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(a) Identify and Set Up:  EMBED Equation.DSMT4  where  EMBED Equation.DSMT4  is the angle between the normal to the sheet  EMBED Equation.DSMT4  and the electric field  EMBED Equation.DSMT4  Execute: In this problem E and  EMBED Equation.DSMT4  are constant over the surface so  EMBED Equation.DSMT4  (b) Evaluate:  EMBED Equation.DSMT4  is independent of the shape of the sheet as long as  EMBED Equation.DSMT4  and E are constant at all points on the sheet. (c) Execute: (i)  EMBED Equation.DSMT4  is largest for  EMBED Equation.DSMT4  (ii)  EMBED Equation.DSMT4  is smallest for  EMBED Equation.DSMT4  Evaluate:  EMBED Equation.DSMT4  is 0 when the surface is parallel to the field so no electric field lines pass through the surface. 22.2. Identify: The field is uniform and the surface is flat, so use  EMBED Equation.DSMT4 . Set Up:  EMBED Equation.DSMT4  is the angle between the normal to the surface and the direction of  EMBED Equation.DSMT4 , so  EMBED Equation.DSMT4 . Execute:  EMBED Equation.DSMT4  Evaluate: If the field were perpendicular to the surface the flux would be  EMBED Equation.DSMT4  The flux in this problem is much less than this because only the component of  EMBED Equation.DSMT4 perpendicular to the surface contributes to the flux. 22.3. Identify: The electric flux through an area is defined as the product of the component of the electric field perpendicular to the area times the area. (a) Set Up: In this case, the electric field is perpendicular to the surface of the sphere, so  EMBED Equation.DSMT4 . Execute: Substituting in the numbers gives  EMBED Equation.DSMT4  (b) Identify: We use the electric field due to a point charge. Set Up:  EMBED Equation.DSMT4  Execute: Solving for q and substituting the numbers gives  EMBED Equation.DSMT4  Evaluate: The flux would be the same no matter how large the circle, since the area is proportional to r2 while the electric field is proportional to 1/r2. 22.4. Identify: Use Eq.(22.3) to calculate the flux for each surface. Use Eq.(22.8) to calculate the total enclosed charge. Set Up:  EMBED Equation.DSMT4 . The area of each face is  EMBED Equation.DSMT4 , where  EMBED Equation.DSMT4 . Execute:  EMBED Equation.DSMT4 .  EMBED Equation.DSMT4 .  EMBED Equation.DSMT4 .  EMBED Equation.DSMT4 .  EMBED Equation.DSMT4   EMBED Equation.DSMT4   EMBED Equation.DSMT4   EMBED Equation.DSMT4  (b)Total flux:  EMBED Equation.DSMT4  Therefore,  EMBED Equation.DSMT4  Evaluate: Flux is positive when  EMBED Equation.DSMT4 is directed out of the volume and negative when it is directed into the volume. 22.5. Identify: The flux through the curved upper half of the hemisphere is the same as the flux through the flat circle defined by the bottom of the hemisphere because every electric field line that passes through the flat circle also must pass through the curved surface of the hemisphere. Set Up: The electric field is perpendicular to the flat circle, so the flux is simply the product of E and the area of the flat circle of radius r. Execute: (E = EA = E( EMBED Equation.DSMT4 ) =  EMBED Equation.DSMT4 E Evaluate: The flux would be the same if the hemisphere were replaced by any other surface bounded by the flat circle. 22.6. Identify: Use Eq.(22.3) to calculate the flux for each surface. Set Up:  EMBED Equation.DSMT4 . Execute: (a)  EMBED Equation.DSMT4 .  EMBED Equation.DSMT4   EMBED Equation.DSMT4 .  EMBED Equation.DSMT4 .  EMBED Equation.DSMT4 .  EMBED Equation.DSMT4 .  EMBED Equation.DSMT4 .  EMBED Equation.DSMT4 .  EMBED Equation.DSMT4 .  EMBED Equation.DSMT4 .  EMBED Equation.DSMT4 .  EMBED Equation.DSMT4 . Evaluate: (b) The total flux through the cube must be zero; any flux entering the cube must also leave it, since the field is uniform. Our calculation gives the result; the sum of the fluxes calculated in part (a) is zero. 22.7. (a) Identify: Use Eq.(22.5) to calculate the flux through the surface of the cylinder. Set Up: The line of charge and the cylinder are sketched in Figure 22.7.  INCLUDEPICTURE "FIG22-007.tif" \* MERGEFORMAT  Figure 22.7 Execute: The area of the curved part of the cylinder is  EMBED Equation.DSMT4  The electric field is parallel to the end caps of the cylinder, so  EMBED Equation.DSMT4  for the ends and the flux through the cylinder end caps is zero. The electric field is normal to the curved surface of the cylinder and has the same magnitude  EMBED Equation.DSMT4  at all points on this surface. Thus  EMBED Equation.DSMT4  and  EMBED Equation.DSMT4  (b) In the calculation in part (a) the radius r of the cylinder divided out, so the flux remains the same,  EMBED Equation.DSMT4  (c)  EMBED Equation.DSMT4  (twice the flux calculated in parts (b) and (c)). Evaluate: The flux depends on the number of field lines that pass through the surface of the cylinder. 22.8. Identify: Apply Gauss s law to each surface. Set Up:  EMBED Equation.DSMT4 is the algebraic sum of the charges enclosed by each surface. Flux out of the volume is positive and flux into the enclosed volume is negative. Execute: (a)  EMBED Equation.DSMT4  (b)  EMBED Equation.DSMT4  (c)  EMBED Equation.DSMT4  (d)  EMBED Equation.DSMT4  (e)  EMBED Equation.DSMT4  Evaluate: (f) All that matters for Gauss s law is the total amount of charge enclosed by the surface, not its distribution within the surface. 22.9. Identify: Apply the results in Example 21.10 for the field of a spherical shell of charge. Set Up: Example 22.10 shows that  EMBED Equation.DSMT4 inside a uniform spherical shell and that  EMBED Equation.DSMT4  outside the shell. Execute: (a)  EMBED Equation.DSMT4  (b)  EMBED Equation.DSMT4  and  EMBED Equation.DSMT4  (c)  EMBED Equation.DSMT4 and  EMBED Equation.DSMT4  Evaluate: Outside the shell the electric field is the same as if all the charge were concentrated at the center of the shell. But inside the shell the field is not the same as for a point charge at the center of the shell, inside the shell the electric field is zero. 22.10. Identify: Apply Gauss s law to the spherical surface. Set Up:  EMBED Equation.DSMT4 is the algebraic sum of the charges enclosed by the sphere. Execute: (a) No charge enclosed so  EMBED Equation.DSMT4 . (b) EMBED Equation.DSMT4  (c)  EMBED Equation.DSMT4  Evaluate: Negative flux corresponds to flux directed into the enclosed volume. The net flux depends only on the net charge enclosed by the surface and is not affected by any charges outside the enclosed volume. 22.11. Identify: Apply Gauss s law. Set Up: In each case consider a small Gaussian surface in the region of interest. Execute: (a) Since  EMBED Equation.DSMT4  is uniform, the flux through a closed surface must be zero. That is:  EMBED Equation.DSMT4  But because we can choose any volume we want,  EMBED Equation.DSMT4  must be zero if the integral equals zero. (b) If there is no charge in a region of space, that does NOT mean that the electric field is uniform. Consider a closed volume close to, but not including, a point charge. The field diverges there, but there is no charge in that region. Evaluate: The electric field within a region can depend on charges located outside the region. But the flux through a closed surface depends only on the net charge contained within that surface. 22.12. Identify: Apply Gauss s law. Set Up: Use a small Gaussian surface located in the region of question. Execute: (a) If  EMBED Equation.DSMT4  and uniform, then  EMBED Equation.DSMT4 inside any closed surface is greater than zero. This implies  EMBED Equation.DSMT4 , so  EMBED Equation.DSMT4  and so the electric field cannot be uniform. That is, since an arbitrary surface of our choice encloses a non-zero amount of charge,  EMBED Equation.DSMT4  must depend on position. (b) However, inside a small bubble of zero charge density within the material with density  EMBED Equation.DSMT4 , the field can be uniform. All that is important is that there be zero flux through the surface of the bubble (since it encloses no charge). (See Problem 22.61.) Evaluate: In a region of uniform field, the flux through any closed surface is zero. 22.13. (a) Identify and Set Up: It is rather difficult to calculate the flux directly from  EMBED Equation.DSMT4  since the magnitude of  EMBED Equation.DSMT4  and its angle with  EMBED Equation.DSMT4  varies over the surface of the cube. A much easier approach is to use Gauss's law to calculate the total flux through the cube. Let the cube be the Gaussian surface. The charge enclosed is the point charge. Execute:  EMBED Equation.DSMT4  By symmetry the flux is the same through each of the six faces, so the flux through one face is  EMBED Equation.DSMT4  (b) Evaluate: In part (a) the size of the cube did not enter into the calculations. The flux through one face depends only on the amount of charge at the center of the cube. So the answer to (a) would not change if the size of the cube were changed. 22.14. Identify: Apply the results of Examples 22.9 and 22.10. Set Up:  EMBED Equation.DSMT4  outside the sphere. A proton has charge +e. Execute: (a)  EMBED Equation.DSMT4  (b) For  EMBED Equation.DSMT4 ,  EMBED Equation.DSMT4  (c)  EMBED Equation.DSMT4 , inside a spherical shell. Evaluate: The electric field in an atom is very large. 22.15. Identify: The electric fields are produced by point charges. Set Up: We use Coulomb s law,  EMBED Equation.DSMT4 , to calculate the electric fields. Execute: (a)  EMBED Equation.DSMT4  (b)  EMBED Equation.DSMT4  (c) Every field line that enters the sphere on one side leaves it on the other side, so the net flux through the surface is zero. Evaluate: The flux would be zero no matter what shape the surface had, providing that no charge was inside the surface. 22.16. Identify: Apply the results of Example 22.5. Set Up: At a point 0.100 m outside the surface,  EMBED Equation.DSMT4 . Execute: (a)  EMBED Equation.DSMT4  (b)  EMBED Equation.DSMT4  inside of a conductor or else free charges would move under the influence of forces, violating our electrostatic assumptions (i.e., that charges aren t moving). Evaluate: Outside the sphere its electric field is the same as would be produced by a point charge at its center, with the same charge. 22.17. Identify: The electric field required to produce a spark 6 in. long is 6 times as strong as the field needed to produce a spark 1 in. long. Set Up: By Gauss s law,  EMBED Equation.DSMT4  and the electric field is the same as for a point-charge,  EMBED Equation.DSMT4 . Execute: (a) The electric field for 6-inch sparks is EMBED Equation.DSMT4  The charge to produce this field is  EMBED Equation.DSMT4 . (b) Using Coulomb s law gives  EMBED Equation.DSMT4 . Evaluate: It takes only about  EMBED Equation.DSMT4  to produce a field this strong. 22.18. Identify: According to Exercise 21.32, the Earth s electric field points towards its center. Since Mars s electric field is similar to that of Earth, we assume it points toward the center of Mars. Therefore the charge on Mars must be negative. We use Gauss s law to relate the electric flux to the charge causing it. Set Up: Gauss s law is  EMBED Equation.DSMT4  and the electric flux is  EMBED Equation.DSMT4 . Execute: (a) Solving Gauss s law for q, putting in the numbers, and recalling that q is negative, gives  EMBED Equation.DSMT4 . (b) Use the definition of electric flux to find the electric field. The area to use is the surface area of Mars.  EMBED Equation.DSMT4  (c) The surface charge density on Mars is therefore  EMBED Equation.DSMT4  Evaluate: Even though the charge on Mars is very large, it is spread over a large area, giving a small surface charge density. 22.19. Identify and Set Up: Example 22.5 derived that the electric field just outside the surface of a spherical conductor that has net charge q is  EMBED Equation.DSMT4  Calculate q and from this the number of excess electrons. Execute:  EMBED Equation.DSMT4  Each electron has a charge of magnitude  EMBED Equation.DSMT4  so the number of excess electrons needed is  EMBED Equation.DSMT4  Evaluate: The result we obtained for q is a typical value for the charge of an object. Such net charges correspond to a large number of excess electrons since the charge of each electron is very small. 22.20. Identify: Apply Gauss s law. Set Up: Draw a cylindrical Gaussian surface with the line of charge as its axis. The cylinder has radius 0.400 m and is 0.0200 m long. The electric field is then 840 N/C at every point on the cylindrical surface and is directed perpendicular to the surface. Execute:  EMBED Equation.DSMT4  The field is parallel to the end caps of the cylinder, so for them  EMBED Equation.DSMT4  From Gauss s law,  EMBED Equation.DSMT4  Evaluate: We could have applied the result in Example 22.6 and solved for  EMBED Equation.DSMT4 . Then  EMBED Equation.DSMT4  22.21. Identify: Add the vector electric fields due to each line of charge. E(r) for a line of charge is given by Example 22.6 and is directed toward a negative line of chage and away from a positive line. Set Up: The two lines of charge are shown in Figure 22.21.  INCLUDEPICTURE "FIG22-021.tif" \* MERGEFORMAT  EMBED Equation.DSMT4 Figure 22.21Execute: (a) At point a,  EMBED Equation.DSMT4  are in the  EMBED Equation.DSMT4  (toward negative charge, away from positive charge).  EMBED Equation.DSMT4   EMBED Equation.DSMT4   EMBED Equation.DSMT4  in the y-direction. (b) At point b,  EMBED Equation.DSMT4  is in the  EMBED Equation.DSMT4   EMBED Equation.DSMT4   EMBED Equation.DSMT4   EMBED Equation.DSMT4  Evaluation: At point a the two fields are in the same direction and the magnitudes add. At point b the two fields are in opposite directions and the magnitudes subtract. 22.22. Identify: Apply the results of Examples 22.5, 22.6 and 22.7. Set Up: Gauss s law can be used to show that the field outside a long conducting cylinder is the same as for a line of charge along the axis of the cylinder. Execute: (a) For points outside a uniform spherical charge distribution, all the charge can be considered to be concentrated at the center of the sphere. The field outside the sphere is thus inversely proportional to the square of the distance from the center. In this case,  EMBED Equation.DSMT4  (b) For points outside a long cylindrically symmetrical charge distribution, the field is identical to that of a long line of charge:  EMBED Equation.DSMT4 that is, inversely proportional to the distance from the axis of the cylinder. In this case  EMBED Equation.DSMT4  (c) The field of an infinite sheet of charge is  EMBED Equation.DSMT4  i.e., it is independent of the distance from the sheet. Thus in this case  EMBED Equation.DSMT4  Evaluate: For each of these three distributions of charge the electric field has a different dependence on distance. 22.23. Identify: The electric field inside the conductor is zero, and all of its initial charge lies on its outer surface. The introduction of charge into the cavity induces charge onto the surface of the cavity, which induces an equal but opposite charge on the outer surface of the conductor. The net charge on the outer surface of the conductor is the sum of the positive charge initially there and the additional negative charge due to the introduction of the negative charge into the cavity. (a) Set Up: First find the initial positive charge on the outer surface of the conductor using  EMBED Equation.DSMT4 where A is the area of its outer surface. Then find the net charge on the surface after the negative charge has been introduced into the cavity. Finally use the definition of surface charge density. Execute: The original positive charge on the outer surface is  EMBED Equation.DSMT4  After the introduction of  EMBED Equation.DSMT4  into the cavity, the outer charge is now  EMBED Equation.DSMT4  The surface charge density is now EMBED Equation.DSMT4  (b) Set Up: Using Gauss s law, the electric field is  EMBED Equation.DSMT4  Execute: Substituting numbers gives  EMBED Equation.DSMT4  (c) Set Up: We use Gauss s law again to find the flux.  EMBED Equation.DSMT4 . Execute: Substituting numbers gives  EMBED Equation.DSMT4 . Evaluate: The excess charge on the conductor is still  EMBED Equation.DSMT4  as it originally was. The introduction of the  EMBED Equation.DSMT4  inside the cavity merely induced equal but opposite charges (for a net of zero) on the surfaces of the conductor. 22.24. Identify: We apply Gauss s law, taking the Gaussian surface beyond the cavity but inside the solid. Set Up: Because of the symmetry of the charge, Gauss s law gives us  EMBED Equation.DSMT4 , where A is the surface area of a sphere of radius R = 9.50 cm centered on the point-charge, and qtotal is the total charge contained within that sphere. This charge is the sum of the  EMBED Equation.DSMT4  point charge at the center of the cavity plus the charge within the solid between r = 6.50 cm and R = 9.50 cm. The charge within the solid is qsolid = (V = (( EMBED Equation.DSMT4 ) = ( EMBED Equation.DSMT4 ()(R3 – r3) Execute: First find the charge within the solid between r = 6.50 cm and R = 9.50 cm:  EMBED Equation.DSMT4  Now find the total charge within the Gaussian surface:  EMBED Equation.DSMT4  Now find the magnitude of the electric field from Gauss s law:  EMBED Equation.DSMT4 . The fact that the charge is negative means that the electric field points radially inward. Evaluate: Because of the uniformity of the charge distribution, the charge beyond 9.50 cm does not contribute to the electric field. 22.25. Identify: The magnitude of the electric field is constant at any given distance from the center because the charge density is uniform inside the sphere. We can use Gauss s law to relate the field to the charge causing it. (a) Set Up: Gauss s law tells us that  EMBED Equation.DSMT4 , and the charge density is given by  EMBED Equation.DSMT4 . Execute: Solving for q and substituting numbers gives  EMBED Equation.DSMT4 . Using the formula for charge density we get  EMBED Equation.DSMT4  (b) Set Up: Take a Gaussian surface of radius r = 0.200 m, concentric with the insulating sphere. The charge enclosed within this surface is  EMBED Equation.DSMT4 , and we can treat this charge as a point-charge, using Coulomb s law  EMBED Equation.DSMT4 . The charge beyond r = 0.200 m makes no contribution to the electric field. Execute: First find the enclosed charge:  EMBED Equation.DSMT4  Now treat this charge as a point-charge and use Coulomb s law to find the field:  EMBED Equation.DSMT4  Evaluate: Outside this sphere, it behaves like a point-charge located at its center. Inside of it, at a distance r from the center, the field is due only to the charge between the center and r. 22.26. Identify: Apply Gauss s law and conservation of charge. Set Up: Use a Gaussian surface that lies wholly within he conducting material. Execute: (a) Positive charge is attracted to the inner surface of the conductor by the charge in the cavity. Its magnitude is the same as the cavity charge:  EMBED Equation.DSMT4  since  EMBED Equation.DSMT4  inside a conductor and a Gaussian surface that lies wholly within the conductor must enclose zero net charge. (b) On the outer surface the charge is a combination of the net charge on the conductor and the charge “left behind” when the  EMBED Equation.DSMT4  moved to the inner surface:  EMBED Equation.DSMT4  Evaluate: The electric field outside the conductor is due to the charge on its surface. 22.27. Identify: Apply Gauss s law to each surface. Set Up: The field is zero within the plates. By symmetry the field is perpendicular to the plates outside the plates and can depend only on the distance from the plates. Flux into the enclosed volume is positive. Execute:  EMBED Equation.DSMT4  enclose no charge, so the flux is zero, and electric field outside the plates is zero. Between the plates,  EMBED Equation.DSMT4  shows that  EMBED Equation.DSMT4 and  EMBED Equation.DSMT4  Evaluate: Our result for the field between the plates agrees with the result stated in Example 22.8. 22.28. Identify: Close to a finite sheet the field is the same as for an infinite sheet. Very far from a finite sheet the field is that of a point charge. Set Up: For an infinite sheet,  EMBED Equation.DSMT4 . For a point charge,  EMBED Equation.DSMT4 . Execute: (a) At a distance of 0.1 mm from the center, the sheet appears “infinite,” so  EMBED Equation.DSMT4 . (b) At a distance of 100 m from the center, the sheet looks like a point, so:  EMBED Equation.DSMT4  (c) There would be no difference if the sheet was a conductor. The charge would automatically spread out evenly over both faces, giving it half the charge density on either face as the insulator but the same electric field. Far away, they both look like points with the same charge. Evaluate: The sheet can be treated as infinite at points where the distance to the sheet is much less than the distance to the edge of the sheet. The sheet can be treated as a point charge at points for which the distance to the sheet is much greater than the dimensions of the sheet. 22.29. Identify: Apply Gauss s law to a Gaussian surface and calculate E. (a) Set Up: Consider the charge on a length l of the cylinder. This can be expressed as  EMBED Equation.DSMT4  But since the surface area is  EMBED Equation.DSMT4  it can also be expressed as  EMBED Equation.DSMT4  These two expressions must be equal, so  EMBED Equation.DSMT4  and  EMBED Equation.DSMT4  (b) Apply Gauss’s law to a Gaussian surface that is a cylinder of length l, radius r, and whose axis coincides with the axis of the charge distribution, as shown in Figure 22.29.  INCLUDEPICTURE "FIG22-029.tif" \* MERGEFORMAT Execute:  EMBED Equation.DSMT4   EMBED Equation.DSMT4 Figure 22.29 EMBED Equation.DSMT4   EMBED Equation.DSMT4  (c) Evaluate: Example 22.6 shows that the electric field of an infinite line of charge is  EMBED Equation.DSMT4   EMBED Equation.DSMT4  so  EMBED Equation.DSMT4  the same as for an infinite line of charge that is along the axis of the cylinder. 22.30. Identify: The net electric field is the vector sum of the fields due to each of the four sheets of charge. Set Up: The electric field of a large sheet of charge is  EMBED Equation.DSMT4 . The field is directed away from a positive sheet and toward a negative sheet. Execute: (a) At  EMBED Equation.DSMT4 .  EMBED Equation.DSMT4  (b)  EMBED Equation.DSMT4 .  EMBED Equation.DSMT4  (c)  EMBED Equation.DSMT4 .  EMBED Equation.DSMT4  Evaluate: The field at C is not zero. The pieces of plastic are not conductors. 22.31. Identify: Apply Gauss s law and conservation of charge. Set Up:  EMBED Equation.DSMT4 in a conducting material. Execute: (a) Gauss s law says +Q on inner surface, so EMBED Equation.DSMT4  inside metal. (b) The outside surface of the sphere is grounded, so no excess charge. (c) Consider a Gaussian sphere with the  Q charge at its center and radius less than the inner radius of the metal. This sphere encloses net charge –Q so there is an electric field flux through it; there is electric field in the cavity. (d) In an electrostatic situation  EMBED Equation.DSMT4  inside a conductor. A Gaussian sphere with the  EMBED Equation.DSMT4  charge at its center and radius greater than the outer radius of the metal encloses zero net charge (the  EMBED Equation.DSMT4  charge and the  EMBED Equation.DSMT4 on the inner surface of the metal) so there is no flux through it and  EMBED Equation.DSMT4 outside the metal. (e) No,  EMBED Equation.DSMT4  there. Yes, the charge has been shielded by the grounded conductor. There is nothing like positive and negative mass (the gravity force is always attractive), so this cannot be done for gravity. Evaluate: Field lines within the cavity terminate on the charges induced on the inner surface. 22.32. Identify and Set Up: Eq.(22.3) to calculate the flux. Identify the direction of the normal unit vector  EMBED Equation.DSMT4  for each surface. Execute: (a)  EMBED Equation.DSMT4   EMBED Equation.DSMT4   EMBED Equation.DSMT4   EMBED Equation.DSMT4   EMBED Equation.DSMT4   EMBED Equation.DSMT4   EMBED Equation.DSMT4   EMBED Equation.DSMT4   EMBED Equation.DSMT4   EMBED Equation.DSMT4   EMBED Equation.DSMT4   EMBED Equation.DSMT4   EMBED Equation.DSMT4   EMBED Equation.DSMT4   EMBED Equation.DSMT4   EMBED Equation.DSMT4   EMBED Equation.DSMT4  EMBED Equation.DSMT4   EMBED Equation.DSMT4  (b) Add the flux through each of the six faces:  EMBED Equation.DSMT4  The total electric flux through all sides is zero. Evaluate: All electric field lines that enter one face of the cube leave through another face. No electric field lines terminate inside the cube and the net flux is zero. 22.33. Identify: Use Eq.(22.3) to calculate the flux through each surface and use Gauss s law to relate the net flux to the enclosed charge. Set Up: Flux into the enclosed volume is negative and flux out of the volume is positive. Execute: (a)  EMBED Equation.DSMT4  (b) Since the field is parallel to the surface,  EMBED Equation.DSMT4  (c) Choose the Gaussian surface to equal the volume s surface. Then  EMBED Equation.DSMT4  and  EMBED Equation.DSMT4 , in the positive x-direction. Since  EMBED Equation.DSMT4  we must have some net flux flowing in so the flux is EMBED Equation.DSMT4  on second face. Evaluate: (d)  EMBED Equation.DSMT4  but we have E pointing away from face I. This is due to an external field that does not affect the flux but affects the value of E. 22.34. Identify: Apply Gauss s law to a cube centered at the origin and with side length 2L. Set Up: The total surface area of a cube with side length 2L is  EMBED Equation.DSMT4 . Execute: (a) The square is sketched in Figure 22.34. (b) Imagine a charge q at the center of a cube of edge length 2L. Then:  EMBED Equation.DSMT4  Here the square is one 24th of the surface area of the imaginary cube, so it intercepts 1/24 of the flux. That is,  EMBED Equation.DSMT4  Evaluate: Calculating the flux directly from Eq.(22.5) would involve a complicated integral. Using Gauss s law and symmetry considerations is much simpler.  INCLUDEPICTURE "FIG22-034.tif" \* MERGEFORMAT  Figure 22.34 22.35. (a) Identify: Find the net flux through the parallelepiped surface and then use that in Gauss s law to find the net charge within. Flux out of the surface is positive and flux into the surface is negative. Set Up:  EMBED Equation.DSMT4  gives flux out of the surface. See Figure 22.35a.  INCLUDEPICTURE "FIG22-035a.tif" \* MERGEFORMAT Execute:  EMBED Equation.DSMT4   EMBED Equation.DSMT4   EMBED Equation.DSMT4   EMBED Equation.DSMT4 Figure 22.35a EMBED Equation.DSMT4  Set Up:  EMBED Equation.DSMT4  gives flux into the surface. See Figure 22.35b.  INCLUDEPICTURE "FIG22-035b.tif" \* MERGEFORMAT Execute:  EMBED Equation.DSMT4   EMBED Equation.DSMT4   EMBED Equation.DSMT4   EMBED Equation.DSMT4 Figure 22.35b EMBED Equation.DSMT4  The net flux is  EMBED Equation.DSMT4  The net flux is negative (inward), so the net charge enclosed is negative. Apply Gauss’s law:  EMBED Equation.DSMT4   EMBED Equation.DSMT4  (b) Evaluate: If there were no charge within the parallelpiped the net flux would be zero. This is not the case, so there is charge inside. The electric field lines that pass out through the surface of the parallelpiped must terminate on charges, so there also must be charges outside the parallelpiped. 22.36. Identify: The  EMBED Equation.DSMT4  particle feels no force where the net electric field due to the two distributions of charge is zero. Set Up: The fields can cancel only in the regions A and B shown in Figure 22.36, because only in these two regions are the two fields in opposite directions. Execute:  EMBED Equation.DSMT4 gives  EMBED Equation.DSMT4 and  EMBED Equation.DSMT4 . The fields cancel 16 cm from the line in regions A and B. Evaluate: The result is independent of the distance between the line and the sheet. The electric field of an infinite sheet of charge is uniform, independent of the distance from the sheet.  INCLUDEPICTURE "FIG22-036.tif" \* MERGEFORMAT  Figure 22.36 22.37. (a) Identify: Apply Gauss s law to a Gaussian cylinder of length l and radius r, where  EMBED Equation.DSMT4  and calculate E on the surface of the cylinder. Set Up: The Gaussian surface is sketched in Figure 22.37a.  INCLUDEPICTURE "FIG22-037a.tif" \* MERGEFORMAT Execute:  EMBED Equation.DSMT4   EMBED Equation.DSMT4  (the charge on the length l of the inner conductor that is inside the Gaussian surface).Figure 22.37a EMBED Equation.DSMT4   EMBED Equation.DSMT4  The enclosed charge is positive so the direction of  EMBED Equation.DSMT4  is radially outward. (b) Set Up: Apply Gauss s law to a Gaussian cylinder of length l and radius r, where r > c, as shown in Figure 22.37b.  INCLUDEPICTURE "FIG22-037b.tif" \* MERGEFORMAT Execute:  EMBED Equation.DSMT4   EMBED Equation.DSMT4  (the charge on the length l of the inner conductor that is inside the Gaussian surface; the outer conductor carries no net charge).Figure 22.37b EMBED Equation.DSMT4   EMBED Equation.DSMT4  The enclosed charge is positive so the direction of  EMBED Equation.DSMT4  is radially outward. (c) E = 0 within a conductor. Thus E = 0 for r < a;  EMBED Equation.DSMT4   EMBED Equation.DSMT4  The graph of E versus r is sketched in Figure 22.37c.  INCLUDEPICTURE "FIG22-037c.tif" \* MERGEFORMAT  Figure 22.37c Evaluate: Inside either conductor E = 0. Between the conductors and outside both conductors the electric field is the same as for a line of charge with linear charge density  EMBED Equation.DSMT4  lying along the axis of the inner conductor. (d) Identify and Set Up: inner surface: Apply Gauss s law to a Gaussian cylinder with radius r, where  EMBED Equation.DSMT4  We know E on this surface; calculate  EMBED Equation.DSMT4  Execute: This surface lies within the conductor of the outer cylinder, where  EMBED Equation.DSMT4  Thus by Gauss’s law  EMBED Equation.DSMT4  The surface encloses charge  EMBED Equation.DSMT4  on the inner conductor, so it must enclose charge  EMBED Equation.DSMT4  on the inner surface of the outer conductor. The charge per unit length on the inner surface of the outer cylinder is  EMBED Equation.DSMT4  outer surface: The outer cylinder carries no net charge. So if there is charge per unit length  EMBED Equation.DSMT4  on its inner surface there must be charge per until length  EMBED Equation.DSMT4  on the outer surface. Evaluate: The electric field lines between the conductors originate on the surface charge on the outer surface of the inner conductor and terminate on the surface charges on the inner surface of the outer conductor. These surface charges are equal in magnitude (per unit length) and opposite in sign. The electric field lines outside the outer conductor originate from the surface charge on the outer surface of the outer conductor. 22.38. Identify: Apply Gauss s law. Set Up: Use a Gaussian surface that is a cylinder of radius r, length l and that has the line of charge along its axis. The charge on a length l of the line of charge or of the tube is  EMBED Equation.DSMT4 . Execute: (a) (i) For  EMBED Equation.DSMT4 , Gauss s law gives EMBED Equation.DSMT4  and  EMBED Equation.DSMT4 . (ii) The electric field is zero because these points are within the conducting material. (iii) For  EMBED Equation.DSMT4 , Gauss s law gives  EMBED Equation.DSMT4  and  EMBED Equation.DSMT4 . The graph of E versus r is sketched in Figure 22.38. (b) (i) The Gaussian cylinder with radius r, for  EMBED Equation.DSMT4 , must enclose zero net charge, so the charge per unit length on the inner surface is  EMBED Equation.DSMT4  (ii) Since the net charge per length for the tube is  EMBED Equation.DSMT4 and there is  EMBED Equation.DSMT4 on the inner surface, the charge per unit length on the outer surface must be  EMBED Equation.DSMT4  Evaluate: For  EMBED Equation.DSMT4 the electric field is due to the charge on the outer surface of the tube.  INCLUDEPICTURE "FIG22-038.tif" \* MERGEFORMAT  Figure 22.38 22.39. (a) Identify: Use Gauss s law to calculate E(r). (i) Set Up:  EMBED Equation.DSMT4  Apply Gauss s law to a cylindrical Gaussian surface of length l and radius r, where  EMBED Equation.DSMT4 as sketched in Figure 22.39a.  INCLUDEPICTURE "FIG22-039a.tif" \* MERGEFORMAT Execute:  EMBED Equation.DSMT4   EMBED Equation.DSMT4  (the charge on the length l of the line of charge)Figure 22.39a EMBED Equation.DSMT4   EMBED Equation.DSMT4  The enclosed charge is positive so the direction of  EMBED Equation.DSMT4  is radially outward. (ii)  EMBED Equation.DSMT4  Points in this region are within the conducting tube, so E = 0. (iii) Set Up:  EMBED Equation.DSMT4  Apply Gauss s law to a cylindrical Gaussian surface of length l and radius r, where  EMBED Equation.DSMT4  as sketched in Figure 22.39b.  INCLUDEPICTURE "FIG22-039b.tif" \* MERGEFORMAT Execute:  EMBED Equation.DSMT4   EMBED Equation.DSMT4  (the charge on length l of the line of charge)  EMBED Equation.DSMT4  (the charge on length l of the tube) Thus  EMBED Equation.DSMT4 Figure 22.39b EMBED Equation.DSMT4  The graph of E versus r is sketched in Figure 22.39c.  INCLUDEPICTURE "FIG22-039c.tif" \* MERGEFORMAT  Figure 22.39c (b) Identify: Apply Gauss s law to cylindrical surfaces that lie just outside the inner and outer surfaces of the tube. We know E so can calculate  EMBED Equation.DSMT4  (i) Set Up: inner surface Apply Gauss s law to a cylindrical Gaussian surface of length l and radius r, where  EMBED Equation.DSMT4  Execute: This surface lies within the conductor of the tube, where E = 0, so  EMBED Equation.DSMT4  Then by Gauss s law  EMBED Equation.DSMT4  The surface encloses charge  EMBED Equation.DSMT4 on the line of charge so must enclose charge  EMBED Equation.DSMT4  on the inner surface of the tube. The charge per unit length on the inner surface of the tube is  EMBED Equation.DSMT4  (ii) outer surface The net charge per unit length on the tube is  EMBED Equation.DSMT4  We have shown in part (i) that this must all reside on the inner surface, so there is no net charge on the outer surface of the tube. Evaluate: For r < a the electric field is due only to the line of charge. For r > b the electric field of the tube is the same as for a line of charge along its axis. The fields of the line of charge and of the tube are equal in magnitude and opposite in direction and sum to zero. For r < a the electric field lines originate on the line of charge and terminate on the surface charge on the inner surface of the tube. There is no electric field outside the tube and no surface charge on the outer surface of the tube. 22.40. Identify: Apply Gauss s law. Set Up: Use a Gaussian surface that is a cylinder of radius r and length l, and that is coaxial with the cylindrical charge distributions. The volume of the Gaussian cylinder is  EMBED Equation.DSMT4 and the area of its curved surface is  EMBED Equation.DSMT4 . The charge on a length l of the charge distribution is  EMBED Equation.DSMT4 , where  EMBED Equation.DSMT4 . Execute: (a) For  EMBED Equation.DSMT4 ,  EMBED Equation.DSMT4 and Gauss s law gives  EMBED Equation.DSMT4  and  EMBED Equation.DSMT4 , radially outward. (b) For  EMBED Equation.DSMT4 ,  EMBED Equation.DSMT4 and Gauss’s law gives  EMBED Equation.DSMT4 and  EMBED Equation.DSMT4 , radially outward. (c) At  EMBED Equation.DSMT4  the electric field for BOTH regions is  EMBED Equation.DSMT4 so they are consistent. (d) The graph of E versus r is sketched in Figure 22.40. Evaluate: For  EMBED Equation.DSMT4 the field is the same as for a line of charge along the axis of the cylinder.  INCLUDEPICTURE "FIG22-040.tif" \* MERGEFORMAT  Figure 22.40 22.41. Identify: First make a free-body diagram of the sphere. The electric force acts to the left on it since the electric field due to the sheet is horizontal. Since it hangs at rest, the sphere is in equilibrium so the forces on it add to zero, by Newton’s first law. Balance horizontal and vertical force components separately. Set Up: Call T the tension in the thread and E the electric field. Balancing horizontal forces gives T sin ( = qE. Balancing vertical forces we get T cos ( = mg. Combining these equations gives tan ( = qE/mg, which means that ( = arctan(qE/mg). The electric field for a sheet of charge is  EMBED Equation.DSMT4  Execute: Substituting the numbers gives us  EMBED Equation.DSMT4 . Then  EMBED Equation.DSMT4  Evaluate: Increasing the field, or decreasing the mass of the sphere, would cause the sphere to hang at a larger angle. 22.42. Identify: Apply Gauss s law. Set Up: Use a Gaussian surface that is a sphere of radius r and that is concentric with the conducting spheres. Execute: (a) For  EMBED Equation.DSMT4 since these points are within the conducting material. For  EMBED Equation.DSMT4  since there is +q inside a radius r. For  EMBED Equation.DSMT4 0, since since these points are within the conducting material For  EMBED Equation.DSMT4  since again the total charge enclosed is +q. (b) The graph of E versus r is sketched in Figure 22.42a. (c) Since the Gaussian sphere of radius r, for  EMBED Equation.DSMT4 , must enclose zero net charge, the charge on inner shell surface is –q. (d) Since the hollow sphere has no net charge and has charge  EMBED Equation.DSMT4 on its inner surface, the charge on outer shell surface is +q. (e) The field lines are sketched in Figure 22.42b. Where the field is nonzero, it is radially outward. Evaluate: The net charge on the inner solid conducting sphere is on the surface of that sphere. The presence of the hollow sphere does not affect the electric field in the region  EMBED Equation.DSMT4 .  INCLUDEPICTURE "FIG22-042a.tif" \* MERGEFORMAT   INCLUDEPICTURE "FIG22-042b.tif" \* MERGEFORMAT  Figure 22.42 22.43. Identify: Apply Gauss s law. Set Up: Use a Gaussian surface that is a sphere of radius r and that is concentric with the charge distributions. Execute: (a) For  EMBED Equation.DSMT4  since these points are within the conducting material. For  EMBED Equation.DSMT4   EMBED Equation.DSMT4  since the charge enclosed is Q. For  EMBED Equation.DSMT4 ,  EMBED Equation.DSMT4  since the charge enclosed is 2Q. (b) The graph of E versus r is sketched in Figure 22.43. Evaluate: For  EMBED Equation.DSMT4  the electric field is unaffected by the presence of the charged shell.  INCLUDEPICTURE "FIG22-043.tif" \* MERGEFORMAT  Figure 22.43 22.44. Identify: Apply Gauss s law and conservation of charge. Set Up: Use a Gaussian surface that is a sphere of radius r and that has the point charge at its center. Execute: (a) For  EMBED Equation.DSMT4 ,  EMBED Equation.DSMT4  radially outward, since the charge enclosed is Q, the charge of the point charge. For  EMBED Equation.DSMT4 ,  EMBED Equation.DSMT4  since these points are within the conducting material. For  EMBED Equation.DSMT4 ,  EMBED Equation.DSMT4 , radially inward, since the total enclosed charge is –2Q. (b) Since a Gaussian surface with radius r, for  EMBED Equation.DSMT4 , must enclose zero net charge, the total charge on the inner surface is  EMBED Equation.DSMT4  and the surface charge density on inner surface is  EMBED Equation.DSMT4 . (c) Since the net charge on the shell is  EMBED Equation.DSMT4  and there is  EMBED Equation.DSMT4  on the inner surface, there must be  EMBED Equation.DSMT4 on the outer surface. The surface charge density on the outer surface is  EMBED Equation.DSMT4  (d) The field lines and the locations of the charges are sketched in Figure 22.44a. (e) The graph of E versus r is sketched in Figure 22.44b.  INCLUDEPICTURE "FIG22-044.tif" \* MERGEFORMAT   INCLUDEPICTURE "FIG22-044b.tif" \* MERGEFORMAT  Figure 22.44 Evaluate:  For  EMBED Equation.DSMT4  the electric field is due solely to the point charge Q. For  EMBED Equation.DSMT4  the electric field is due to the charge  EMBED Equation.DSMT4  that is on the outer surface of the shell. 22.45. Identify: Apply Gauss s law to a spherical Gaussian surface with radius r. Calculate the electric field at the surface of the Gaussian sphere. (a) Set Up: (i)  EMBED Equation.DSMT4  The Gaussian surface is sketched in Figure 22.45a.  INCLUDEPICTURE "FIG22-045a.tif" \* MERGEFORMAT Execute:  EMBED Equation.DSMT4   EMBED Equation.DSMT4  no charge is enclosed  EMBED Equation.DSMT4  says  EMBED Equation.DSMT4 Figure 22.45a(ii)  EMBED Equation.DSMT4  Points in this region are in the conductor of the small shell, so E = 0. (iii) Set Up:  EMBED Equation.DSMT4  The Gaussian surface is sketched in Figure 22.45b. Apply Gauss s law to a spherical Gaussian surface with radius  EMBED Equation.DSMT4   INCLUDEPICTURE "FIG22-045b.tif" \* MERGEFORMAT Execute:  EMBED Equation.DSMT4  The Gaussian surface encloses all of the small shell and none of the large shell, so  EMBED Equation.DSMT4 Figure 22.45b EMBED Equation.DSMT4  Since the enclosed charge is positive the electric field is radially outward. 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INCLUDEPICTURE "FIG22-045c.tif" \* MERGEFORMAT Execute:  EMBED Equation.DSMT4  The Gaussian surface encloses all of the small shell and all of the large shell, so  EMBED Equation.DSMT4 Figure 22.45c EMBED Equation.DSMT4   EMBED Equation.DSMT4  Since the enclosed charge is positive the electric field is radially outward. The graph of E versus r is sketched in Figure 22.45d.  INCLUDEPICTURE "FIG22-045d.tif" \* MERGEFORMAT  Figure 22.45d (b) Identify and Set Up: Apply Gauss s law to a sphere that lies outside the surface of the shell for which we want to find the surface charge. Execute: (i) charge on inner surface of the small shell: Apply Gauss s law to a spherical Gaussian surface with radius  EMBED Equation.DSMT4  This surface lies within the conductor of the small shell, where E = 0, so  EMBED Equation.DSMT4  Thus by Gauss s law  EMBED Equation.DSMT4  so there is zero charge on the inner surface of the small shell. (ii) charge on outer surface of the small shell: The total charge on the small shell is  EMBED Equation.DSMT4  We found in part (i) that there is zero charge on the inner surface of the shell, so all  EMBED Equation.DSMT4  must reside on the outer surface. (iii) charge on inner surface of large shell: Apply Gauss s law to a spherical Gaussian surface with radius  EMBED Equation.DSMT4  The surface lies within the conductor of the large shell, where E = 0, so  EMBED Equation.DSMT4  Thus by Gauss s law  EMBED Equation.DSMT4  The surface encloses the  EMBED Equation.DSMT4  on the small shell so there must be charge  EMBED Equation.DSMT4  on the inner surface of the large shell to make the total enclosed charge zero. (iv) charge on outer surface of large shell: The total charge on the large shell is  EMBED Equation.DSMT4  We showed in part (iii) that the charge on the inner surface is  EMBED Equation.DSMT4  so there must be  EMBED Equation.DSMT4  on the outer surface. Evaluate: The electric field lines for  EMBED Equation.DSMT4  originate from the surface charge on the outer surface of the inner shell and all terminate on the surface charge on the inner surface of the outer shell. These surface charges have equal magnitude and opposite sign. The electric field lines for  EMBED Equation.DSMT4  originate from the surface charge on the outer surface of the outer sphere. 22.46. Identify: Apply Gauss s law. Set Up: Use a Gaussian surface that is a sphere of radius r and that is concentric with the charged shells. Execute: (a) (i) For  EMBED Equation.DSMT4  since the charge enclosed is zero. (ii) For  EMBED Equation.DSMT4  since the points are within the conducting material. (iii) For  EMBED Equation.DSMT4 outward, since charge enclosed is  EMBED Equation.DSMT4  (iv) For  EMBED Equation.DSMT4  since the points are within the conducting material. (v) For  EMBED Equation.DSMT4 since the net charge enclosed is zero. The graph of E versus r is sketched in Figure 22.46. (b) (i) small shell inner surface: Since a Gaussian surface with radius r, for  EMBED Equation.DSMT4 , must enclose zero net charge, the charge on this surface is zero. (ii) small shell outer surface:  EMBED Equation.DSMT4 . (iii) large shell inner surface: Since a Gaussian surface with radius r, for  EMBED Equation.DSMT4 , must enclose zero net charge, the charge on this surface is  EMBED Equation.DSMT4 . (iv) large shell outer surface: Since there is  EMBED Equation.DSMT4 on the inner surface and the total charge on this conductor is  EMBED Equation.DSMT4 , the charge on this surface is zero. Evaluate: The outer shell has no effect on the electric field for  EMBED Equation.DSMT4 . For  EMBED Equation.DSMT4 the electric field is due only to the charge on the outer surface of the larger shell.  INCLUDEPICTURE "FIG22-046.tif" \* MERGEFORMAT  Figure 22.46 22.47. Identify: Apply Gauss s law Set Up: Use a Gaussian surface that is a sphere of radius r and that is concentric with the charged shells. Execute: (a) (i) For EMBED Equation.DSMT4  since charge enclosed is zero. (ii)  EMBED Equation.DSMT4  since the points are within the conducting material. (iii) For EMBED Equation.DSMT4  outward, since charge enclosed is +2q. (iv) For EMBED Equation.DSMT4  since the points are within the conducting material. (v) For  EMBED Equation.DSMT4  inward, since charge enclosed is  2q. The graph of the radial component of the electric field versus r is sketched in Figure 22.47, where we use the convention that outward field is positive and inward field is negative. (b) (i) small shell inner surface: Since a Gaussian surface with radius r, for a < r < b, must enclose zero net charge, the charge on this surface is zero. (ii) small shell outer surface: +2q. (iii) large shell inner surface: Since a Gaussian surface with radius r, for c < r < d, must enclose zero net charge, the charge on this surface is  2q. (iv) large shell outer surface: Since there is  2q on the inner surface and the total charge on this conductor is  4q, the charge on this surface is  2q. Evaluate: The outer shell has no effect on the electric field for  EMBED Equation.DSMT4 . For  EMBED Equation.DSMT4 the electric field is due only to the charge on the outer surface of the larger shell.  INCLUDEPICTURE "FIG22-047.tif" \* MERGEFORMAT  Figure 22.47 22.48. Identify: Apply Gauss s law. Set Up: Use a Gaussian surface that is a sphere of radius r and that is concentric with the sphere and shell. The volume of the insulating shell is  EMBED Equation.DSMT4 . Execute: (a) Zero net charge requires that  EMBED Equation.DSMT4 , so  EMBED Equation.DSMT4 . (b) For  EMBED Equation.DSMT4 since this region is within the conducting sphere. For EMBED Equation.DSMT4  since the net charge enclosed by the Gaussian surface with this radius is zero. For  EMBED Equation.DSMT4 , Gauss s law gives  EMBED Equation.DSMT4  and  EMBED Equation.DSMT4 . Substituting  EMBED Equation.DSMT4  from part (a) gives  EMBED Equation.DSMT4  The net enclosed charge for each r in this range is positive and the electric field is outward. (c) The graph is sketched in Figure 22.48. We see a discontinuity in going from the conducting sphere to the insulator due to the thin surface charge of the conducting sphere. But we see a smooth transition from the uniform insulator to the surrounding space. Evaluate: The expression for E within the insulator gives  EMBED Equation.DSMT4 at  EMBED Equation.DSMT4 .  INCLUDEPICTURE "FIG22-048.tif" \* MERGEFORMAT  Figure 22.48 22.49. Identify: Use Gauss s law to find the electric field  EMBED Equation.DSMT4  produced by the shell for  EMBED Equation.DSMT4  and then use  EMBED Equation.DSMT4  to find the force the shell exerts on the point charge. (a) Set Up: Apply Gauss s law to a spherical Gaussian surface that has radius  EMBED Equation.DSMT4  and that is concentric with the shell, as sketched in Figure 22.49a.  INCLUDEPICTURE "FIG22-049a.tif" \* MERGEFORMAT Execute:  EMBED Equation.DSMT4   EMBED Equation.DSMT4 Figure 22.49a EMBED Equation.DSMT4  The magnitude of the field is  EMBED Equation.DSMT4  and it is directed toward the center of the shell. Then  EMBED Equation.DSMT4  directed toward the center of the shell. (Since q is positive,  EMBED Equation.DSMT4  are in the same direction.) (b) Set Up: Apply Gauss s law to a spherical Gaussian surface that has radius  EMBED Equation.DSMT4  and that is concentric with the shell, as sketched in Figure 22.49b.  INCLUDEPICTURE "FIG22-049b.tif" \* MERGEFORMAT Execute:  EMBED Equation.DSMT4   EMBED Equation.DSMT4 Figure 22.49b EMBED Equation.DSMT4  Then  EMBED Equation.DSMT4  Evaluate: Outside the shell the electric field and the force it exerts is the same as for a point charge  EMBED Equation.DSMT4  located at the center of the shell. Inside the shell  EMBED Equation.DSMT4  and there is no force. 22.50. Identify: The method of Example 22.9 shows that the electric field outside the sphere is the same as for a point charge of the same charge located at the center of the sphere. Set Up: The charge of an electron has magnitude  EMBED Equation.DSMT4 . Execute: (a)  EMBED Equation.DSMT4 . For  EMBED Equation.DSMT4 ,  EMBED Equation.DSMT4 so  EMBED Equation.DSMT4 . The number of excess electrons is  EMBED Equation.DSMT4 . (b)  EMBED Equation.DSMT4 .  EMBED Equation.DSMT4 . Evaluate: The magnitude of the electric field decreases according to the square of the distance from the center of the sphere. 22.51. Identify: The net electric field is the vector sum of the fields due to the sheet of charge on each surface of the plate. Set Up: The electric field due to the sheet of charge on each surface is  EMBED Equation.DSMT4 and is directed away from the surface. Execute: (a) For the conductor the charge sheet on each surface produces fields of magnitude  EMBED Equation.DSMT4 and in the same direction, so the total field is twice this, or  EMBED Equation.DSMT4 . (b) At points inside the plate the fields of the sheets of charge on each surface are equal in magnitude and opposite in direction, so their vector sum is zero. At points outside the plate, on either side, the fields of the two sheets of charge are in the same direction so their magnitudes add, giving  EMBED Equation.DSMT4 . Evaluate: Gauss s law can also be used directly to determine the fields in these regions. 22.52. Identify: Example 22.9 gives the expression for the electric field both inside and outside a uniformly charged sphere. Use  EMBED Equation.DSMT4 to calculate the force on the electron. Set Up: The sphere has charge  EMBED Equation.DSMT4 . Execute: (a) Only at  EMBED Equation.DSMT4 is  EMBED Equation.DSMT4 for the uniformly charged sphere. (b) At points inside the sphere,  EMBED Equation.DSMT4 . The field is radially outward.  EMBED Equation.DSMT4 . The minus sign denotes that  EMBED Equation.DSMT4 is radially inward. For simple harmonic motion,  EMBED Equation.DSMT4 , where  EMBED Equation.DSMT4 .  EMBED Equation.DSMT4  so  EMBED Equation.DSMT4  and  EMBED Equation.DSMT4 . (c) If  EMBED Equation.DSMT4  then  EMBED Equation.DSMT4  The atom radius in this model is the correct order of magnitude. (d) If  EMBED Equation.DSMT4 ,  EMBED Equation.DSMT4  and  EMBED Equation.DSMT4 . The electron would still oscillate because the force is directed toward the equilibrium position at  EMBED Equation.DSMT4 . But the motion would not be simple harmonic, since  EMBED Equation.DSMT4 is proportional to  EMBED Equation.DSMT4 and simple harmonic motion requires that the restoring force be proportional to the displacement from equilibrium. Evaluate: As long as the initial displacement is less than R the frequency of the motion is independent of the initial displacement. 22.53. Identify: There is a force on each electron due to the other electron and a force due to the sphere of charge. Use Coulomb s law for the force between the electrons. Example 22.9 gives E inside a uniform sphere and Eq.(21.3) gives the force. Set Up: The positions of the electrons are sketched in Figure 22.53a.  INCLUDEPICTURE "FIG22-053a.tif" \* MERGEFORMAT If the electrons are in equilibrium the net force on each one is zero.Figure 22.53aExecute: Consider the forces on electron 2. There is a repulsive force  EMBED Equation.DSMT4  due to the other electron, electron 1.  EMBED Equation.DSMT4  The electric field inside the uniform distribution of positive charge is  EMBED Equation.DSMT4  (Example 22.9), where  EMBED Equation.DSMT4  At the position of electron 2, r = d. The force  EMBED Equation.DSMT4  exerted by the positive charge distribution is  EMBED Equation.DSMT4  and is attractive. The force diagram for electron 2 is given in Figure 22.53b.  INCLUDEPICTURE "FIG22-053b.tif" \* MERGEFORMAT  Figure 22.53b Net force equals zero implies  EMBED Equation.DSMT4  and  EMBED Equation.DSMT4  Thus  EMBED Equation.DSMT4  Evaluate: The electric field of the sphere is radially outward; it is zero at the center of the sphere and increases with distance from the center. The force this field exerts on one of the electrons is radially inward and increases as the electron is farther from the center. The force from the other electron is radially outward, is infinite when d = 0 and decreases as d increases. It is reasonable therefore for there to be a value of d for which these forces balance. 22.54. Identify: Use Gauss s law to find the electric field both inside and outside the slab. Set Up: Use a Gaussian surface that has one face of area A in the  EMBED Equation.DSMT4 z plane at  EMBED Equation.DSMT4  and the other face at a general value  EMBED Equation.DSMT4  The volume enclosed by such a Gaussian surface is Ax. Execute: (a) The electric field of the slab must be zero by symmetry. There is no preferred direction in the  EMBED Equation.DSMT4 z plane, so the electric field can only point in the x-direction. But at the origin, neither the positive nor negative x-directions should be singled out as special, and so the field must be zero. (b) For  EMBED Equation.DSMT4 , Gauss s law gives EMBED Equation.DSMT4  and  EMBED Equation.DSMT4 , with direction given by  EMBED Equation.DSMT4  (away from the center of the slab). Note that this expression does give  EMBED Equation.DSMT4  at  EMBED Equation.DSMT4  Outside the slab, the enclosed charge does not depend on  EMBED Equation.DSMT4 and is equal to  EMBED Equation.DSMT4 . For  EMBED Equation.DSMT4 , Gauss s law gives  EMBED Equation.DSMT4 and  EMBED Equation.DSMT4 , again with direction given by  EMBED Equation.DSMT4  Evaluate: At the surfaces of the slab,  EMBED Equation.DSMT4 . For these values of x the two expressions for E (for inside and outside the slab) give the same result. The charge per unit area  EMBED Equation.DSMT4 of the slab is given by  EMBED Equation.DSMT4 and  EMBED Equation.DSMT4 . The result for E outside the slab can therefore be written as  EMBED Equation.DSMT4 and is the same as for a thin sheet of charge. 22.55. (a) Identify and Set Up: Consider the direction of the field for x slightly greater than and slightly less than zero. The slab is sketched in Figure 22.55a.  INCLUDEPICTURE "FIG22-055a.tif" \* MERGEFORMAT  EMBED Equation.DSMT4 Figure 22.55aExecute: The charge distribution is symmetric about x = 0, so by symmetry  EMBED Equation.DSMT4  But for x > 0 the field is in the  EMBED Equation.DSMT4  direction and for x < 0 the field is in the  EMBED Equation.DSMT4  direction. At x = 0 the field can t be both in the  EMBED Equation.DSMT4  directions so must be zero. That is,  EMBED Equation.DSMT4  At point x = 0 this gives  EMBED Equation.DSMT4  and this equation is satisfied only for  EMBED Equation.DSMT4  (b) Identify and Set Up:  EMBED Equation.DSMT4  (outside the slab) Apply Gauss s law to a cylindrical Gaussian surface whose axis is perpendicular to the slab and whose end caps have area A and are the same distance  EMBED Equation.DSMT4  from x = 0, as shown in Figure 22.55b.  INCLUDEPICTURE "FIG22-055b.tif" \* MERGEFORMAT Execute:  EMBED Equation.DSMT4 Figure 22.55b INCLUDEPICTURE "FIG22-055c.tif" \* MERGEFORMAT To find  EMBED Equation.DSMT4  consider a thin disk at coordinate x and with thickness dx, as shown in Figure 22.55c. The charge within this disk is  EMBED Equation.DSMT4 Figure 22.55cThe total charge enclosed by the Gaussian cylinder is  EMBED Equation.DSMT4  Then  EMBED Equation.DSMT4   EMBED Equation.DSMT4   EMBED Equation.DSMT4  is directed away from  EMBED Equation.DSMT4  Identify and Set Up:  EMBED Equation.DSMT4  (inside the slab) Apply Gauss s law to a cylindrical Gaussian surface whose axis is perpendicular to the slab and whose end caps have area A and are the same distance  EMBED Equation.DSMT4  from  EMBED Equation.DSMT4  as shown in Figure 22.55d.  INCLUDEPICTURE "FIG22-055d.tif" \* MERGEFORMAT Execute:  EMBED Equation.DSMT4 Figure 22.55d EMBED Equation.DSMT4  is found as above, but now the integral on dx is only from 0 to x instead of 0 to d.  EMBED Equation.DSMT4  Then  EMBED Equation.DSMT4   EMBED Equation.DSMT4   EMBED Equation.DSMT4  is directed away from  EMBED Equation.DSMT4  Evaluate: Note that E = 0 at x = 0 as stated in part (a). Note also that the expressions for  EMBED Equation.DSMT4  agree for  EMBED Equation.DSMT4  22.56. Identify: Apply  EMBED Equation.DSMT4 to relate the force on q to the electric field at the location of q. Set Up: Flux is negative if the electric field is directed into the enclosed volume. Execute: (a) We could place two charges  EMBED Equation.DSMT4 on either side of the charge  EMBED Equation.DSMT4 , as shown in Figure 22.56. (b) In order for the charge to be stable, the electric field in a neighborhood around it must always point back to the equilibrium position. (c) If  EMBED Equation.DSMT4  is moved to infinity and we require there to be an electric field always pointing in to the region where  EMBED Equation.DSMT4  had been, we could draw a small Gaussian surface there. We would find that we need a negative flux into the surface. That is, there has to be a negative charge in that region. However, there is none, and so we cannot get such a stable equilibrium. (d) For a negative charge to be in stable equilibrium, we need the electric field to always point away from the charge position. The argument in (c) carries through again, this time implying that a positive charge must be in the space where the negative charge was if stable equilibrium is to be attained. Evaluate: If q is displaced to the left or right in Figure 22.56, the net force is directed back toward the equilibrium position. But if q is displaced slightly in a direction perpendicular to the line connecting the two charges Q, then the net force on q is directed away from the equilibrium position and the equilibrium is not stable for such a displacement.  INCLUDEPICTURE "FIG22-056.tif" \* MERGEFORMAT  Figure 22.56 22.57.  EMBED Equation.DSMT4   EMBED Equation.DSMT4  (a) Identify: The charge density varies with r inside the spherical volume. Divide the volume up into thin concentric shells, of radius r and thickness dr. Find the charge dq in each shell and integrate to find the total charge. Set Up: The thin shell is sketched in Figure 22.57a.  INCLUDEPICTURE "FIG22-057a.tif" \* MERGEFORMAT Execute: The volume of such a shell is  EMBED Equation.DSMT4  The charge contained within the shell is  EMBED Equation.DSMT4 Figure 22.57aThe total charge Q in the charge distribution is obtained by integrating dq over all such shells into which the sphere can be subdivided:  EMBED Equation.DSMT4   EMBED Equation.DSMT4  as was to be shown. (b) Identify: Apply Gauss s law to a spherical surface of radius r, where r > R. Set Up: The Gaussian surface is shown in Figure 22.57b.  INCLUDEPICTURE "FIG22-057b.tif" \* MERGEFORMAT Execute:  EMBED Equation.DSMT4   EMBED Equation.DSMT4 Figure 22.57b EMBED Equation.DSMT4  same as for point charge of charge Q. (c) Identify: Apply Gauss s law to a spherical surface of radius r, where r < R: Set Up: The Gaussian surface is shown in Figure 22.57c.  INCLUDEPICTURE "FIG22-057c.tif" \* MERGEFORMAT Execute:  EMBED Equation.DSMT4   EMBED Equation.DSMT4 Figure 22.57cThe calculate the enclosed charge  EMBED Equation.DSMT4  use the same technique as in part (a), except integrate dq out to r rather than R. (We want the charge that is inside radius r.)  EMBED Equation.DSMT4   EMBED Equation.DSMT4   EMBED Equation.DSMT4  Thus Gauss s law gives  EMBED Equation.DSMT4   EMBED Equation.DSMT4  (d) The graph of E versus r is sketched in Figure 22.57d.  INCLUDEPICTURE "FIG22-057d.tif" \* MERGEFORMAT  Figure 22.57d (e) Where the electric field is a maximum,  EMBED Equation.DSMT4  Thus  EMBED Equation.DSMT4  At this value of r,  EMBED Equation.DSMT4  Evaluate: Our expressions for  EMBED Equation.DSMT4  for  EMBED Equation.DSMT4  and for  EMBED Equation.DSMT4  agree at  EMBED Equation.DSMT4  The results of part (e) for the value of r where  EMBED Equation.DSMT4  is a maximum agrees with the graph in part (d). 22.58. Identify: Apply Gauss s law. Set Up: Use a Gaussian surface that is a sphere of radius r and that is concentric with the spherical distribution of charge. The volume of a thin spherical shell of radius r and thickness dr is  EMBED Equation.DSMT4 . Execute: (a)  EMBED Equation.DSMT4   EMBED Equation.DSMT4 . The total charge is zero. (b) For  EMBED Equation.DSMT4 ,  EMBED Equation.DSMT4 , so  EMBED Equation.DSMT4 . (c) For  EMBED Equation.DSMT4 ,  EMBED Equation.DSMT4 .  EMBED Equation.DSMT4  and  EMBED Equation.DSMT4 . (d) The graph of E versus r is sketched in Figure 22.58. (e) Where E is a maximum,  EMBED Equation.DSMT4 . This gives EMBED Equation.DSMT4  and  EMBED Equation.DSMT4 . At this r,  EMBED Equation.DSMT4 . Evaluate: The result in part (b) for  EMBED Equation.DSMT4 gives  EMBED Equation.DSMT4 at  EMBED Equation.DSMT4 ; the field is continuous at the surface of the charge distribution.  INCLUDEPICTURE "FIG22-058.tif" \* MERGEFORMAT  Figure 22.58 22.59. Identify: Follow the steps specified in the problem. Set Up: In spherical polar coordinates  EMBED Equation.DSMT4 .  EMBED Equation.DSMT4 . Execute: (a)  EMBED Equation.DSMT4  (b) For any closed surface, mass OUTSIDE the surface contributes zero to the flux passing through the surface. Thus the formula above holds for any situation where  EMBED Equation.DSMT4  is the mass enclosed by the Gaussian surface. That is, EMBED Equation.DSMT4  Evaluate: The minus sign in the expression for the flux signifies that the flux is directed inward. 22.60. Identify: Apply  EMBED Equation.DSMT4 . Set Up: Use a Gaussian surface that is a sphere of radius r, concentric with the mass distribution. Let  EMBED Equation.DSMT4 denote  EMBED Equation.DSMT4  Execute: (a) Use a Gaussian sphere with radius  EMBED Equation.DSMT4 , where R is the radius of the mass distribution. g is constant on this surface and the flux is inward. The enclosed mass is M. Therefore,  EMBED Equation.DSMT4 and  EMBED Equation.DSMT4 , which is the same as for a point mass. (b) For a Gaussian sphere of radius  EMBED Equation.DSMT4 , where R is the radius of the shell,  EMBED Equation.DSMT4  (c) Use a Gaussian sphere of radius  EMBED Equation.DSMT4 , where R is the radius of the planet. Then  EMBED Equation.DSMT4 . This gives  EMBED Equation.DSMT4 and  EMBED Equation.DSMT4 , which is linear in  EMBED Equation.DSMT4  Evaluate:  The spherically synmetric distribution of mass results in an acceleration due to gravity  EMBED Equation.DSMT4  that is radical and that depends only on r, the distance from the center of the mass distribution. 22.61. (a) Identify: Use  EMBED Equation.DSMT4  from Example (22.9) (inside the sphere) and relate the position vector of a point inside the sphere measured from the origin to that measured from the center of the sphere. Set Up: For an insulating sphere of uniform charge density  EMBED Equation.DSMT4  and centered at the origin, the electric field inside the sphere is given by  EMBED Equation.DSMT4  (Example 22.9), where  EMBED Equation.DSMT4  is the vector from the center of the sphere to the point where E is calculated. But  EMBED Equation.DSMT4  so this may be written as  EMBED Equation.DSMT4  And  EMBED Equation.DSMT4  is radially outward, in the direction of  EMBED Equation.DSMT4  For a sphere whose center is located by vector  EMBED Equation.DSMT4  a point inside the sphere and located by  EMBED Equation.DSMT4  is located by the vector  EMBED Equation.DSMT4  relative to the center of the sphere, as shown in Figure 22.61.  INCLUDEPICTURE "FIG22-061.tif" \* MERGEFORMAT Execute: Thus  EMBED Equation.DSMT4 Figure 22.61Evaluate: When b = 0 this reduces to the result of Example 22.9. When  EMBED Equation.DSMT4  this gives  EMBED Equation.DSMT4  which is correct since we know that  EMBED Equation.DSMT4  at the center of the sphere. (b) Identify: The charge distribution can be represented as a uniform sphere with charge density  EMBED Equation.DSMT4  and centered at the origin added to a uniform sphere with charge density  EMBED Equation.DSMT4  and centered at  EMBED Equation.DSMT4  Set Up:  EMBED Equation.DSMT4  is the field of a uniformly charged sphere with charge density  EMBED Equation.DSMT4  and  EMBED Equation.DSMT4  is the field of a sphere located at the hole and with charge density  EMBED Equation.DSMT4  (Within the spherical hole the net charge density is  EMBED Equation.DSMT4 ) Execute:  EMBED Equation.DSMT4  where  EMBED Equation.DSMT4  is a vector from the center of the sphere.  EMBED Equation.DSMT4  at points inside the hole. Then  EMBED Equation.DSMT4  Evaluate:  EMBED Equation.DSMT4  is independent of  EMBED Equation.DSMT4  so is uniform inside the hole. The direction of  EMBED Equation.DSMT4  inside the hole is in the direction of the vector  EMBED Equation.DSMT4  the direction from the center of the insulating sphere to the center of the hole. 22.62. Identify: We first find the field of a cylinder off-axis, then the electric field in a hole in a cylinder is the difference between two electric fields: that of a solid cylinder on-axis, and one off-axis, at the location of the hole. Set Up: Let  EMBED Equation.DSMT4 locate a point within the hole, relative to the axis of the cylinder and let  EMBED Equation.DSMT4 locate this point relative to the axis of the hole. Let  EMBED Equation.DSMT4 locate the axis of the hole relative to the axis of the cylinder. As shown in Figure 22.62,  EMBED Equation.DSMT4 . Problem 23.48 shows that at points within a long insulating cylinder,  EMBED Equation.DSMT4 . Execute:  EMBED Equation.DSMT4 .  EMBED Equation.DSMT4 . Note that  EMBED Equation.DSMT4 is uniform. Evaluate: If the hole is coaxial with the cylinder,  EMBED Equation.DSMT4 and  EMBED Equation.DSMT4 .  INCLUDEPICTURE "FIG22-062.tif" \* MERGEFORMAT  Figure 22.62 22.63. Identify: The electric field at each point is the vector sum of the fields of the two charge distributions. Set Up: Inside a sphere of uniform positive charge,  EMBED Equation.DSMT4   EMBED Equation.DSMT4  directed away from the center of the sphere. Outside a sphere of uniform positive charge,  EMBED Equation.DSMT4  directed away from the center of the sphere. Execute: (a)  EMBED Equation.DSMT4  This point is inside sphere 1 and outside sphere 2. The fields are shown in Figure 22.63a.  INCLUDEPICTURE "FIG22-063a.tif" \* MERGEFORMAT  EMBED Equation.DSMT4 Figure 22.63a EMBED Equation.DSMT4  Thus  EMBED Equation.DSMT4  (b)  EMBED Equation.DSMT4  This point is inside sphere 1 and outside sphere 2. Each field is directed away from the center of the sphere that produces it. The fields are shown in Figure 22.63b.  INCLUDEPICTURE "FIG22-063b.tif" \* MERGEFORMAT  EMBED Equation.DSMT4   EMBED Equation.DSMT4 Figure 22.63b EMBED Equation.DSMT4   EMBED Equation.DSMT4  (c) x = R. This point is at the surface of each sphere. The fields have equal magnitudes and opposite directions, so E = 0. (d) x = 3R. This point is outside both spheres. Each field is directed away from the center of the sphere that produces it. The fields are shown in Figure 22.63c.  INCLUDEPICTURE "FIG22-063c.tif" \* MERGEFORMAT  EMBED Equation.DSMT4   EMBED Equation.DSMT4 Figure 22.63c EMBED Equation.DSMT4   EMBED Equation.DSMT4  Evaluate: The field of each sphere is radially outward from the center of the sphere. We must use the correct expression for E(r) for each sphere, depending on whether the field point is inside or outside that sphere. 22.64. Identify: The net electric field at any point is the vector sum of the fields at each sphere. Set Up: Example 22.9 gives the electric field inside and outside a uniformly charged sphere. For the positively charged sphere the field is radially outward and for the negatively charged sphere the electric field is radially inward. Execute:  (a) At this point the field of the left-hand sphere is zero and the field of the right-hand sphere is toward the center of that sphere, in the +x-direction. This point is outside the right-hand sphere, a distance  EMBED Equation.DSMT4 from its center.  EMBED Equation.DSMT4 . (b) This point is inside the left-hand sphere, at  EMBED Equation.DSMT4 , and is outside the right-hand sphere, at  EMBED Equation.DSMT4 . Both fields are in the +x-direction.  EMBED Equation.DSMT4 . (c) This point is outside both spheres, at a distance  EMBED Equation.DSMT4 from their centers. Both fields are in the +x-direction.  EMBED Equation.DSMT4 . (d) This point is outside both spheres, a distance  EMBED Equation.DSMT4 from the center of the left-hand sphere and a distance  EMBED Equation.DSMT4 from the center of the right-hand sphere. The field of the left-hand sphere is in the +x-direction and the field of the right-hand sphere is in the  EMBED Equation.DSMT4 .  EMBED Equation.DSMT4 . Evaluate: At all points on the x-axis the net field is parallel to the x-axis. 22.65. Identify: Let  EMBED Equation.DSMT4  be the electron charge contained within a spherical shell of radius  EMBED Equation.DSMT4 and thickness  EMBED Equation.DSMT4 . Integrate  EMBED Equation.DSMT4 from 0 to r to find the electron charge within a sphere of radius r. Apply Gauss s law to a sphere of radius r to find the electric field  EMBED Equation.DSMT4 . Set Up: The volume of the spherical shell is  EMBED Equation.DSMT4 . Execute: (a)  EMBED Equation.DSMT4 .  EMBED Equation.DSMT4   EMBED Equation.DSMT4 ; the total net charge of the atom is zero. (b) The electric field is radially outward. Gauss s law gives  EMBED Equation.DSMT4  and  EMBED Equation.DSMT4 . (c) The graph of E versus r is sketched in Figure 22.65. What is plotted is the scaled E, equal to  EMBED Equation.DSMT4 , versus scaled r, equal to  EMBED Equation.DSMT4 .  EMBED Equation.DSMT4 is the field of a point charge. Evaluate: As  EMBED Equation.DSMT4 , the field approaches that of the positive point charge (the proton). For increasing r the electric field falls to zero more rapidly than the  EMBED Equation.DSMT4 dependence for a point charge.  INCLUDEPICTURE "FIG22-065.tif" \* MERGEFORMAT  Figure 22.65 22.66. Identify: The charge in a spherical shell of radius r and thickness dr is  EMBED Equation.DSMT4 . Apply Gauss s law. Set Up: Use a Gaussian surface that is a sphere of radius r. Let  EMBED Equation.DSMT4  be the charge in the region  EMBED Equation.DSMT4 and let  EMBED Equation.DSMT4 be the charge in the region where  EMBED Equation.DSMT4 . Execute: (a) The total charge is  EMBED Equation.DSMT4 , where  EMBED Equation.DSMT4  and  EMBED Equation.DSMT4 . Therefore,  EMBED Equation.DSMT4  and  EMBED Equation.DSMT4 . (b) For  EMBED Equation.DSMT4 , Gauss s law gives EMBED Equation.DSMT4  and  EMBED Equation.DSMT4 . For  EMBED Equation.DSMT4 ,  EMBED Equation.DSMT4  and  EMBED Equation.DSMT4  For  EMBED Equation.DSMT4 ,  EMBED Equation.DSMT4 and  EMBED Equation.DSMT4 . (c)  EMBED Equation.DSMT4  (d) For  EMBED Equation.DSMT4 ,  EMBED Equation.DSMT4 , so the restoring force depends upon displacement to the first power, and we have simple harmonic motion. (e) Comparing to EMBED Equation.DSMT4 . Then  EMBED Equation.DSMT4  and  EMBED Equation.DSMT4  Evaluate: (f) If the amplitude of oscillation is greater than  EMBED Equation.DSMT4  the force is no longer linear in  EMBED Equation.DSMT4  and is thus no longer simple harmonic. 22.67. Identify: The charge in a spherical shell of radius r and thickness dr is  EMBED Equation.DSMT4 . Apply Gauss s law. Set Up: Use a Gaussian surface that is a sphere of radius r. Let  EMBED Equation.DSMT4  be the charge in the region  EMBED Equation.DSMT4 and let  EMBED Equation.DSMT4 be the charge in the region where  EMBED Equation.DSMT4 . Execute: (a) The total charge is  EMBED Equation.DSMT4 , where  EMBED Equation.DSMT4  and  EMBED Equation.DSMT4  Therefore,  EMBED Equation.DSMT4  and  EMBED Equation.DSMT4 . (b) For  EMBED Equation.DSMT4 , Gauss s law gives  EMBED Equation.DSMT4  and  EMBED Equation.DSMT4 . For  EMBED Equation.DSMT4 ,  EMBED Equation.DSMT4 .  EMBED Equation.DSMT4  and  EMBED Equation.DSMT4 . For  EMBED Equation.DSMT4 ,  EMBED Equation.DSMT4 , since all the charge is enclosed. (c) The fraction of  EMBED Equation.DSMT4  between  EMBED Equation.DSMT4  is  EMBED Equation.DSMT4  (d)  EMBED Equation.DSMT4  using either of the electric field expressions above, evaluated at  EMBED Equation.DSMT4  Evaluate: (e) The force an electron would feel never is proportional to  EMBED Equation.DSMT4  which is necessary for simple harmonic oscillations. It is oscillatory since the force is always attractive, but it has the wrong power of  EMBED Equation.DSMT4  to be simple harmonic motion.     22- PAGE 10 Chapter 22 Gauss s Law  22- PAGE 11 22-1 22 ¶HøH¾HĄHśHüH\I^I`IbIdItIvIxI¦IØIŖI¬IVJXJ†JˆJŠJŒJŽJJ¬J®J°J²JąJāJäJóčóčąÜąŌąÜŹŗąÜ­ ąÜąÜ“†ąÜ€Üy€qm`Sjóhmj]hll2EHč’UjåėÅI hmj]CJUVhmj]jhmj]U h"Ę5: h"ĘCJj:ļhmj]h"ĘEHö’UjāėÅI h"ĘCJUVjąėhmj]h"ĘEHņ’UjąėÅI h"ĘCJUVhÓµh"ĘOJPJQJ^Jo(hÓµh"Ę5:jÉ“hll2Uh"Ęjh"ĘUh"Ęhmj]@ˆž’aJh"Ęhmj]6@ˆž’aJ ®J°JčJ¼K(L’L®LĪM¤PŗRŽV¬Xī[8\]ä`Ōe§¢¢¢™”¢¢¢¢¢¢¢¢¢gdÓµgd"ʤx¤xgd²#ägdmj]WkdØņ$$IfT–lÖ0”’×Q€C €z tæö6ööÖ’’Ö’’Ö’’Ö’’4Ö4Ö laöŠTäJęJčJźJKKKK|K¼KÖKŲKčKźK(L*LŠLŒLŽLL¬L®L¶LĘLČLŠLŽLąLĪMŽMąM>N@N¼N¾NģNīNšNņNvOxOŠOŒOŗO÷ó÷óęŁ÷óÕóŠóŠóČÄČ¼ČÕó·­·ó­ó­ó•ó÷óˆ{÷óŠó÷ójK*hmj]hmj]EHś’UjéėÅI hmj]CJUVjh¤P’UhÓµhÓµOJPJQJ^Jo(hÓµhÓµ5: hmj]5jŃśhķrŽUh8äjh8äU hmj]6h"ĘjM÷hmj]hll2EHč’UjēėÅI hmj]CJUVhmj]jhmj]U+ŗO¼O¾OĄOźOģOPPP PQQTQVQ„Q†QˆQŠQ>R@RnRpRrRtRSS’S”SĀSÄSĘSČSJTLTPTRT^T`TņåŻŁŻŁĢæŻŁ·ŁŻŁŖŻŁŻŁƒŻŁ·ŁŻŁviŻŁdŁ`ŁŻhTSq hmj]6j%8hmj]hmj]EHś’UjóėÅI 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Equation MathType EFEquation.DSMT4ō9²q Į“ ļ,ŲųīDSMT5WinAllBasicCodePagesTimes New RomanSymbolCourier NewMT Extra)„ōXōBōAōuō!šAPōAōEō%ōB_AōC_AōEō*_HōAō@ōAHōA*_D_Eō_Eō_A  ƒE†== …¦F ƒE ƒA†== ƒqWinANSIWWdoc08PP ˆ0 ƒA†== ƒqPP ˆ0 ˆ4„Ąpƒr ˆ2ž’ ’’’’ĪĄFMathType 5.0 Equation MathType EFEquation.DSMT4ō9²qOĀYØč Ķ€čDSMT5WinAllBasicCodePagesTimes New RomanSymbolCourier NewMT Extra)„ōXōBōA_1237727343’’’’’’’’ÖĪĄFP­¬æ‘ĒP­¬æ‘ĒOle ’’’’’’’’’’’’CompObjÕ×’’’’iObjInfo’’’’Ų’’’’Equation Native ’’’’’’’’’’’’u_1239104373’’’’’’’’ŪĪĄFP­¬æ‘ĒP­¬æ‘ĒOle ’’’’’’’’’’’’ CompObjŚÜ’’’’!iōuō!šAPōAōEō%ōB_AōC_AōEō*_HōAō@ōAHōA*_D_Eō_Eō_A  ƒE†== ˆ4‚.ˆ5ˆ0†×“ˆ1ˆ0 †"-ˆ6  C‚(ˆ8‚.ˆ8ˆ5†×“ˆ1ˆ0 †"-ˆ1ˆ2  C 2 /N†Å"ׁm 2 ‚)‚(ˆ4„Ąp‚)‚(ˆ0‚.ˆ2ˆ5ˆ0 m‚) ˆ2 †==ˆ6‚.ˆ4ˆ7†×“ˆ1ˆ0 ˆ5  N/C.ž’ ’’’’ĪĄFMathType 5.0 Equation MathType EFEquation.DSMT4ō9²q Į! ļ,ŲųīDSMT5WinAllBasicCodePagesTimes New RomanSymbolCourier NewMT Extra)„ōXōBōAōuō!šAPōAōEō%ōB_AōC_AōEō*_HōAō@ōAHōA*_D_Eō_Eō_A  …¦F ƒE †== ObjInfo’’’’Ż’’’’#Equation Native 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