Your approach is on the right track, but let's clarify and streamline the derivation step by step to ensure clarity and correctness.

Step-by-Step Derivation of the Electric Field in a Conical Capacitor

1. Understanding Geometry and Symmetry:

   • You correctly noted that the system exhibits azimuthal symmetry (around the z-axis, ϕ) but lacks radial (r) and angular (θ) symmetry. This means the electric field will depend on r and θ.

2. Using Gauss's Law:

   • Gauss's Law states: $\Phi_E = \frac{Q_{\text{enc}}}{\epsilon_0}$
   • The electric flux through a closed surface is given by: $\Phi_E = \oint \vec{E} \cdot d\vec{A}$
   • For the conical capacitor, consider a conical Gaussian surface between the electrodes.

3. Electric Field Representation:

   • Due to the symmetry, assume: $\vec{E} = E(r, \theta) \hat{r}$

4. Surface Area Element:

   • The area element on the conical surface can be expressed as: $dA = r^2 \sin(\theta) d\theta d\phi$
   • Thus, the electric flux through the Gaussian surface becomes: $\Phi_E = E(r, \theta) \oint dA = E(r, \theta) \int_0^{\phi_0} \int_{r_a}^{r_b} r^2 \sin(\theta) d\theta d\phi$

5. Charge Enclosed:

   • The charge enclosed by the Gaussian surface can be calculated as: $Q_{\text{enc}} = \sigma_s \times \text{(Area of the positive electrode)}$
   • The area of the conical surface of the positive electrode is: $A = \int_0^{\phi_0} \int_{r_a}^{r_b} r \sin(\theta_+) dr d\phi = \sin(\theta_+) \left( \frac{(r_b^2 - r_a^2)}{2} \right) \phi_0$

6. Combining Equations:

   • Therefore, we have: $Q_{\text{enc}} = \sigma_s \sin(\theta_+) \left( \frac{(r_b^2 - r_a^2)}{2} \right) \phi_0$

7. Setting Up Gauss's Law:

   • Now substitute into Gauss's Law: $E(r, \theta) \int_0^{\phi_0} \int_{r_a}^{r_b} r^2 \sin(\theta) d\theta d\phi = \frac{\sigma_s \sin(\theta_+) \left( \frac{(r_b^2 - r_a^2)}{2} \right) \phi_0}{\epsilon_0}$
   • The integral on the left gives: $\int_0^{\phi_0} d\phi \int_{r_a}^{r_b} r^2 \sin(\theta) dr = \phi_0 \left( \frac{(r_b^3 - r_a^3)}{3} \right) \sin(\theta)$

8. Final Expression:

   • Rearranging gives: $E(r, \theta) = \frac{\sigma_s \sin(\theta_+) (r_b^2 - r_a^2) / (2 \epsilon_0)}{\phi_0 (r^2 \sin(\theta) \frac{(r_b^3 - r_a^3)}{3})}$
   • This indicates the electric field is dependent on both r and θ, consistent with your assertion.

Conclusion

Your fundamental understanding was correct, but be mindful of the integration limits and the form of the charge density. The final expression for $E(r, \theta)$ can often be simplified further depending on the specific geometry and boundary conditions. Make sure to verify that you've considered the physical limits and any approximations.

Questions for Further Consideration:

1. How does the electric field vary if the angle θ_+ is changed?
2. What happens to the electric field if the radius $r_a$ approaches $r_b$?
3. Can you analyze how the dielectric material affects the electric field?
4. How would the electric field change if the capacitor were to operate under different voltage conditions?
5. What is the impact of edge effects at the boundaries of the conical capacitor?

Tip

Visualizing the setup using diagrams can significantly enhance your understanding of how the electric field behaves in non-standard geometries like conical capacitors.