Question 4.7: Find and plot the potential distribution between two long co...
Find and plot the potential distribution between two long concentric cylinders. The length of the cylinders is \mathscr{L} . The boundary conditions are: V(ρ = a) = V_{0} and V(ρ ≥ b) = 0.
Because of the symmetry, we should employ Laplace’s equation in cylindrical coordinates (4.22). We can assume that the cylinders are of infinite length and let \partial V / \partial z = 0 . In addition, there is no variation in the \phi direction, hence \partial V / \partial \phi = 0. We have to solve the one-dimensional Laplace’s equation in cylindrical coordinates, which becomes an ordinary differential equation in the independent variable ρ.
∇^{2} V = \frac{1}{\rho } \frac{\partial}{\partial \rho }\left\lgroup\rho \frac{\partial V }{\partial\rho } \right\rgroup + \frac{1}{\rho ^{2}} \frac{\partial ^{2} V}{\partial \phi^{2}} + \frac{\partial ^{2} V}{\partial z^{2} } (4.22)
\frac{1}{ρ} \frac{d}{dρ} \left\lgroup\rho \frac{dV}{d\rho } \right\rgroup = 0After multiplying both sides of this equation by ρ, we find the first integration yields
\frac{dV}{d\rho } = \frac{C_{1}}{\rho}which is the radial component of the electric field with a minus sign. The second integration yields
V =C_{1} \ln\rho + C_{2}Applying the boundary conditions, we write
V(\rho =a) \Rightarrow V_{0} = C_{1} \ln a+ C_{2}\\V(\rho = b) \Rightarrow 0 = C_{1} \ln b + C_{2}The constants of integration C_{1} and C_{2} are found from the simultaneous solution of these two equations.
C_{1} = \frac{V_{0}}{\ln(a/b)} and C_{2} = -C_{1} \ln b
The potential variation between the two cylinders is finally written as
V = V_{0} \left\lgroup\frac{\ln(\rho /b)}{\ln(a /b)} \right\rgroupwhere a < ρ < b. The variation of the potential for the radius b = 4a is shown below. Note that the potential within the inner cylinder is a constant.
