Apositive charge q is distributed uniformly along the positive y-axis between y=0 and y=a. a negative point charge, -q, lies on the positive x-axis, a distance x from the origin. calculate the x and y components of the e-field produced by the charge distribution q at points on the positive x axis.

This can be done using integration and vector analysis. By considering a differential element of the line charge distribution dq = Q/a*dy, we could calculate the dFx and dFy and solve for Fx and Fy through integration.

dFx = -xkQdq / (y^2 + x^2)^3/2 = -xkQ^2/a/(y^2+x^2)^(3/2)*dy
Fx = integral from 0 to a -xkQ^2/a/(y^2+x^2)^(3/2)*dy
Fx = -xkQ^2/a*int(0, a, dy/(y^2+x^2)^(3/2))
Fx = -kQ^2/x(a^2+x^2)^(1/2)

dFy = kQy/(y^2+x^2)^(3/2) * dq
dFy = kQ^2/a*y/(y^2+x^2)^(3/2)*dy
Fy = int(0, a, kQ^2/a*y/(y^2+x^2)^(3/2)*dy)
Fy = kQ^2/a int(0, a, dy y/(y^2+x^2)^(3/2))
Fy = kQ^2/a * (1/x - 1/(a^2+x^2)^(1/2))

He can play two games. 13/4=3.253.25*2=6.5