To derive the Laplacian in the context of a conical capacitor defined in cylindrical coordinates, we first note the geometry described:

1. Geometry:
   • The positive electrode is defined at a cone angle $\theta = \theta_+$.
   • The negative electrode is at $\theta = \theta_-$.
   • The radial distance $r$ is bounded by $r_a$ and $r_b$.
   • The azimuthal angle $\phi$ ranges from $0$ to $\phi_0$.

In cylindrical coordinates $(r, \phi, z)$, the Laplacian is given by:

$\nabla^2 V = \frac{1}{r} \frac{\partial}{\partial r} \left( r \frac{\partial V}{\partial r} \right) + \frac{1}{r^2} \frac{\partial^2 V}{\partial \phi^2} + \frac{\partial^2 V}{\partial z^2}$

Step 1: Convert to cylindrical coordinates

The conical shape can be represented in cylindrical coordinates, where $z$ can be expressed in terms of $r$ and $\theta$:

$z = r \tan(\theta)$

Step 2: Substitute into the Laplacian

Since $V$ is a function of $r$, $\phi$, and $z$, we express $z$ in terms of $r$:

• Given $z = r \tan(\theta)$, we find that $\frac{\partial V}{\partial z}$ needs to be expressed in terms of $r$ and $\phi$.

The Laplacian in this conical setup requires the following considerations:

1. Radial component: We can write the radial term as it is since it depends on $r$.
2. Azimuthal component: The dependence on $\phi$ remains as it is.
3. Vertical component: To include the vertical (or axial) term, we need to recognize the conical coordinate system's effect on the z-coordinate.

The expression becomes:

$\nabla^2 V = \frac{1}{r} \frac{\partial}{\partial r} \left( r \frac{\partial V}{\partial r} \right) + \frac{1}{r^2} \frac{\partial^2 V}{\partial \phi^2} + \tan^2(\theta) \frac{\partial^2 V}{\partial r^2}$

Step 3: Applying the boundary conditions

For the conical capacitor:

• Boundary Conditions:
  • At $\theta = \theta_+$, $V = V_+$ (positive potential)
  • At $\theta = \theta_-$, $V = V_-$ (negative potential)

The Laplacian must equal zero in the regions defined between the electrodes, as given:

$\nabla^2 V = 0$

This leads to the conclusion that in the volume defined between the electrodes, the potential $V$ satisfies Laplace's equation.

Final Laplacian Expression

Thus, in the conical capacitor setup, assuming the potential $V$ only depends on $r$ and $\phi$, we express it as:

$\frac{1}{r} \frac{\partial}{\partial r} \left( r \frac{\partial V}{\partial r} \right) + \frac{1}{r^2} \frac{\partial^2 V}{\partial \phi^2} + \tan^2(\theta) \frac{\partial^2 V}{\partial r^2} = 0$

This equation encapsulates the essence of the problem, reflecting the dependence of the electric potential on the conical geometry.

Summary

To summarize:

• We derived the Laplacian in a conical coordinate system.
• The boundary conditions were outlined for the potential between the electrodes.

If you have further questions or need details on specific steps, let me know!

Further Questions

1. How does the shape of the electrodes affect the electric field distribution?
2. What assumptions are made about the potential function $V$?
3. Can you derive the electric field from the potential $V$?
4. How would the Laplacian change if the cone was truncated?
5. What physical interpretations can be drawn from the solution to Laplace's equation here?

Tip

Understanding boundary conditions is crucial for solving problems involving Laplace's equation, as they define the behavior of the solution in the given geometry.