Chapter 24: 61 - Excercises And Problems (page 658)

| A spherical ball of charge has radius R and total charge Q. The electric field strength inside the ball 1r … R2 is E1r2 = r4 Emax /R4 . a. What is Emax in terms of Q and R? b. Find an expression for the volume charge density r1r2 inside the ball as a function ofr.c. Verify that your charge density gives the total charge Q when integrated over the volume of the ball

Short Answer

Expert verified

The maximum electric field is

Emax=Q4πε0R2

So the charge density is

ρ=6Q4πR3r3R3

We can see that the charge density does equal Q when integrated over the volume of the sphere, given that's how it was derived.

Step by step solution

01

part (a) step 1: given information

We use Gauss' Law to find the electric field inside the slab of a given thickness.

Φ=E·da=Qε0

Where

Φis the Electric Flux

E is the Electric Field

d a is the area

Q is the total charge

ε0 is the permittivity

The Emax or the maximum electric field is equal to the electric field at the surface of the sphere of radius R. It can be evaluated by Gauss law for the total charge density Q and spherical area4πR2

Substituting all the values in Gauss Law:

Emax4πR2=Qε0

This reduces to

Emax=Q4πε0R2

02

part (b) step 1: given information

From part(a) the maximum electric field is

Emax=Q4πε0R2

Emax is the maximum Electric Field

R is the radius of sphere

Q is the total charge

ε0 is the permittivity

Let us assume a charge density

ρ=ρ0r3R3

Where

r is the radius

R is the radius of sphere

The differential charge d q inside the sphere with charge distribution \rho and differential volume d v is given by the expression:

dq=ρdv

The total charge in the sphere can be evaluated as

dq=ρ0r3R34πr2dr

Integrating we get:

Q=dq=ρ0r3R34πr2dr

Evaluating the integral between 0 and Rwe get:

Q=4πρ0R66R3

From the above equation we can confirm that

ρ0=6Q4πR3

To evaluate the total charge inside the sphere, we have to repeat the process above except that we must replace R with r

Q(r)=4πρ0r66R3

The electric field inside the sphere can be evaluated as

E=Q(r)4πr2ε0R2r6R4

Which is nothing but

E=Emaxr4R4

03

part (c) step 1: given information

From part(b) the charge density is

ρ=6Q4πR3r3R3

Where

r is the radius

R is the radius of sphere

Q is the total charge

ε0 is the permittivity

The total charge is given by

Q=dq=ρdv

Where

d v is the differential volume

Q is the total charge

ρis the charge density

From the above equation, substitute

ρ=6Q4πR3r3R3

and the differential volume dv=4πr2drand limits 0 and R. The total charge is:

Q=0Rρdv=0R6Q4πR3r3R34πr2dr

Q=0R6Q4πR3r3R34πr2dr=0R6Qr5R6dr=6QR66R6=Q

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Most popular questions from this chapter

A tetrahedron has an equilateral triangle base with20-cm-long edges and three equilateral triangle sides. The base is parallel to the ground, and a vertical uniform electric field of strength 200N/C passes upward through the tetrahedron. a. What is the electric flux through the base? b. What is the electric flux through each of the three sides?

FIGURE EX24.17shows three charges. Draw these charges on your paper four times. Then draw two-dimensional cross sections of three-dimensional closed surfaces through which the electric flux is (a) 2q/ϵ0, (b) q/ϵ0, (c) 0,and (d) 5q/ϵ0.

The conducting box in FIGURE EX24.26 has been given an excess negative charge. The surface density of excess electrons at the center of the top surface is 5.0×1010electrons/m2 . What are the electric field strengths E1to E3at points 1 to 3?

A small, metal sphere hangs by an insulating thread within the larger, hollow conducting sphere of FIGURE Q24.10. A conducting wire extends from the small sphere through, but not touching, a small hole in the hollow sphere. A charged rod is used to transfer positive charge to the protruding wire. After the charged rod has touched the wire and been removed, are the following surfaces positive, negative, or not charged? Explain. a. The small sphere. b. The inner surface of the hollow sphere. c. The outer surface of the hollow sphere.

The electric field is constant over each face of the cube shown in FIGURE EX24.4. Does the box contain positive charge, negative charge, or no charge? Explain.

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