Electric Field at the Center of a Circular Ring

Given:
- Volume charge density ρ(τ, φ, 2) = ροδ(z)6(-)cos(φ)
- Circular ring of radius R centered at the origin

To find:
- Electric field at the center of the ring

Solution:

1. Introduction
The electric field at a point due to a charged object can be calculated using Coulomb's law or Gauss's law. In this case, we will use Gauss's law to find the electric field at the center of the circular ring.

2. Gauss's Law
Gauss's law states that the electric flux through any closed surface is proportional to the total charge enclosed by that surface. Mathematically, Gauss's law can be written as:
∮E⃗ · dA⃗ = (1/ε₀) ∫ρdτ,

where E⃗ is the electric field, dA⃗ is a vector element of the surface, ε₀ is the vacuum permittivity, and ρ is the charge density.

3. Applying Gauss's Law
To apply Gauss's law, we consider a Gaussian surface in the form of a sphere centered at the origin and passing through the center of the circular ring. The radius of the sphere is chosen such that it encloses the entire ring.

4. Symmetry Considerations
Due to the symmetry of the problem, the electric field at any point on the Gaussian surface will have a radial component only. This is because the charge distribution is symmetrical about the z-axis and does not depend on the azimuthal angle φ.

5. Electric Field Calculation
By Gauss's law, the electric flux through the Gaussian surface is:
∮E⃗ · dA⃗ = E ∮dA = EA,

where E is the magnitude of the electric field and A is the area of the Gaussian surface.

6. Charge Enclosed by the Gaussian Surface
The charge enclosed by the Gaussian surface is the total charge of the circular ring. We can calculate it by integrating the charge density over the volume of the ring.

7. Integrating the Charge Density
The charge density ρ(τ, φ, 2) can be expressed as ροδ(z)6(-)cos(φ), where ρο is a constant. We integrate this charge density over the volume of the ring to find the total charge enclosed.

8. Simplifying the Integral
Since the charge density does not depend on the azimuthal angle φ, the integral over φ will give a factor of 2π. The integral over τ will give the volume of the ring, which is πR².

9. Electric Field Calculation (continued)
Substituting the calculated charge enclosed and the area of the Gaussian surface into Gauss's law equation, we have:
EA = (1/ε₀) ∫ρdτ.

10. Solving for Electric Field
We can solve for the electric field E by dividing both sides of the equation by the area A:
E = (1/ε₀) ∫ρdτ / A.

11. Final Result
Finally, we evaluate the integral, substitute the appropriate values, and solve for the electric field E. The electric field at the center of the circular ring can then be determined.

Remember to double-check all calculations and units to obtain an accurate result.