1. The problem statement, all variables and given/known data The figure below shows a finite line charge with linear charge density of λ and total length L. The point P shown is a distance s away from its end. Please calculate a formula for the electric field at point P, in terms of λ, L and s. Then use the following values to find it numerically. λ = +3 μC/m, L = 5 m, s = 4 m 2. Relevant equations Electric field to point P in the x direction is going to be dE*cosΘ y direction is going to be -dE*sinΘ the equation for is E=∫(k*λdy/r^2) , where λ=Q/L, r=(s^2+L^2)^(1/2) limit of integral is from 0 to 5 3. The attempt at a solution i did get the answer for +x direction, but i couldnt get the answer for -y direction for the x-dir, my eq start from ∫(Kλdy/r^2)cosθ. where cos=s/r so it become ∫(kλdy/r^2)(s/r) since, k,λ,s, is a constant, the integration become kλs∫dy/(s^2+y^2)^(3/2) using table of integration for ∫dy/(s^2+L^2)^(3/2), then kλs(1/s^2)(y/(y^2+s^2)^(1/2)) s on the numerator and denominator cancels each other (kλ/s)(5/(5^2+4^2)^(1/2)) ((9*E9)(3*E-6)(5)/4)*(5/(5^2+4^2)^(1/2))=5270.8644 for the y-dir, its negative. limit of integration is 0 to 5 of L value my eq start from ∫(Kλdy/r^2)sinθ. where sin=L/r so it become -∫(kλdy/r^2)(L/r) since, k,λ,L, is a constant, the integration become -kλL∫dy/(s^2+y^2)^(3/2) using table of integration for ∫dy/(s^2+L^2)^(3/2), then -kλL(1/s^2)(y/(y^2+s^2)^(1/2) -(kλL)(5/(5^2+4^2)^(1/2)) -((9*e9)(3*E-6)(5)/(4^2))*(5/(5^2+4^2)^(1/2))= -6588.58058 but for the y-dir value is wrong. can someone help me?