ANSWER TO CHAPTER 23 HOMEWORK

1. All field lines enter and leave the cube, so the net flux is

         F = 0.

    We find the flux through a face from

         F = ò E×dA.

     There are two faces with the field lines perpendicular to the face, say one at x = 0 and one at x = l.

     Thus for these two faces we have

          Fx=0 = -EA = -(6.50 x 103N/c)l2 

              Fx=l = +EA  = + (6.50 x103N/)l2

      For all other faces, the field is parallel to the face, so we have

          Fall others = 0.

 2. The total electric flux through the surface depends only on the enclosed charge:

             F = ò E×dA = Q/eo.

          The only contributions to the integral are from the faces perpendicular to the electric filed. Over each of these two surfaces, the magnitude of the field is constant, so we have

              F = ElA - EoA = (El - Eo)A = Q/eo;

                  (410N/c - 560N/C)(30m)2 = Q/(8.85 x10-12C2/N.m2),

which gives Q = -1.2 x 10-6C = -1.2 mC.

 

    3. From Example 22-5, the field from a long thin wire is radial with a magnitude given by

                 E = l/2peor2

          (a) At a distance of 5.0m the field is

                 E = (-2.8 x 10-6C/m)/2p(8.85 x 10-12C2/N.m2)(5.0m) = -1.0 x 104N/C (toward the wire).

          (b) At a distance of 2.0m from the field is

                 E = (-2.8 x 10-6C/m)/2p(8.85 x 10-12C2/N.m2)(2.0m) = -2.5 x 104N/C (toward the wire).

 4. From the symmetry of the charge distribution, we know that the electric field must be radial, with a magnitude independent of the direction.

    (a) For a spherical Gaussian surface within the spherical cavity, we have

          òE×dA = E4pr2 = Qenclosed/eo, so we have

          E = (1/4peo)Q/r2

             = (9.0 x 109N.m2/C2)(5.50 x 10-6C)/(0.030m)2

             = 5.5 x 107N/C (away from center).

     (b) the point 6.0cm from the center is inside the conductor, thus the electric field is 0.l

      Note that there must be a negative charge of -5.50mC on the surface of the cavity and a positive charge of +5.50mC on the outer surface of the sphere.

 5. From  the symmetry of the charge distribution, for points far from the ends and not too far from the shell, we know that the electric field must be radial, away from the axis of the cylinder, with a magnitude independent of the direction. For a Gaussian surface we choose a cylinder of length l and radius r, centered on the axis. On the ends of this surface, the electric field is not constant but E and dA are perpendicular, so we have E×dA = 0. On the curved side, the field has a constant magnitude and E and dA are parallel, so we have E× dA = E dA.

(a) For the region where r > Ro, the charge inside the Gaussian surface is Q = rEpRo2l.

For Gauss's Law we have

        òE×dA = òendsE×dA + òsideE×dA = Qenclosed/eo;

           0 + E2prl = rEpRo2l/eo, or E = rERo2/2eor, r > Ro.

(b) For the region where r < Ro, the charge inside the Gaussian surface is Q = rEpr2l, so we have

         ò E×dA = òendsE×dA + òsideE×dA = Qenclosed/eo;

         0 + E2prl = rEpr2l/eo, or E = rEr/2eo; r < Ro

 

 

6.* On the ends of the cylinder the electric field will vary in magnitude and direction. Thus we must integrate to find the flux through the ends. We choose a circular ring of radius y and thickness dy. From the diagram we see that

              R0 = rcosq,  

              y =  R0tanq,

             dy = R0sec2qdq = (R0/cos2q)dq.

The flux through one end is

            Fend = òE·dA = òEcosqdA

                     = ò(Q/4peor2)cosq2pydy

                     = Q/2eoò0p/4(cosq)(R0tanq)(cos2q)d/(R0/cosq)2dq

                     = Q/2eoò0p/4/4sinq dq

                     = Q/2eo(-cosq)|0p/4

                     = Q/2eo(1 - 1/Ö2)

The total flux through the closed surface is Q/eo, so the flux through the curved sides is

             Fsides  = Ftotal  - 2Fend  

             = (Q/eo) -2(Q/eo)(1 - 1/Ö2)

             = Q/eoÖ2.

7.* (a) The field from a large plate, not near the edge, is perpendicular to the plate and uniform: E = r/eo. for regions outside the slab, it can be considered an infinite number of plates. We can find the equivalent surface density by considering the charge in the slab with an area A.

              Qslab = sslabA = rEAd, or sslab = rEd.

the electric field to the left of the plate is

              Ea = Eplate + Eslab

                   = s/2eo + sslab/2eo

                  = (s  + rE)/2eo (left).

(b) The electric field to the right of the plate is

              Eb = Eplate + Eslab

                   = s/2eo + sslab/2eo

                  = (s  + rE)/2eo (right).

(c) To find the field inside the slab, we choose a cylinder for the Gaussian surface with one end of area A inside the slab parallel to the plate and the other end of area A to the left of the plate.

The cylinder is a distance x inside the slab. when we apply Gauss's law, we have

             òE·dA =  òend E·dA + òsideE·dA = Q/eo;

             EaA + EcA + 0 = (sA + rExA)/eo;

             (s + rEd)2eo + Ec = (s + rEx)/eo,

so         Ec  = [s + rE(2x - d)]/2eo (right).