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The potential inside and outside of an infinitely long cylinder, where σ=a*sin(5θ)

  1. Oct 19, 2012 #1
    1. The problem statement, all variables and given/known data

    Charge density: σ(θ)=w*sin(5θ)

    (where a is a constant) is glued over the surface of an infinite cylinder of radius R. Find the potential inside and outside the cylinder.


    2. Relevant equations
    s is a point inside or outside of the cylinder, and θ is the angle between s and the x-axis.
    Laplace equation in cylindrical coordinates:

    V(s,θ)=a0+b0*ln(s)+Ʃ(s^k(a*cos(kθ)+b*sin(kθ)+s^-k(c*cos(kθ)+d*sin(kθ)))

    (*)Charge density: σ=-ε0(∂V(outside)/∂s-∂V(inside)/∂s) evaluated at s=R

    3. The attempt at a solution

    So I solved inside and outside potential, the answers are given as follows:

    Inside: V(s,θ)=a0+Ʃ(s^k(a*cos(kθ)+b*sin(kθ)) This is because ln(0) is undefined so b0 must equal 0 AND 0^-k is undefined so s^-k(c*cos(kθ)+d*sin(kθ) must equal 0.

    Outside: V(s,θ)=a0+Ʃ(s^-k(c*cos(kθ)+d*sin(kθ)))

    Now using the charge density equation (*):

    σ=ε0*Ʃ(K*R^(-k-1)*(c*cos(kθ)+d*sin(kθ)-K*R^(k-1)*(a*cos(kθ)+b*sin(kθ))
    Thus:
    w*sin(5θ)=ε0*Ʃ(K*R^(-k-1)*(c*cos(kθ)+d*sin(kθ)-K*R^(k-1)*(a*cos(kθ)+b*sin(kθ))

    Now this is where I begin to struggle. The answer manual assumes that a=c=0 and b=d=0 except when k=5. It then proceeds to say that w=5*ε0(d5/R^6+R^4*b5). I am missing the connection as to why these claims are true.

    Any help is appreciated, thank you
     
  2. jcsd
  3. Oct 19, 2012 #2

    gabbagabbahey

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    Re: The potential inside and outside of an infinitely long cylinder, where σ=a*sin(5θ

    This is a consequence of the orthogonality of the sines and cosines. Basically, on your left you have only a single Fourier sine mode, w*sin(5θ). On your right you have an infinite sum of Fourier sine and cosine modes, in other words, you have a full Fourier series. To see explicitly why the only nonzero terms are the sin(5θ) terms, try multiplying both sides of the equation by sin(kθ) and integrating over a full period, then try the same thing with cos(kθ) instead.
     
  4. Oct 19, 2012 #3
    Re: The potential inside and outside of an infinitely long cylinder, where σ=a*sin(5θ

    I forgot to mention that the sum:
    Ʃ(K*R^(-k-1)*(c*cos(kθ)+d*sin(kθ)-K*R^(k-1)*(a*cos(kθ)+b*sin(kθ))

    is from k=1 to infinity.

    Did you mean divide both sides by sin(5θ)? If I multiply then I would have w*sin^2(5θ)= the integral with respect to theta on the right side?
     
  5. Oct 20, 2012 #4

    gabbagabbahey

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    Re: The potential inside and outside of an infinitely long cylinder, where σ=a*sin(5θ

    Multiply by sin(nθ) (where n is an integer) and integrate over theta for a full period (you will have a definite integral)....when is the integral on the LHS non-zero? What does that tell you about the terms on the RHS?
     
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