Figure 23-27 shows four solid spheres, each with charge Quniformly distributed through its volume. (a) Rank the spheres according to their volume charge density, greatest first. The figure also shows a point for each sphere, all at the same distance from the center of the sphere. (b) Rank the spheres according to the magnitude of the electric field they produce at point P, greatest first.

Short Answer

Expert verified
  1. The rank of the spheres according to their volume charge densities is.a>b>c>d
  2. The rank of the spheres according to the magnitude of the electric field at point P is.Ea>Eb>Ec>Ed

Step by step solution

01

The given data:

Figure 23-27 showing four solid spheres, each with charge Q uniformly distributed through its volume are given.

02

Understanding the concept of volume charge density and electric field:

The volume charge density of a body is given by the amount of charge carries by the body divided by its volume. Now, using this volume charge density relation with the electric field of the sphere, the rank of the spheres can be determined for different positions of the point P.

Formulae:

The volume charge density of a sphere,

ρ=q43πR3….. (i)

The electric field at point inside the sphere,

E=ρr3ε0(r<R) ….. (ii)

03

(a) Calculation of the rank of spheres according to volume charge densities:

From equation (i), it can see that the volume charge density of a sphere(ρ) is inversely proportional to the radius of the sphere(R). That is given as:

ρ1R3

Thus, as the radius of the sphere increases the volume charge density decreases.

The rank of the radius of the spheres from the figure can be given as:

d>c>b>a

Hence, the rank of the spheres according to charge densities isa>b>c>d.

04

(b) Calculation of the rank of the spheres according to the magnitude of electric field:

For situation a and b, the points location is outside the sphere.

Thus, their electric field outside the sphere is same as that at the surface of the sphere can be given using equation (ii) for the same charge spread:

E=q4πε0r2 …… (iii)

Here, r is the distance of point P from the center of the sphere,

Again, the radius of the sphere is small than that of sphere b, thus the electric field value can be given as:

Ea>Eb

The sphere a and b has the same magnitude of the electric field, because for both the sphere, the charge q and distance r are the same.

Now, for the spheres and , the points are located inside the sphere, thus, the electric field is given by equation (ii).

Therefore,

Eρr

If you assume that the point to be equally at distance from the center, then the electric field would be given by:

ErR3 ….. (iv)

Thus, the rank of electric fields for c and d would be

Because the sphere is having the volume charge density greater than the sphere d .

Now, considering equations (iii) and (iv), the final rank of the spheres according to fields is given as:

Ea>Eb>Ec>Ed

Hence, the require rank is.Ea>Eb>Ec>Ed

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 Vaia!

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 long, straight wire has fixed negative charge with a linear charge density of magnitude 3.6nC/m . The wire is to be enclosed by a coaxial, thin-walled non-conducting cylindrical shell of radius 1.5 cm . The shell is to have positive charge on its outside surface with a surface charge density s that makes the net external electric field zero. Calculate s.

A charged particle causes an electric flux of -750 N.m2/Cto pass through a spherical Gaussian surface of 10.0 cmradius centered on the charge.

(a) If the radius of the Gaussian surface were doubled, how much flux would pass through the surface?

(b) What is the charge of the particle?

The chocolate crumb mystery. Explosions ignited by electrostatic discharges (sparks) constitute a serious danger in facilities handling grain or powder. Such an explosion occurred in chocolate crumb powder at a biscuit factory in the 1970 s. Workers usually emptied newly delivered sacks of the powder into a loading bin, from which it was blown through electrically grounded plastic pipes to a silo for storage. Somewhere along this route, two conditions for an explosion were met: (1) The magnitude of an electric field became3.0×106N/Cor greater, so that electrical breakdown and thus sparking could occur. (2) The energy of a spark was150mJor greater so that it could ignite the powder explosively. Let us check for the first condition in the powder flow through the plastic pipes. Suppose a stream of negatively charged powder was blown through a cylindrical pipe of radiusR=5.0cm. Assume that the powder and its charge were spread uniformly through the pipe with a volume charge density r.

(a) Using Gauss’ law, find an expression for the magnitude of the electric fieldin the pipe as a function of radial distance r from the pipe center.

(b) Does E increase or decrease with increasing r?

(c) IsEdirected radially inward or outward?

(d) Forρ=1.1×103C/m3(a typical value at the factory), find the maximum E and determine where that maximum field occurs.

(e) Could sparking occur, and if so, where? (The story continues with Problem 70 in Chapter 24.)

A uniformly charged conducting sphere of1.2 mdiameter has surface charge density 8.1 mC/m2 . Find (a) the net charge on the sphere and (b) the total electric flux leaving the surface.

Space vehicles traveling through Earth’s radiation belts can intercept a significant number of electrons. The resulting charge buildup can damage electronic components and disrupt operations. Suppose a spherical metal satellite1.3min diameter accumulates2.4μCof charge in one orbital revolution. (a) Find the resulting surface charge density. (b) Calculate the magnitude of the electric field just outside the surface of the satellite, due to the surface charge.

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.

Sign-up for free