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Fundamentals of Physics, 2011 Assignment

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1 Chap24/11P, 33,44P; Chap25/8P,9P

Problem 1 (Chap 24/11P) A point charge q is placed at one corner of a cube of edgea. What is

the flux through each of the cube faces?

[Solution]

Note that when a point charge is placed at one corner of a cube, the field lines from the charge is

tangential to the three faces meeting at the point charge (See the above diagram). So the flux

through the these faces must be zero.

To find the flux through the other faces, we can build a large cube composed of 8 identical smaller

cubes and place the charge at the center of the large one (See the above diagram). According to

the Gauss’ law,

0encEq, the flux flowing out of the large cube is

0Eq.

Note that the faces of the large cube is composed of 24 faces of the smaller cube, and the fluxes

through each smaller face have the same value due to the symmetry, namely

024eachsmallfaceq.

This is just the flux through each of the other three faces of a cube when a point charge qis

placed at its corner.

[Problem 2] Chap24/25P. Charge is distributed uniformly throughout the volume of an infinitely

long cylinder of radius R. (a) Show that, at a distance r from the cylinder axis (forrR),

02rE

whereis the volume charge density. (b)Write an expression for E when rR .

[Solution] Due to the cylindrical symmetry of the charge distribution, we choose a circular

cylinder of radius rand height h, coaxial with the charged cylinder, as the Gaussian surface (as Fundamentals of Physics, 2011 Assignment

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2 shown in the diagram). Two end caps are included to ensure the Gaussian surface is closed. Then

the electric field lines will flow throught out of the cylindrical part of the Gaussian surface

uniformly and perpendicularly no matter rR or rR if the charge is positive.

: volume charge densityRrR?EhE: volume charge densityRrR?h

Accordingly, the electric flux through the Gaussian surface is given by

(2)EErh

where E is the field magnitude at the cylindrical surface of radius r and 2rh is the area of this

side surface.

On the other hand, the net charge enclosed by the Gaussian surface is given by

22(), for

(), for encrhrRqRhrR

Applying the Gauss’s law

0encEq, we have the magnitude of the electric field at a distance

r from the cylinder axis as

020, for 2, for 2rrRERrRr

The direction of the field is radially outward.

[Problem 3] Chap24/33. A planar slab of thickness d and volume charge density .

[Solution]

(1) The magnitude of the electric field inside the slab, at a distance x from the central plane of the

slab.

dxxEE Fundamentals of Physics, 2011 Assignment

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3 The charge distribution has a planar symmetry, so we choose a closed cylinder perpendicular to

the slab as the Gaussian surface. The end caps of area A are located at a distance x from the

central plane of the slab, above and below respectively.

From the symmetry, we know that the electric field E at any point on the end caps must have

the same magnitude and be perpendicular to the caps in an either outward or inward direction.( In

the above diagram, the electric field lines are assumed to flow out of the end caps. ) So the flux

through the end caps is determined by

12capcapEA

For the cylindrical part, the electric field is parallel to the surface at any point so that there is no

flux through it, namely,

0cyl

The total flux through the Gauss surface is given by

122clycapcapEA

And the charge enclosed by this Gauss surface is

2encxA

According the the Gauss’ law0/enc , we have

022/EAxA

So the magnitude of the electric field at a distance x

0/Ex

(2) The magnitude of the electric field outside the slab, at a distance x from the central plane of

the slab.

dxxEE Fundamentals of Physics, 2011 Assignment

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4

Similarly, we choose a closed cylinder as the Gaussian surface. The distance from the end caps

to the central plane now satisfies x>d/2, as shown in the above diagram.

The total flux through the Gaussian surface is still

122clycapcapEA,

But the enclosed charge is given by encAd.

Then the Gauss’ law 0/enc reduces to

02/EAAd

So that the magnitude of electric field outside the slab is always

0/2Ed,

independent of the distance x from the central plane.

[Problem 4] Chap.24/44P The figure shows a spherical shell of charge with uniform volme

charge density . Plot Edue to the shell for distances r from the center of t he shell ranging