Chapter 24
24.17 (a) Consider the front surface of the trapezoidal box in Figure 24.22a, detached from the rest of the trapezoidal box. Does the field line flux through that surface increase, decrease, or stay the same if any of the following quantities is increased: (i) the area of the front surface, (ii) the magnitude of the electric field (keeping the area constant), (iii) the slope of the front surface (that is to say, the angle between the surface and the direction of the electric field is increased)? (b) Does the electric flux through the front surface increase, decrease, or stay the same if any of these quantities is changed?
(a) (i) Increases (more field lines intercepted); (ii) increases (greater field line density so more field lines intercepted); (iii) increases (more field lines intercepted). (b) (i) Increases (greater A yields greater ΦE , by Eq. 24.1); (ii) increases (greater E S yields greater ΦE); (iii) increases (greater slope means smaller u, which yields a greater cos u).
24.3 Imagine a hollow sphere enclosing the charged object in Figure 24.5, centered on the object. (a) Given that 26 field lines emanate from the charged object, how many field lines cross the surface of the hollow sphere? (b) If the radius of the hollow sphere is R, what is the number of field line crossings per unit surface area? (c) Now consider a second sphere with radius 2R, also centered on the charged object. How many field lines cross the surface of this second sphere? (d) How does the number of field line crossings per unit area on the second sphere compare with that on the first sphere? (e) How does the electric field at a location on the second sphere compare with the field at a location on the first sphere?
(a) As illustrated in Figure S24.3, the same 26 field lines pass through the surface of the hollow sphere. (b) The surface area of a sphere of radius R is 4pR2 , so the number of field lines per unit area is (26)>(4pR2 ). (c) Again 26. (d) The number of field lines per unit area on the second sphere is (26)>[4p(2R) 2 ] = (26)>(16pR2 ), which is reduced by a factor of 4 from that for the sphere of radius R. (e) The electric field decreases as 1>r 2 , so doubling the distance reduces the electric field by a factor of 4.
24.19 Consider a spherical Gaussian surface of radius r with a particle that carries a charge +q at its center. (a) What is the magnitude of the electric field due to the particle at the Gaussian surface? (b) What is the electric flux through the sphere due to the charged particle? (c) Combining your answers to parts a and b, what is the relationship between the electric flux through the sphere and the enclosed charge qenc? (d) Would this relationship change if you doubled the radius r of the sphere?
(a) At a distance r from a particle carrying a charge +q, the magnitude of the electric field is kq>r 2 (see Eq. 23.4). (b) Because the electric field is perpendicular to the sphere at all points and has the same constant magnitude, the electric flux is given by the product of the electric field and the area of the sphere: ΦE = EA = E(4pr 2 ) = (kq>r 2 )(4pr 2 ) = 4pkq. (c) The enclosed charge is +q, so ΦE = 4pkqenc. (d) No. If we double the radius r, the field decreases by a factor of 4 but the area of the sphere increases by a factor of 4. Thus the electric flux and its relationship to qenc remain the same.
24.7 (a) If more than one field line reenters the donut in Figure 24.9a, what happens to the field line flux? (b) Are there any closed surfaces enclosing a charged particle through which the field line flux is different from that through a simple sphere around that particle?
(a) Each field line that reenters the donut must also exit, contributing a value of (+1) + (-1) = 0 to the field line flux. Thus, regardless of how many field lines reenter the donut, the field line flux remains 6. (b) No. Regardless of the shape of the surface, each field line contributes no more and no less than +1 to the field line flux, so the field line flux is always equal to 6.
24.10 Consider the two-dimensional field line diagram in Figure 24.11, part of which is hidden from view. (a) If the object in the top left carries a charge of +1 C, what is the charge enclosed in the region that is hidden? (b) What is the field line flux through a surface that encloses the entire area represented by the diagram?
(a) Fifteen field lines go into the surface and three come out of it. The field line flux is thus (-15) + (+3) = -12. Twelve field lines emanate from the object carrying a charge of +1 C, so the charge inside the hidden region must be -1 C. Figure S24.10 shows the entire field line diagram; as you can see, the charge enclosed by the dashed line is indeed (+3 C) + (-4 C) = -1 C. (b) Going along the perimeter of the illustration, we note that four field lines leave the edge of the illustration and four enter it. The field line flux is thus (+4) + (-4) = 0. This makes sense because the charge enclosed within the diagram is (+1 C) + (-1 C) = 0.
24.5 Imagine moving the hollow sphere of radius R of Checkpoint 24.3a sideways so that the charged object is no longer at the center of the sphere (but still within it). (a) How does the number of field line crossings through the surface of the sphere change as it is moved? (b) How does the average number of field line crossings per unit surface area of the sphere change? (c) Does the electric field at a fixed position on the surface of the sphere change or remain the same as the sphere is moved? (d) Are your answers to parts b and c in contradiction, given the relationship between the electric field magnitude and field line crossings per unit area?
(a) It remains the same—each field line still passes through the surface of the sphere. (b) Because neither the number of field lines nor the surface area of the sphere changes, the average number of field line crossings per unit surface area remains the same. (c) The electric field strength increases as the distance to the charged object decreases, so moving the sphere off-center increases the electric field strength on one side and decreases it on the opposite side. (d) No. The answer to part b gives the average field line density. Moving the sphere off-center increases the field line density on one side and decreases it on the other, so the average field line density can remain the same.
24.13 Consider a point on the right surface of the Gaussian surface in Figure 24.15. Does the magnitude of the electric field at that point increase, decrease, or stay the same if you (a) change the location of the point on the right surface, or (b) increase the height h of the Gaussian surface? (c) Is the field line flux through the right surface of the Gaussian surface positive, negative, or zero? (d) How does the field line flux through the right surface compare to that through the left surface? (e) Is the field line flux through the curved surface of the Gaussian surface positive, negative, or zero?
(a) It stays the same because the field line density is the same everywhere along a plane parallel to the charged sheet. (b) It stays the same because the field line spacing doesn't change. (c) Positive, because the field lines cross from inside the surface to outside. (d) They are the same in magnitude and in sign. (e) Zero. The field lines are all perpendicular to the sheet, so no field lines pass through the curved surface.
24.12 Consider a point on the curved part of the Gaussian surface in Figure 24.14. Does the magnitude of the electric field at that point increase, decrease, or stay the same if you (a) change the location of the point on the curved surface or (b) increase the radius of the Gaussian surface? (c) What is the field line flux through the left and right surfaces of the Gaussian surface?
(a) It stays the same because the field line density is the same everywhere on that surface. (b) It decreases because the field line spacing increases. (c) Zero. The field lines are all perpendicular to the wire, so no field lines pass through those surfaces.
24.4 (a) In Figure 24.6, for what orientation is the number of field lines that cross the surface a minimum? (b) How many field lines cross a plane surface of area 0.5 m2 placed perpendicular to the field lines in Figure 24.6? (c) Using your answer to part b, what is the number of field line crossings per unit area through the 0.5@m2 surface? (d) How does this compare to the number of field line crossings per unit area for the 1@m2 surface in Figure 24.6a?
(a) No field lines cross the surface when it is parallel to the field lines. (b) If 16 field lines pass through an area of 1 m2 , then 8 field lines pass through an area of 0.5 m2 . (c) The number of field lines per unit area is (8)>(0.5 m2 ) = 16 m-2 , or 16 field lines per square meter. (d) (16)>(1 m2 ) = 16 m-2 , as above. The field line density is the same because the field is uniform.
24.2 (a) Is it possible for two electric field lines to cross? (Hint: What is the direction of the electric field at the point of intersection?) (b) Can two electric field lines touch?
(a) No, because at the point of intersection the direction of the electric field would not be unique—it cannot be tangent to both intersecting lines at the same time. (b) No. Although two field lines that touch have the same tangent at the point where they touch, the two field lines would have different tangents on either side of that point, and so the direction of the electric field would again not be unique.
24.6 Suppose eight field lines emanate from an object carrying a charge +q. How many field lines pierce the surface of a hollow sphere if the sphere contains (a) a single object carrying a charge +2q and (b) two separate objects, each carrying a charge +q? (c) If the sphere is pierced by 20 field lines, what can you deduce about the combined charge on objects inside the sphere?
(a) Sixteen field lines emanate from the object, so 16 field lines pass through the surface of the sphere. (b) Eight field lines emanate from each of the objects, so 16 field lines pass through the surface of the sphere. (c) The amount of charge enclosed by the sphere is (20 field lines)>(8>q field lines per unit charge) = 2.5q.
24.18 Let the area of the back surface of the trapezoidal box in Figure 24.22 be 1.0 m2 , the magnitude of the electric field be 1.0 N>C, and u = 30° for the front surface. (a) What are the magnitudes of the area vectors for the front and back surfaces of the trapezoidal box? (b) What are the electric fluxes through the front and back surfaces?
(a) The area of the back surface is 1.0 m2 , so the magnitude of the corresponding area vector A S back is 1.0 m2 . From Figure S24.18 on the next page we see that h>hfront = cos u = cos 30° = 0.87, so the area of the front surface is larger by a factor of hfront>h = 1.2. The magnitude of the area vector A S front is thus 1.2 m2 . (b) We use Eq. 24.2 to calculate the electric fluxes. For the back surface we get ΦE = EAback cos (180°) = (1.0 N>C)(1.0 m2 )(-1) = -1.0 N # m2>C; for the front surface we have ΦE = EAfront cos u = (1.0 N>C)(1.2 m2 ) (0.87) = 1.0 N # m2>C.
24.24 (a) A very large metal plate of surface area A carries a positive charge q. What is the surface charge density of the plate? What is the magnitude of the field created by the plate? (b) A very large, thin nonconducting sheet of surface area A carries a fixed, uniformly distributed positive charge q. What is the surface charge density of the sheet? What is the magnitude of the field created by the sheet?
(a) The charge spreads out over the outside surface area of the plate. If we ignore the edges of the plate, 1 2q spreads out over each side of surface area A, and so the surface charge density (charge per unit surface area) is 1 2q>A on each side. The magnitude of the field is given by Eq. 24.17: E = s>P0 = ( 1 2q>A)>P0 = q>(2P0A). (b) The surface charge density is q>A; the magnitude of the electric field is given by the solution of Exercise 24.8: E = s>(2P0) = (q>A)>(2P0) = q>(2P0A), which is the same result as in part a. The point to remember therefore is that the difference between the solutions of Exercise 24.8 and Eq. 24.17 arises solely from the difference in the surface charge that should be used in each of them.
24.8 (a) What is the field line flux through the closed spherical surface in Figure 24.9b due to a charged particle outside the sphere? (b) Does your answer to part a change if we move the particle around (but keep it outside the volume enclosed by the surface)?
(a) The field line flux is zero because each field line that enters the sphere also exits the sphere, and so does not contribute to the field line flux. (b) No. Moving the particle changes the number of field lines that enter the sphere, but regardless of how many field lines enter the sphere, each of them must also leave the sphere, so the field line flux remains zero.
24.14 Consider a spherical Gaussian surface inside a positively charged conducting object that has reached electrostatic equilibrium. (a) Is the field line flux through the Gaussian surface positive, negative, or zero? (b) What can you conclude from your answer to part a about the charge enclosed by the Gaussian surface?
(a) The field line flux is zero because the field is zero everywhere inside the conducting object. (b) According to the relationship between field line flux and enclosed charge derived in Section 24.3, the charge enclosed by the surface must be zero.
24.20 Suppose Coulomb's law showed a 1>r 2.00001 dependence instead of a 1>r 2 dependence. (a) Calculate the electric flux through a spherical Gaussian surface of radius R centered on a particle carrying a charge +q. (b) Substitute your result in Eq. 24.8. What do you notice?
(a) With the modified Coulomb's law, the magnitude of the electric field at the surface of the sphere would be kq>R2.00001. The electric flux thus becomes ΦE = EA = E(4pR2 ) = (kq>R2.00001)(4pR2 ) = 4pkqR-0.00001. (b) 4pkqR-0.00001 = qenc >P0 = 4pkqenc or qR-0.00001 = qenc, which is not an equality. The cancellation of the radius R happens only when the dependence on r in Coulomb's law is exactly 1>r 2 .
24.1 Draw several field lines representing the electric field of an isolated positively charged particle. Repeat for a negatively charged particle.
(they're all pointing out on the positive one and all pointing in on the negative one)
24.21 Suppose q is outside the irregularly shaped surface in Figure 24.26. Show that the electric flux due to q through the closed surface is zero. (Hint: Draw a small wedge from q through the surface and determine the electric flux through the two intersections between the wedge and the surface.)
Figure S24.21 shows how a wedge from q intersects the surface. The electric flux through intersection A1 is equal to the electric flux through the intersection of the wedge with a sphere of radius r1 centered on q. Likewise, the electric flux through A2 is equal to the electric flux through the intersection of the wedge with a sphere of radius r2 centered on q. As we saw in Checkpoint 24.19, the magnitudes of the electric flux through a sphere are independent of the radius. Because the intersections of the wedge with the two concentric spheres represent a fixed fraction of the total surface area of the spheres, the electric fluxes through these two intersections must also be the same. Consequently, the magnitudes of the electric fluxes through A1 and A2 are the same. Because the electric field points out of the sphere at A1 and into the sphere at A2, however, the algebraic signs of the electric fluxes are opposite: ΦE1 = -ΦE2, so ΦE1 + ΦE2 = 0. Because this holds for any wedge, we learn that the electric flux through the surface due to q is zero when q is outside the surface
24.9 What is the field line flux through the surface of a rectangular box that encloses both ends of an electric dipole?
Five field lines emanating from the positively charged particle contribute a flux of +5 (the field line that points from the positive end to the negative end doesn't pass through the surface; see Figure S24.9). Five field lines terminating on the negatively charged particle contribute a flux of -5. This gives a flux of zero, which makes sense because the amount of charge enclosed by the box is +q + (-q) = 0. [Notice that this answer remains valid even if we make the box much larger so that all the curved field lines fit inside it. Then only two field lines pass through the box—one emanating from the positively charged particle, the other terminating on the negatively charged one. The flux is then still zero: (+1) + (-1) = 0.]
24.11 There are two reasons the field line flux through a closed surface may be zero: because the field is zero everywhere or because the outward flux is balanced by an equal inward flux. Why can't the latter situation be true for the Gaussian surface in Figure 24.12c?
If there were a flux into the shell from the left and out of the shell on the right, then the situation would no longer have the required symmetry—if you were to rotate the sphere 180° about the vertical axis, the field line flux would be reversed and therefore the situation would be different. Because the situation is spherically symmetrical, this case is not possible.
24.23 The direct integration procedure also yields an expression for a rod of finite length (see Example 23.4). Can you use Gauss's law to derive this expression as well? Why or why not?
No, because such a charge configuration does not have sufficient symmetry to make the calculation practical. A finite rod still has rotational symmetry (you can rotate it about its axis without changing the configuration), but it cannot be translated about its axis without changing the configuration. For a finite rod the field changes as you move up and down parallel to the y axis in Figure 24.30. With only rotational symmetry, we cannot easily find a closed surface such that for the separate regions of the surface either the magnitude of the field is constant or the electric flux is zero
24.16 In Example 24.4, is the electric field inside the cavity zero?
No. Field lines emanate from the positively charged particle and terminate on the negative charge carriers that line the cavity wall. Thus, "E = 0 everywhere inside a conducting object" means everywhere inside the bulk of the object. Cavities don't count!
24.15 Suppose the charged conducting object in Figure 24.17 contains an empty cavity. Does any surplus charge reside on the inner surface of the cavity?
No. Imagine a Gaussian surface around the cavity, just inside the conducting object (Figure S24.15). Because E = 0 everywhere inside the object, the field line flux through this Gaussian surface is zero and so the charge enclosed by it must also be zero. So, the surplus charge on the object cannot reside on an inner surface.
24.22 What is the electric field outside a solid sphere carrying a charge +q uniformly distributed throughout its volume?
The electric field is also given by Eq. 24.15 because none of the arguments leading up to that result change: The field is still spherically symmetrical, and the enclosed charge is still +q.
