To simplify the expression $E(r,\theta) = \frac{\rho_s \left( \frac{r_b^2 - r_a^2}{2} \right) \phi_0 \sin(\theta_+)}{\epsilon_0 r^2 \sin(\theta) d\theta d\phi}$, we can take several steps. First, let’s analyze the expression itself:

Steps to Simplify the Expression

1. Factor Out Constants: Identify any constants that can be factored out. Here, $\rho_s$, $\frac{r_b^2 - r_a^2}{2}$, and $\phi_0$ are constants with respect to $r$ and $\theta$.

2. Rewrite the Expression: $E(r,\theta) = \frac{\rho_s \cdot \frac{r_b^2 - r_a^2}{2} \cdot \phi_0 \sin(\theta_+)}{\epsilon_0} \cdot \frac{1}{r^2 \sin(\theta) d\theta d\phi}$

3. Substitute Variables: If you have specific values or relationships for $r_a$, $r_b$, and $\phi_0$, substitute those to see if further simplification is possible.

4. Focus on Angular Terms: Note that $d\theta d\phi$ is a differential area element in spherical coordinates. If you need $E$ in terms of a surface area or related quantity, further context from the setup may allow additional simplifications.

Considering the Context of the Problem

In relation to the conical capacitor described:

1. Gauss' Law: You need to apply Gauss' Law, which states: $\Phi_E = \oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\epsilon_0}$ where $\Phi_E$ is the electric flux, and $Q_{\text{enc}}$ is the charge enclosed.

2. Choosing a Gaussian Surface: For a conical capacitor, a conical Gaussian surface is appropriate. The symmetry suggests that the electric field $E$ will have a constant magnitude at any radius $r$, but will vary with $\theta$.

3. Calculating Charge Enclosed: The charge $Q_{\text{enc}}$ can be computed using the surface charge density $\rho_s$ over the surface area of the conical electrode.

Derivation of the Electric Field

1. Surface Area of the Conical Electrode: $A = 2\pi r^2 (1 - \cos(\theta)) \text{ (for a conical section)}$ This accounts for the azimuthal symmetry.

2. Applying Gauss' Law: For the conical surface: $\Phi_E = E(r,\theta) \cdot A$ Thus: $E(r,\theta) \cdot A = \frac{Q_{\text{enc}}}{\epsilon_0}$

3. Substituting $Q_{\text{enc}}$: $Q_{\text{enc}} = \rho_s \cdot A$

4. Final Expression for $E$: Combining the above, you will derive $E(r,\theta)$ in terms of the parameters provided.

Conclusion

Thus, the expression can be simplified and restructured based on the physical context. The crucial step is to express it in relation to the charge enclosed and the area through which it acts.

Questions to Explore Further

1. How does the electric field change as you move from $r_a$ to $r_b$?
2. What are the boundary conditions at $\theta_+$ and $\theta_-$?
3. How can the angle $\theta$ affect the field distribution?
4. What assumptions are made about the symmetry of the electric field?
5. How would the electric field behave if the conical surfaces were not uniform?

Tip

When simplifying expressions, always look for common factors and consider the physical significance of the variables involved to guide your simplification.