5) Find the electrostatic Potential on the axis of a hollow Cylinder. § (r=0,2), (0&rse the lateral surface of which is maintained at 8(r=a, 7) =/ ooLZLC CEZ LL and the ends have potentials , Vp = V Vo=0 and v=v { c С Voor
To find the electrostatic potential on the axis of a hollow cylinder, we can use Laplace's equation and the method of separation of variables. Let's go through the steps:
1) Set up the problem:
We have a hollow cylinder with a radius of "a" and a uniform potential Vp on the ends of the cylinder (where the surface is perpendicular to the axis), and the lateral surface of the cylinder is maintained at a constant potential Vo. We want to find the potential at a point on the axis of the cylinder, denoted by r.
2) Apply the method of separation of variables:
Assuming that the potential is separable, we can write the potential as a product of two functions: V(r, θ) = R(r)Θ(θ).
3) Solve the radial equation:
Plug in the separable form of the potential into Laplace's equation (ΔV = 0) in cylindrical coordinates. We get:
1/r d/dr (r dR/dr) + (1/r^2)((d^2R/dr^2) + (k^2)R = 0
where k^2 is the separation constant.
To solve this equation, we look for solutions of the form R(r) = AR1(r) + BR2(r), where R1 and R2 are linearly independent solutions. We can assume that B = 0 to satisfy the boundary condition at r = 0 (in order to avoid a singularity at the origin).
Thus, we are left with the equation:
1/r d/dr (r dR1/dr) + (1/r^2)(k^2 - l(l+1)/r^2)R1 = 0
where l(l+1) is another separation constant.
From this equation, we can find two solutions R1 and R2 which will be linearly independent Bessel functions of order l: R1 = Jl(kr) and R2 = Yl(kr).
4) Solve the angular equation:
Plug in the separable form of the potential into the angular equation in Laplace's equation. We can write Θ(θ) = e^ilθ, where l is an integer.
5) Apply boundary conditions:
Next, apply the boundary conditions to determine the values of the constants in the solutions. At r = a, the potential is Vo, so we get:
R(a)Θ(θ) = Vo
Substituting R = Jl(kr) and Θ = e^ilθ, we get:
Jl(ka)e^ilθ = Vo
6) Evaluate the potential on the axis:
To evaluate the potential on the axis (r = 0), we need to take the limit as r approaches 0 of the radial solution. In this case, R1 = Jl(kr), so we have:
lim (r→0) Jl(kr) = lim (r→0) (kr)^l / (2l+1)! = 0
Therefore, for the potential to be finite at r = 0, the coefficient A in the radial solution must be zero.
7) Find the values of k and l:
To find the values of k and l, we need to examine the equation Jl(ka)e^ilθ = Vo. The product of Bessel functions and exponential term must be real, so we can write:
Jl(ka)e^ilθ = Vo = Vp - V0
Solving for ka and l:
ka = αn (where αn is the nth zero of Jl(x))
l = l (integer)
8) Evaluate the potential on the axis:
Finally, we can write the potential on the axis as:
V(r) = ∑ (Vo - V0) Jl(ka)e^ilθ
where the summation is over all possible values of l.
This equation gives you the expression for the electrostatic potential on the axis of a hollow cylinder. To find the potential at a specific value of r, substitute the appropriate values for l, ka, and θ into the equation and calculate the sum.