Electric Field at the Origin due to a Segment of a Circular Wire

To find the electric field at the origin (point O) due to a segment of a circular wire, we can use the principles of Coulomb's law and integration.

1. Setting up the Problem
- We have a circular wire with radius R.
- The segment of the wire extends from 0 to Φ/2, where Φ is the angle in radians.
- The wire carries a constant linear charge density λ.

2. Electric Field Due to a Small Element of the Wire
- Consider a small element of the wire with length dℓ at an angle θ from the positive x-axis.
- The charge of this element can be calculated as dq = λdℓ.
- The electric field due to this element at point O can be found using Coulomb's law: dE = (k * dq) / r^2, where k is the electrostatic constant and r is the distance from the element to point O.
- Since r is constant for all elements of the wire, we can write dE = (k * dq) / R^2.

3. Integration to Find the Total Electric Field
- To find the total electric field at point O, we need to integrate the contributions from all the small elements of the wire.
- The total electric field at point O is given by E = ∫ dE.
- Substituting the expression for dE, we have E = ∫ (k * dq) / R^2.
- Since λ is constant along the wire, we can write dq = λdℓ.
- The limits of integration are from 0 to Φ/2.
- Therefore, E = ∫ (k * λdℓ) / R^2.

4. Solving the Integral
- Integrating the expression, we get E = (k * λ / R^2) * ∫ dℓ.
- The integral of dℓ is simply the length of the wire segment, which is R * Φ/2.
- Therefore, E = (k * λ / R^2) * (R * Φ/2).
- Simplifying, E = (k * λ * Φ) / (2R).

5. Final Result
- The electric field at the origin due to the segment of the circular wire is given by E = (k * λ * Φ) / (2R).

Conclusion
The electric field at the origin due to a segment of a circular wire of radius R, extending from 0 to Φ/2 and carrying a constant linear charge density λ, is given by (k * λ * Φ) / (2R). This result can be obtained by considering the contribution of each small element of the wire and integrating over the entire segment.