Electric potential in cylindrical coordinates using separation of variables

[itwong]

Hi, I am new to the forum. I've encountered a couple problems with separation of variables in cylindrical coordinates.

Problem #1: a clindrical surface of radius R is oriented along the z-axis, and is split into two
conducting half-cylinders. The potential satisfies the boundary conditions
V(R,phi) = V0 , for 0<phi<pi; V(R,phi) = -V0, for pi<phi<2*pi. Find the potential
inside and outside the cylinder and the capacitance per unit length.

For this problem, I know the general solution is independent of z, namely V(r,phi) and it contains sin(n*phi), cos(n*phi) and r^n terms. But how to find the coefficients using the boundary condition confuses me. Is it necessary to assume V->0 as r -> infinity? Please
help!

Problem #2: an empty can haas the shape of a sector of a cylinder. The volume of the can
is defined by the cylindrical coordinates: 0<= r <= R, 0<= phi <= pi/2, -inf<z<inf
Basically one quarter of a cylinder in the first quadrant with boundary conditions
(i) V= 0 when phi = 0; (ii) V = 0 when phi = pi/2; (iii) V= V0 when r = R. Find the
potential everywhere inside the can.

For this problem, I know it's similar to the first problem except the boundary conditions.
maybe even simpler because of the zero potential. but I still have no idea how to approach this problem. Thank you for your help.

 

[eljose79]

Hello and welcome to the forum! Separation of variables in cylindrical coordinates can definitely be tricky, but with some practice and guidance, it can become easier to understand.

For Problem #1, we can start by looking at the general solution you mentioned: V(r,phi) = sum of (A_n*r^n + B_n*r^(-n))*(C_n*cos(n*phi) + D_n*sin(n*phi)). Since we are dealing with a cylindrical surface, we can assume that the potential is independent of z, as you mentioned. This means that the coefficients A_n and B_n will be zero. So our solution becomes V(r,phi) = sum of C_n*cos(n*phi) + D_n*sin(n*phi).

Now, we can use the boundary conditions to find the coefficients C_n and D_n. For 0 < phi < pi, we know that the potential is V0, so we can set phi = 0 and phi = pi in our solution and equate them to V0. This will give us two equations: C_n = V0 and -C_n = V0. Since these two equations contradict each other, we can conclude that there is no solution that satisfies both boundary conditions. This means that V(r,phi) = V0 is not a valid solution for 0 < phi < pi. A similar argument can be made for pi < phi < 2*pi.

For the potential inside the cylinder, we can use the boundary condition at r = R to find the coefficients. We know that V(R,phi) = V0 for 0 < phi < pi and V(R,phi) = -V0 for pi < phi < 2*pi. This means that C_n = V0 and -C_n = -V0, so C_n = V0. For the potential outside the cylinder, we can use the boundary condition at r = infinity to find the coefficients. Since we assume that V -> 0 as r -> infinity, this means that D_n = 0. Therefore, our final solution for the potential becomes V(r,phi) = sum of V0*cos(n*phi).

To find the capacitance per unit length, we can use the formula C = Q/V, where Q is the charge on each half-cylinder and V is the potential difference between them. Since V(R,phi) = V0 and V(infinity,phi) = 0,

 

[astrokreng]

Hello and welcome to the forum! Separation of variables is a useful technique for solving potential problems in cylindrical coordinates. In this technique, we assume that the potential can be expressed as a product of functions that depend only on one variable each. For cylindrical coordinates, the potential can be expressed as V(r,phi,z) = R(r)*Φ(phi)*Z(z). Then, we can use the Laplace equation to solve for each of these functions separately.

For problem #1, we can start by solving for the potential inside the cylinder, where r < R. Using the boundary condition V(R,phi) = V0, we can set the potential function equal to V0 and solve for the coefficients of the sine and cosine terms. Similarly, for the outside of the cylinder, where r > R, we can use the boundary condition V(R,phi) = -V0 to solve for the coefficients. As for the capacitance per unit length, we can use the formula C = Q/V, where Q is the charge on one half-cylinder and V is the potential difference between the two half-cylinders.

For problem #2, we can use the same approach as problem #1. However, since the boundary conditions are different, we will get different coefficients for the sine and cosine terms. We can use the boundary condition V= 0 when phi = 0 to solve for the coefficients of the sine terms, and V = 0 when phi = pi/2 to solve for the coefficients of the cosine terms. Then, for the boundary condition V= V0 when r = R, we can set the potential function equal to V0 and solve for the coefficients of the r^n terms. This will give us the potential everywhere inside the can.

It is not necessary to assume V->0 as r->infinity in these problems. The boundary conditions are sufficient to determine the coefficients and the potential function within the specified region. I hope this helps in solving these problems. Let me know if you have any further questions. Happy problem solving!