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

[heycoa]

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

 

[gabbagabbahey]

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.

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.

 

[heycoa]

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?

 

[gabbagabbahey]

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?

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?