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heycoa
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Homework Statement
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.
Homework 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
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