Page 354 - Intro to Tensor Calculus
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348



              I 8.    A homogeneous dielectric is defined by D i and E i having parallel vector fields. Show that for a
                                     j
               homogeneous dielectric    =0.
                                     i,k
              I 9.  Show that for a homogeneous, isotropic dielectric medium that   is a constant.

              I 10.  Show that for a homogeneous, isotropic linear dielectric in Cartesian coordinates

                                                               α e
                                                       P i,i =     ρ f .
                                                             1+ α e
              I 11.  Verify the Maxwell’s equations in Gaussian units for a charge free isotropic homogeneous dielectric.
                                                                       ~
                                                                                 ~
                                       1                      ~     1 ∂B     µ ∂H
                                            ~
                                    ~
                                ∇· E = ∇· D =0            ∇× E = −  c ∂t  = −  c ∂t

                                                                                   ~
                                                                     ~
                                          ~
                                    ~
                                ∇· B =µ∇H =0                  ~   1 ∂D   4π  ~    ∂E   4π  ~
                                                          ∇× H =       +    J =      +   σE
                                                                  c ∂t    c     c ∂t    c
              I 12.    Verify the Maxwell’s equations in Gaussian units for an isotropic homogeneous dielectric with a
               charge.
                                                                            ~
                                                                         1 ∂B
                                                                   ~
                                               ~
                                           ∇· D =4πρ          ∇× E = −
                                                                         c ∂t
                                                                                ~
                                               ~
                                           ∇· B =0                ~   4π  ~  1 ∂D
                                                              ∇× H =     J +
                                                                       c     c ∂t
              I 13.  For a volume charge ρ in an element of volume dτ located at a point (ξ, η, ζ) Coulombs law is
                                                                ZZZ
                                                             1        ρ
                                                 ~
                                                E(x, y, z)=            2  b e r dτ
                                                           4π  0    V  r
                                        2
                              2
                                                 2
                                                          2
                (a) Show that r =(x − ξ) +(y − η) +(z − ζ) .
                                  1
                (b) Show that b e r =  ((x − ξ) b e 1 +(y − η) b e 2 +(z − ζ) b e 3 ) .
                                  r
                (c) Show that
                                   ZZZ                                                ZZZ
                                1        (x − ξ) b e 1 +(y − η) b e 2 +(z − ζ) b e 3  1        b e r
                    ~
                   E(x, y, z)=                 2         2        2 3/2  ρdξdηdζ =         ∇    2  ρdξdηdζ
                              4π  0      [(x − ξ) +(y − η) +(z − ζ) ]            4π  0         r
                                       V                         ZZZ                     V
                                                              1                  ρ(ξ, η, ζ)
                                                    ~
                (d) Show that the potential function for E is V =            2        2         2 1/2  dξdηdζ
                                                            4π  0   V  [(x − ξ) +(y − η) +(z − ζ) ]
                             ~
                (e) Show that E = −∇V.
                                      ρ
                               2
                (f) Show that ∇ V = −    Hint: Note that the integrand is zero everywhere except at the point where

                   (ξ, η, ζ)=(x, y, z). Consider the integral split into two regions. One region being a small sphere
                   about the point (x, y, z) in the limit as the radius of this sphere approaches zero. Observe the identity

                            b e r              b e r
                   ∇ (x,y,z)  2  = −∇(ξ, η, ζ)  2  enables one to employ the Gauss divergence theorem to obtain a
                            r                  r
                                                                       ZZ
                                                                    ρ       b e r      ρ
                   surface integral. Use a mean value theorem to show −      2  · ˆndS =  4π since ˆn = − b e r .
                                                                   4π  0  S  r        4π  0
                                                        ∗
              I 14.   Show that for a point charge in space ρ = qδ(x − x 0 )δ(y − y 0 )δ(z − z 0 ), where δ is the Dirac delta
               function, the equation (2.6.5) can be reduced to the equation (2.6.1).
              I 15.
                                        ~
                                            1
                (a) Show the electric field E =  r 2 b e r is irrotational. Here b e r =  r ~ r  is a unit vector in the direction of r.
                                                       ~
                (b) Find the potential function V such that E = −∇V which satisfies V(r 0 )= 0 for r 0 > 0.
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