<p>@RentonT</p>
<p>You have to integrate (k*dQ/R) rather than assume that all the charge is concentrated at the center. Since it is uniformly distributed over a flat quarter circle, it’s uniform relative to length. So dQ = (σ) dL. The σ value is equal to Q/(π /2 * R), because it’s total charge over total length. </p>
<p>And if you observe a small piece of length dL, you know the relation is dL = R * dΘ. So all in all, your total expression is dQ = [Q/(π /2 * R)] * R * dΘ, which ultimately cancels out into dQ = 2Q/π dΘ. When you integrate k*dQ/R, you substitute this into the expression, getting :</p>
<p>2kQ/πR dΘ. When you integrate theta from -π/4 to π/4 (quarter circle), you get π/2. This multiplied towards 2kQ/πR cancels out into kQ/R.</p>
<p>For part (e), instead you integrate k*dQ/r^2 * cos Θ (Because all the vertical components cancel out due to symmetric property). You still have the same dQ value, so you integrate (2kQ/πR^2 * cosΘ dΘ). The first half is constant and you pull it out, and the integral of cos Θ dΘ is sin Θ. When you subtract, sin(π/4) - sin (-π/4 ), you get sqrt(2). You multiply this to the constant you have on the outside, and get 2kQ *sqrt 2/πR. Now just substitute k = 1/4πe0, and you get the same answer that they provided. Hope that helped!</p>