The polariztion vector in a dielectric tube of inner radius r, and outer radius r 0 is P → = ( a ^ x 2 x + a ^ y 3 y ). The axis of the dielectric tube defined by z 0 &gt; z &gt; 0 coincides with the z-axis.a) Determine the equivalent polarization surface and volume charge densisites

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The polariztion vector in a dielectric tube of inner radius r, and outer radius       r    0   is       P    →    =  (                    a        ^              x    2  x  +                    a        ^              y    3  y  ). The axis of the dielectric tube defined by       z    0    &amp;gt;  z  &amp;gt;  0 coincides with the z-axis.a) Determine the equivalent polarization surface and volume charge densisites

Gia Schaefer · Accepted Answer

From Gauss&#039;s Law, the divergence of the polarization vector is equal to surface bound charge density.  ρ  =  −            ▽      →        ⋅            P      →      Where   ρ is polarization volume charge density and P is polarization vector. The polarization vector is given by :            P      →        =      P    0    (                    a        ^              x    2  x  +                    a        ^              y    3  y  )              ▽      →        =      ∂          ∂      x                          a        ^              x    +      ∂          ∂      y                          a        ^              y  using equation   ρ  =  −            ▽      →        ⋅            P      →          ρ  =  −  (      ∂          ∂      x                          a        ^              x    +      ∂          ∂      y                          a        ^              y    )  ⋅  (      P    0    (                    a        ^              x    )  2  x  +                    a        ^              y    3  y  )    ρ  =  −  (  2      P    0    +  3      P    0    )  =  −  5      P    0  The polarization volume charge density is   −  5      P    0  Inner radius   =      r    i  Outer radius   =      r    o  Normal radial vector to the polarizing surface   =            a      →      For inner surface with radius       r    1              a      →        =  −                    a        r            →      such that,  −                    a        r            →        =  −  (                    a        x            →        cos  ⁡  (  ϕ  )  +                    a        y            →        sin  ⁡  (  ϕ  )  )where   ϕ is the angle that the radial vector makes with vector                     a        x            →       and                     a        y            →        x  =      r    1    cos  ⁡  (  ϕ  )    y  =      r    1    sin  ⁡  (  ϕ  )let the polarizing surface charge density be       ρ    i   for inner surface.      ρ    i    =  −            P      →        ⋅                    a        r            →              ρ    i    =  −      P    0    (                    a        x            →        2      r    i    cos  ⁡  (  ϕ  )  +                    a        →              y    3      r    i    sin  ⁡  (  ϕ  )  )  ⋅  (                    a        x            →        cos  ⁡  (  ϕ  )  +                    a        y            →        sin  ⁡  (  ϕ  )  )        ρ    i    =      P    0        r    i    (  2  +      sin    2    ⁡  ϕ  )The polarization surface charge density in the inner surface is        ρ    i    =      P    0        r    i    (  2  +      sin    2    ⁡  ϕ  )

The polariztion vector in a dielectric tube of inner radius r, and outer radius r 0 </

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