 \mathrm{nC} / \mathrm{m}\) runs down the center of a long, thin-walled, pipe with radius \(1.0 \mathrm{cm}\) carrying \(-4.2 \mathrm{nC} / \mathrm{m}\) spread uniformly over its surface. Find the electric field (a) \(0.50 \mathrm{cm}\) from the wire and (b) \(1.5 \mathrm{cm}\) from the wire.

Calculate the electric fields in Example 21.2 directly, using the superposition principle and integration. Consider the shell to be composed of charge elements that are coaxial rings, whose axes pass through the field point, which is a distance \(r\) from the center. (Hint: Consult Example \(20.6 .\) You'll have to evaluate the cases \(r< R \text { and } r>R\) separately.)

Can electric field lines ever cross? Why or why not?

You measure the electric field strength at points directly above the center of a square plate carrying charge spread uniformly over its surface. The data are tabulated in the next column, with \(x\) the perpendicular distance from the center of the plate. Use the data to determine (a) the total charge on the plate and (b) the plate's size. Hint: You'll need to consider separately data taken close to the plate and also far away. For the latter, plot \(E\) versus a quantity that should yield a straight line. $$\begin{array}{|l|l|l|l|l|l|l|}\hline x(\mathrm{cm}) & 0.01 & 0.02 & 1.2 & 6.0 & 12.0 & 24.0 \\\\\hline E(\mathrm{N} / \mathrm{C}) & 5870 & 5860 & 4840 & 1960 & 754 & 221 \\\\\hline\end{array}$$ $$\begin{array}{|l|l|l|l|l|l|}\hline x(\mathrm{cm}) & 48.0 & 72.0 & 96.0 & 120 & 240 \\\\\hline E(\mathrm{N} / \mathrm{C}) & 57.6 & 26.7 & 16.1 & 8.45 & 2.34 \\\\\hline\end{array}$$
See all solutions

Recommended explanations on Physics Textbooks

Thermodynamics

Read Explanation

Astrophysics

Read Explanation

Magnetism

Read Explanation

Classical Mechanics

Read Explanation

Scientific Method Physics

Read Explanation

Fundamentals of 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.