Recent content by Sonic98

That's right, so I can do: $$E = - \frac {πKλ} {2R} $$ $$E = 2Kλ * ( 1 - \frac {2y} {\sqrt{4y^2 + L^2}})$$ $$ L = 2 * π * R $$ Which should give me: R = 1.18962 L = 7.4746 Lambda: 8.41483E-10 Now I can pass to the next questions. Thank you :) !

L is equal to 2 * π * R ?

Electric field for the semi-circle $$E = - \frac {πKλ} {2R} $$ In this case E is equals to 10 N/C Electric field for the straighten wire $$E = 2Kλ * ( 1 - \frac {2y} {\sqrt{4y^2 + L^2}})$$ In this case E is equals to 8 N/C What I'm searching is R, λ, and the length of the wire, so I think...