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?

 

[paranormal]

for your time and effort.



Thank you for sharing your solution and the answer manual's solution. It seems like you have correctly solved for the potential inside and outside the cylinder. However, I can see why you are struggling to understand the answer manual's approach.

First, let's take a closer look at the charge density equation (*). This equation is derived from the Laplace equation in cylindrical coordinates, which is a fundamental equation in electrostatics. It relates the potential to the charge density, and it is valid for both inside and outside the cylinder. So, in order to use this equation to solve for the potential, we need to know the charge density at the surface of the cylinder (s=R). This is where the value of w comes into play.

Now, let's look at the answer manual's approach. It assumes that a=c=0 and b=d=0 except when k=5. This means that for all values of k, except when k=5, the coefficients a, b, c, and d are equal to 0. This simplifies the charge density equation to:

w*sin(5θ)=ε0*K*R^(-k-1)*(d5*cos(5θ)-b5*sin(5θ))

Since we know that w is a constant, and the only non-zero coefficients are d5 and b5, we can rewrite the equation as:

w*sin(5θ)=ε0*K*R^(-k-1)*(d5*cos(5θ)+b5*sin(5θ))

Now, when k=5, the equation simplifies further to:

w*sin(5θ)=ε0*K*R^(-6)*(d5*cos(5θ)+b5*sin(5θ))

This is the equation that the answer manual uses to solve for w. By equating the coefficients of sin(5θ) and cos(5θ), they are able to solve for d5 and b5, and ultimately for w.

I hope this explanation helps you understand the answer manual's approach better. Keep up the good work in solving electrostatics problems!