Question

A thin half ring with a radius of R = 10 cm is uniformly charged with a linear density of = 1 Mikrokulon/m and located in a vacuum. Determine the force F of interaction between the half ring and a point charge q = 20 nC located at the center of curvature. (don't use chatgpt please)

Answer 1

Step-by-step explanation:

F = k * q * lambda * R * π * (1 - √2/2)

Substituting the given values of q, lambda, R, and k, we get:

F = (9 x 10^9 N*m^2/C^2) * (20 x 10^-9 C) * (1 x 10^-6 C/m) * (0.1 m) * π * (1 - √2/2)

F ≈ 8.58 x 10^-4 N

Therefore, the force of interaction between the half ring and the point charge is approximately 8.58 x 10^-4 N.