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PHYSICS 505: CLASSICAL ELECTRODYNAMICS HOMEWORK 4 3 b) Let us ﬁnd the magnitude and direction of the electric ﬁeld at the origin. By the azimuthal symmetry of the problem, the electric ﬁeld must point in the direction ˆ z deﬁned by θ = 0. Computing rather directly, we see |E | = − ∂ϕ(r, θ) ∂r ¸ ¸ ¸ ¸ r=0 = − Q 8π 0 ∞ =1 2 + 1 r −1 R +1 [P +1 (cos α) −P −1 (cos α)] P (cos(0)) ¸ ¸ ¸ ¸ ¸ r=0 , = − Q 8π 0 1 3R 2 P 1 (1) [P 2 (cos α) −P 0 (cos α)] , = − Q 8π 0 1 3R 2 _ 1 2 _
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PHYSICS 505: CLASSICAL ELECTRODYNAMICS HOMEWORK 4 3 b)  Let us ﬁnd the magnitude and direction of the electric ﬁeld at the srcin.By the azimuthal symmetry of the problem, the electric ﬁeld must point in the directionˆ z  deﬁned by  θ  = 0. Computing rather directly, we see | E   | = −  ∂ϕ ( r,θ ) ∂r  r =0 = −  Q 8 π 0 ∞   =1  2   + 1 r  − 1 R  +1  [ P   +1 (cos α ) − P   − 1 (cos α )] P   (cos(0))  r =0 , = −  Q 8 π 0 13 R 2 P  1 (1)[ P  2 (cos α ) − P  0 (cos α )] , = −  Q 8 π 0 13 R 2  12  3cos 2 α − 1  − 1  , = −  Q 16 π 0 R 2  cos 2 α − 1 / 3 − 2 / 3  , ∴  E    = − Q sin 2 α 16 π 0 R 2  ˆ z. ‘ ´ oπρ  ’ ´ δι δ            ιξαι c)  Let us brieﬂy discuss the limiting cases of the above results when  α → 0 ,π .Because sin 2 ( α ) =  α 2 −  a 4 3  + O ( α 6 ), and is  π -periodic, it is clear that the electric ﬁeldapproaches E    = −  Qα 2 16 π 20 ˆ z  + O ( α 4 ) , for  α → 0 and E    = − Q ( π − α ) 2 16 π 20 ˆ z  + O (( π − α ) 4 ) , for  α → π.  This is expected. Notice that as the spherical charge distribution closes, α  →  0, we approach the ﬁeld inside a closed sphere, which vanishes by Gauß’ law.The symmetric situation as  α → π  also begins to vanish because the total charge onthe sphere decreases like ( π − α ) 2 near  α ∼ π . Therefore, the ﬁeld will decrease like( π − α ) 2 .Similarly, for  α  →  0, cos α  ∼  1 and so  P   +1 (1)  −  P   − 1 (1) vanishes. Therefore, thepotential will vanish for  α → 0 as expected for the interior of a charged sphere. For α → π , the potential will decrease like ( π − α ) 2 and will approach the potential of apoint charge of magnitude  Q/ 8 πR 2 ( π − α ) 2 . Problem 3.5 Let us consider the potential inside a sphere of radius  a  where the potential at the surface is speciﬁed.We are to demonstrate that ϕ ( r,θ,φ ) =  a ( a 2 − r 2 )4 π    ϕ ( θ  ,φ  ) d Ω  ( r 2 +  a 2 − 2 ar cos γ  ) 3 / 2  = ∞   =0   m = −  A   m  ra   Y  m ( θ,φ ) , where cos γ   = cos θ cos θ  + sin θ sin θ  cos( φ − φ  ) and  A   m  =    d Ω  Y   ∗ m ( θ  ,φ  ) ϕ ( θ  ,φ  ).From our work in Jackson’s second chapter, we know that the ﬁrst expression for the potential isthat obtained from the Green’s function G ( r,r  ) = 1 | r − r  | −  ar  | r −  a 2 r  2 r  | . Therefore, it is suﬃcient for us to show that the second expression for the potential is obtainable fromthe above Green’s function to show that the two expressions for the potential are equivalent.  4 JACOB LEWIS BOURJAILY Let us begin by expressing  G ( x,x  ) in terms of spherical harmonics. Following the now ‘standard’procedure used above, we see that G ( r,r  ) = 4 π ∞   =0   m = −  12   + 1 Y   ∗ m ( θ  ,φ  ) Y  m ( θ,φ )   r  r   +1  −  ar  r   a 2 r  2 r    +1  , = 4 π ∞   =0   m = −  12   + 1 Y   ∗ m ( θ  ,φ  ) Y  m ( θ,φ )   r  r   +1  −  r   r  a 2  +1  . To ﬁnd the potential, we must compute the normal derivative of the Green’s function in the directionof   r  , evaluated at  r  =  a . Let us spend a moment and compute this. ∂G ( r,r  ) ∂r   r  = a = 4 π ∞   =0   m = −  12   + 1 Y   ∗ m ( θ  ,φ  ) Y  m ( θ,φ )  − (   + 1)  r  r   +2  − r   − 1 r  a 2  +1  r  = a , = 4 π ∞   =0   m = −  12   + 1 Y   ∗ m ( θ  ,φ  ) Y  m ( θ,φ )  − (   + 1)  r  a  +2  −  r  a 2  +2  , = − 4 π ∞   =0   m = −  12   + 1 Y   ∗ m ( θ  ,φ  ) Y  m ( θ,φ )(2   + 1)  ra   1 a 2 , = − 4 π ∞   =0   m = −  Y   ∗ m ( θ  ,φ  ) Y  m ( θ,φ )  ra   1 a 2 . Now, we can directly compute the potential  ϕ ( r,θ,φ ) using ϕ ( r,θ,φ ) = −  14 π    ϕ ( θ  ,φ  )  ∂G ( r,r  ) ∂r   r  = a a 2 d Ω  , = 14 π    ϕ ( θ  ,φ  )4 π ∞   =0   m = −  Y   ∗ m ( θ  ,φ  ) Y  m ( θ,φ )  ra   1 a 2 a 2 d Ω  , =    ϕ ( θ  ,φ  ) ∞   =0   m = −  Y   ∗ m ( θ  ,φ  ) Y  m ( θ,φ )  ra   d Ω  , = ∞   =0   m = −  Y  m ( θ,φ )  ra      ϕ ( θ  ,φ  ) Y   ∗ m ( θ  ,φ  ) d Ω  , ∴  ϕ ( r,θ,φ ) = ∞   =0   m = −  A   m  ra   Y  m ( θ,φ ) . ‘ ´ oπρ  ’ ´ δι δ            ιξαι Problem 3.6 Let us consider a system of two point charges of charge  ± q   located at  z  = ± a , respectively. a)  Let us ﬁnd the electrostatic potential as an expansion in spherical harmonics and powers of   r .First notice that the electrostatic potential is trivially given by ϕ ( x ) =  q  4 π 0   1 | x − a | −  1 | x  + a |  , where  a  ≡  ( a, 0 , 0) in spherical coordinates. As before, we can expand the function1 / | x − a |  following Jackson’s equation (3.70). Furthermore, by azimuthal symmetry,it is clear that only  m  = 0 spherical harmonics contribute and so we can substi-tute Legendre polynomials in their place. Azimuthal symmetry also implies that

Jul 22, 2017

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Jul 22, 2017
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