Chapter 22 solutions and Explanations

1. A charge Q is positioned at the center of a sphere of radius R. The flux of the electric field through the sphere is equal to . If the charge Q is now placed at the center of a cube the flux of the electric field through the surface of the cube is equal to

0.

.

/2.

2.

The value of the flux depends on the dimensions of the cube.

Explanation:

The total electric flux through an enclosed surface due to a charge trapped inside is constant regardless of the surface

shape or size.

2. A charge q = 2 C is placed at the origin in a region where there is already a uniform electric

field iCNE )/100(

. Calculate the flux of the net electric field through a Gaussian sphere of radius R = 10

cm centered at the origin.

1.13 105 N m2/C

5.52 105 N m2/C

0.565 105 N m2/C

2.26 105 N m2/C

0

Explanation:

A uniform electric field means that the field direction is well defined and that the spacing between the field lines is

identical. The already existing field would go through the proposed Gaussian surface from one side and exit from the

other side. It will have no net flux at the surface of my Gaussian (remember that flux in is negative while the flux-out

is positive; thus, the two quantities cancel out since we have the same number of lines going in as those going out).

The charge at the origin being inside the gaussian will produce the total net flux. We know that

enclosedq so the

flux is: CmN /1026.21085.8

102 2512

6

. We would have obtained a negative flux if the charge were to be

negative.

3. A non-conducting sphere of radius R = 7 cm carries a charge Q = 4 mC distributed uniformly throughout its volume. At what distance, measured from the center of the sphere does the electric field reach a value equal

to half its maximum value?

9.9 cm only

4.9 cm only

3.5 cm only

3.5 cm and 4.9 cm

3.5 cm and 9.9 cm

Explanation:

Remember how the electric field looks like inside and outside a NON-CONDUCTING or insulating sphere with a net

uniform charge distribution. The field inside the sphere (r R) starts from zero at the center of the sphere and

increases linearly with r using this equation: rR

QkE

3 to reach a maximum value at the surface of the sphere

2max R

QkE . Outside, the field is inversely proportional to r2, i.e.,

2r

QkE .

Looking at the graph of E versus r (check the slides), you will realize that there are two locations where the electric

field would be half the maximum value. One location is inside the sphere and the other is outside.

If r is the distance where E=Emax/2

Then (i) when rR

QkE

3 = Emax/2 = 2R2

Qk r = R/2 = 3.5 cm

and (ii) when 2r

QkE = Emax/2 = 2R2

Qk r = 2 R = 9.9 cm

What would be the electric field inside a conducting sphere? Where would the electric field be zero for either spheres

(conducting or non-conducting)?

4. A positive charge Q is located at the center of an imaginary Gaussian cube of sides a. The flux of the electric field through the surface of the cube is . A second, negative charge -Q is placed next to Q inside the cube. Which of the following statements will be true in this case?

The net flux through the surface of the cube is equal to zero

The net flux through the surface is equal to 2

The electric field on the surface of the cube is perpendicular to the surface

The magnitude of the net electric field is constant on the entire surface of the cube

The net electric field on the surface of the cube is equal to zero

Explanation:

The net charge inside the Gaussian cube is zero. Gauss law states that the net electric flux through any closed surface

equals the net charge inside the surface divided by 0. The net charge is zero so the net flux through the Gaussian

surface must be zero as well.

5. A solid non-conducting sphere of radius R carries a uniform charge density. At a radial distance r1 = R/4 the electric field has a magnitude E0. What is the magnitude of the electric field at a radial distance r2 = 2R?

0

E0/2

E0/4

E0

2E0

Explanation:

This is somewhat Similar to Question 3. We know the electric field equations inside and outside the nonconducting

sphere. Plug the values of r1 and r2 in the appropriate equations and check.

For r1 = R/4 (r R) Eo = 13 rR

Qk =

4

R

R

Qk

3=

2R4

Qk

For r2 = 2R (r R) E = 2)R2(

Qk =

2R4

Qk = Eo

6. An advantage in evaluating surface integrals related to Gauss's law for symmetric charge distributions is

the charge is always on the surface.

the electric field is a constant on any surface.

the flux is inward.

the electric field is of constant magnitude on certain surfaces.

the flux is outward.

Explanation:

Remember that we were trying hard in class to imagine Gaussian surfaces that mimic the charge distribution

symmetry in such a way we render E and the angle between E and dA constant if possible. When we assume

symmetric charge distribution, it is possible to find a simple Gaussian surface over which the surface integral is easily

evaluated.

While its possible to find surfaces on which the electric field may be constant, this is not always possible so option B

must be wrong. Options C and E depend on the sign of the charge enclosed and have nothing to do with symmetries

and the shape of the surface. Option A is obviously wrong because we can imagine the surface anywhere we like and

of any size which means that the charges may be at any location inside or even outside (if we wish to exclude them).

7. An infinitely long cylinder of radius R = 2 cm carries a uniform charge density = 18 C/m3. Calculate the electric field at distance r = 1 cm from the axis of the cylinder.

2.0 103 N/C

0

5.1 103 N/C

10.2 103 N/C

2.5 103 N/C

Explanation:

This problem requires imagining a Gaussian surface inside the cylinder. The best shape to imagine is obviously also

cylindrical. The actual cylinder (lets think of it as a solid think rod to avoid confusion with the imaginary

cylindrical Gaussian surface) and the Gaussian cylinder have the same axis but with different radii and heights.

To find the electric field inside the rod, we will need to solve both versions of gausss law. Lets start with

onlysurfacecurved

AdEAdE

. Notice that the top and bottom caps of my Gaussian cylindrical surface will not have any

flux through them. Therefore, the closed integral changed to a bounded integral over the curved side of the

cylindrical Gaussian. With this choice, E is constant at the surface of my Gaussian and the angle between E and dA is

zero everywhere around the Gaussian. The integral can be easily solved to give )2( rhE (please check the slides

for a similar problem) where r is the radius of my Gaussian and h represents any arbitrary height.

Now the second form of gausss law relates the flux to the charge enclosed inside

enclq . We need to figure out

how much charge is trapped inside my Gaussian surface. The problem doesnt explicitly say it but you should realize

that we have a uniform charge distribution throughout the physical cylinder because we were given the VOLUME

charge density .

The charge enclose within my Gaussian is thus times the volume selected (of my Gaussian). The volume of my

Gaussian cylinder is the base area (cap area) times the cylinder height or (r2)h. Thus,

)hr ( 2 encl

q.

Equating the two flux equations allow me to drop the height h.

2

)hr ( )2(

2 rEorrhE

Then, the electric field inside an infinitely long insulating cylinder (with r R) is E = r/2o

= 18 x 10-6 x (1 x 10-2) / (2 x 8.85 x 10-12) N/C = 10.2 x 103N/C

8. Consider a spherical Gaussian surface of radius R centered at the origin. A charge Q is placed inside the sphere. Where should the charge be located to maximize the magnitude of the flux of the electric field

through the Gaussian surface?

at the origin

at x = 0, y = R/2, z = 0

at x = R/2, y = 0, z = 0

at x = 0, y = 0, z = R/2

The flux does not depend on the position of the charge as long as it is inside the sphere

Explanation:

It doesnt matter where its located as long as it is inside the Gaussian surface. The flux only depends on the net

charge enclosed but not on its position.

9. Consider two oppositely charged, parallel metal plates. The plates are square with sides L and carry charges Q and -Q. What is the magnitude of the electric field in the region between the plates?

2o L

Q2E

2

o L2

QE

0E

2

o L

Q4E

2

o L

QE

Explanation:

Weve seen a related problem in chapter 21. Inside such a device that we call capacitor, the electric field is constant.

E = / o = (Q/L2)/ o = Q/ oL

2 . Remember the surface charged density "" can be calculated by dividing the total

charge Q by the available area A on which its placed (here A = L2).

10. The figure shows four Gaussian surfaces surrounding a distribution of charges. Which Gaussian surfaces have an electric flux of +q/o through them?

a.

b.

c.

b and d.

b and c.

Explanation:

We just need to examine each surface one at a time and count only the charges within. Do not count anything outside

the surface no matter how tempting it might be.

For Gaussian surface a, the total charge enclosed = +2q + q - q = +2q a = +2q/o

For Gaussian surface b, the total charge enclosed = +2q - q = q b = +q/o

For Gaussian surface c, the total charge enclosed = +q - q = o c = 0

For Gaussian surface d, the total charge enclosed = -q d = -q/o

11. Fig xx above shows four Gaussian surfaces surrounding a distribution of charges. Which Gaussian surfaces have no electric flux through them?

a.

b.

c.

b and c.

b and d.

Explanation:

Solved above in problem #10.

12. A uniform electric field with a magnitude of 6 106 N/C is applied to a cube of edge length 0.1 m as shown in the Figure. If the direction of the E-field is along the +x-axis, what is the electric flux passing through the

shaded face of the cube?

60 104 Nm2/C

600 104 Nm2/C

6 104 Nm2/C

600 0 104 Nm2/C

0.6 104 Nm2/C

Explanation:

Notice that the proposed electric field direction is completely parallel to the x-axis. The shaded area is fully

perpendicular to that field so it should intercept the maximum flux. Now, lets think of the angles between vectors E

and A. Remember that vector A represent the area itself and is perpendicular to it. So, it must be parallel to E and the

angle between the two vectors is zero.

We know, the electric flux through a flat surface is

= EA cos = 6 x 106 x (0.1)2 cos(0o) = 6 x 104 Nm2/C

13. Gauss's law can be applied using any surface.

True

False

Explanation:

False, because we know by now that Gausss law only applies to any closed surface regardless of its size or shape.

14. Gauss's law may be applied only to charge distributions that are symmetric.

True

False

Explanation:

False, it can be applied to any charge distribution but its much easier when the distribution exhibits some symmetry.

Then, we can imagine a symmetrical surface as well.

15. Gaussian surfaces A and B enclose the same positive charge +Q. The area of Gaussian surface A is three times larger than that of Gaussian surface B. The flux of electric field through Gaussian surface A is

nine times larger than the flux of electric field through Gaussian surface B.

three times larger than the flux of electric field through Gaussian surface B.

three times smaller than the flux of electric field through Gaussian surface B.

unrelated to the flux of electric field through Gaussian surface B.

equal to the flux of electric field through Gaussian surface B.

Explanation:

Check the solution of problem #1. Gauss law states that the net electric flux through any closed surface equals the

net charge inside the surface divided by 0. No dependence on the surface size or shape.

16. If a charge is located at the center of a spherical volume and the electric flux through the surface of the sphere is o, what is the flux through the surface if the radius of the sphere doubles?

5 o

0.500 o

o

8 o

0.125o

Explanation:

The same, how many times should we say this? see solutions of 1 and 15.

17. If a closed surface surrounds a dipole, the net flux through the surface is zero.

True

False

Explanation:

Absolutely, the dipole consists of two charges that are equal in magnitude but having opposite signs. Therefore, the

net charge of the dipole as a whole would be zero. The next flux through a surface surrounding one or more dipoles is

zero.

18. If a point charge is located at the center of a cube and the electric flux through one face of the cube is 5.0 Nm2/C, what is the total flux leaving the cube?

25 Nm2/C

30 Nm2/C

1 Nm2/C

20 Nm2/C

5.0 Nm2/C

Explanation:

What works in our favor here is that the charge is located at the center of a cube. This is a highly symmetrical

position and the distances from the charge to one face are exactly the same to the distances to all the remaining cube

faces. If we know how much flux went through one face, all I have to do is to multiply this value by the number of

the cube faces (6). Thus the total flux leaving the cube = 6 x 5.0 Nm2/C = 30 Nm

2/C. The problem wouldnt be that

obvious if we were to shift the charge off-center. I would have no clue then how much flux is going through the other

faces.

19. If a point charge is located at the center of a cylinder and the electric flux leaving one end of the cylinder is 20% of the total flux leaving the cylinder, what portion of the flux leaves the curved surface of the cylinder?

20%

60%

100%

40%

80%

Explanation:

This is another example showing you how to use symmetry to your advantage. The charge is located at the center of

the cylinder. We know how much flux goes through one end (cap). The opposite end (cap) must receive the same

flux (because of the symmetrical location of the charge). Therefore, we have a total of 2 x 20% leaving through the

caps. This leaves 100 - 40 = 60% flux leaving through the curved surface of the cylinder.

20. If a rectangular area is rotated in a uniform electric field from the position where the maximum electric flux goes through it to an orientation where only half the flux goes through it, what has been the angle of

rotation?

Explanation:

This is a flat surface intercepting an electric field. The electric flux can be calculated using the now familiar equation:

= EA cos . Maximum flux means that the angle between E and A = 0o or max = EA cos0o = EA. Notice that I do

not need to know the values of E or A in this problem.

When the flat area rotated in such a way it now receives half the maximum flux, I should know that the flux is now

= max/2 = EA/2 = EA cos. This equality shows that cos = or = 60o.

21. If the electric flux through a circular area is 5.0 Nm2/C, what is the electric flux through a circle of double the diameter assuming the orientations of the circles are the same and the electric field is uniform?

Explanation:

We have a flat circular area (disk) of area A. We also have a uniform electric field so the lines are pointing in the

same direction with equal spacing. This is very important to know. Now double the diameter of the disk and

calculate the new area A. Calculate the ratio between the new area A and the old area A. This is how much more

surface area is now available to intercept more flux.

Initial disk area: A = d/2)2 = () d)2

New disk area: A= d/2)2 = d/2)2d)2

The ratio A/A = d)2 / () d)2 = 4

This last result means that I have 4 times more surface area. As such the flux intercepted using the larger disk is 4 x 5

= 20 Nm2/C

22. If the electric flux through a rectangular area is 5.0 Nm2/C, and the electric field is then doubled, what is the resulting flux through the area?

5.0 Nm2/C

20 Nm2/C

2.5 Nm2/C

1 Nm2/C

10 Nm2/C

30

26.6

60

45

90

1.0 Nm2/C

2.5 Nm2/C

5.0 Nm2/C

10.0 Nm2/C

20 Nm2/C

Explanation:

Remember that the flux equation through a flat surface is: = EA cos. The flux would change if we separately

change any of these quantities. Doubling E in this equation means that were doubling how much flux that can go

through the fixed area A at the same angle . As such, = 2 x 5 = 10 Nm2/C.

23. If the net flux through a closed surface is positive, then the net charge enclosed must be positive.

True

False

Explanation:

Please check the slides. We already demonstrated that the flux is positive when the electric field lines exit the surface.

The flux is negative when the electric field lines enter the Gaussian surface. For the flux to be positive through the

proposed surface in this problem, the charge within must then be positive as well (electric field lines emitted to the

outside).

24. If the net flux through a closed surface is zero, then there can be no charge or charges within that surface.

True

False

Explanation:

While its certainly possible that no charges exist inside my chosen closed Gaussian surface, the statement in this

problem is obviously wrong because it read as a general rule or condition that must be met. There are many situations

where we have plenty of charges enclosed within a surface but the net charge = 0; meaning we have as many positive

as negative charges inside the surface. Just think of a Gaussian surface wrapped around an uncharged neutral piece of

metal. Since there is no excess charge on the metal, there would be no net electric flux on the Gaussian surface.

However, the neutral metal itself contains an enormous number of charged electrons and protons but in equal amounts

so the net flux remains zero.

25. Outside a spherically symmetric charge distribution of net charge Q, Gauss's law can be used to show that the electric field at a given distance

must be greater than zero.

must be zero.

acts like it originated in a point charge Q at the center of the distribution.

must be directed outward.

must be directed inward.

Explanation:

Please check the last few slides of Chapter 22. We demonstrated in class that outside ( r R) a nonconducting spherical charge distribution, the electric field can be calculated using the same equation derived for a point-charge,

namely: 2r

QkE . Remember that the distance r is measured from the center of my spherical charge distribution to

the point I am interested in outside the sphere. This means that I am treating this sphere exactly as I would treat a

point-like charge; therefore the two act similarly and the charge appears as if it were completely at the center of the

distribution.

Of course, the stuation inside the spherical charge distribution is completely different. Remember that the electric

field inside increase from 0 at the center of the sphere and increases linearlyas a function of r until I reach the

surface. Outside, we switch to 2r

QkE .

What about a spherical shell of charges or a charged metallic sphere? Please make sure to review those slides

carefully and the related problems in this chapter and in chapter 21.

26. The electric field in a region of space is oriented along the positive y axis. A circle of radius R is placed in the xz-plane. The flux of the electric field through this circle is . The same electric field passing through a second circle of radius 2R parallel to xz-plane would result in a flux equal to

4.

.

0.

2.

3.

Explanation:

This is similar to problem #21. There we doubled the diameter. When d = 2R, doubling d you would get d = 2R =

2d = 4 R or the new radius R is 2R. Therefore, the flux through the larger flat circular area is four times as much as

the small circle.