Calculate The Electric Field Generated By A Quarter Circle I'm just beginning to study electrostatics and have a question about the following problem:

A quarter circle of radius $R$ is uniformly charged with a total charge $Q$. What is the electric field at the origin, which is the center of the arc?


My Idea:
We can use $E =\frac{F_e}{q}$ where $q$ is the point charge at the origin.
First of all, the linear charge density, $\lambda$, would be $\frac{Q}{\frac{\pi R}{2}} = \frac{2Q}{\pi R}.$
If we position the quarter circle such that the two ends of the arc are at $\frac{3 \pi}{4}$ and $\frac{5 \pi}{4}$ radians, then the $y$-forces will cancel out. The only force on the origin will be in the $x-$direction and can be calculated by finding twice the force applied on the semi-circle from $\frac{3 \pi}{4}$ to $\pi$ (due to symmetricity).
Choose a point $X$ on the arc that has an angle of $\theta$ (in polar form) and consider the width of the arc to be $d \theta$. Then, the total charge is $\lambda d \theta = \frac{2Q}{ \pi R} d \theta$ and the distance is $R$. So, by Coulomb's Law, the total electrostatic force felt is
$$\frac{1}{4 \pi \epsilon_0} \frac{ \frac{2Q}{ \pi R} d \theta q}{R^2} = \frac{Q q}{2 \pi^2 \epsilon_0 R^3} d \theta$$
Thus, the force in the $x$-direction is
$$\frac{Q q}{2 \pi^2 \epsilon_0 R^3} ( - \cos(\theta)) d\theta$$
Now, we integrate this:
$$F_e = 2 \int_{\frac{3 \pi}{4}}^\pi \frac{Q q}{2 \pi^2 \epsilon_0 R^3}  (-\cos(\theta)) d\theta = \frac{Qq}{\pi^2 \epsilon_0 R^3} \int_{\frac{3 \pi}{4}}^\pi - \cos(\theta) d\theta = \frac{ \sqrt{2}}{2} \frac{Qq}{\pi^2 \epsilon_0 R^3}$$
So, the electric field is $$\frac{ \sqrt{2}}{2} \frac{Q}{\pi^2 \epsilon_0 R^3}$$
However, apparently, the correct answer should be
$$\frac{ \sqrt{2}}{2} \frac{Q}{\pi^2 \epsilon_0 R^2}$$
Can anyone see why I am off by a factor of $R$? What did I mess up?

From $\vec{E} = \vec{F}/q$ we can arrive to more convenient formula for finding electric field
$$
\vec{E} = \frac{1}{4\pi\epsilon_0} \int \frac{dq}{r^2} ( {\hat{x} \cos\theta + \hat{y} \sin\theta} ).
$$
Then using $dq = \lambda dl = \lambda r d\theta$ as pointed out by Aryan Komarla, where $r = R$ in your case, we can arrive at
\begin{equation}\tag{1}\label{eqn:electric-field-arc}
\begin{array}{rcl}
\vec{E} & = & \displaystyle \frac{1}{4\pi\epsilon_0} \int \frac{R\lambda d\theta}{R^2} ( {\hat{x} \cos\theta + \hat{y} \sin\theta} ) \newline
& = & \displaystyle \frac{1}{4\pi\epsilon_0} \frac{\lambda}{R} \int ( {\hat{x} \cos\theta + \hat{y} \sin\theta} ) d\theta.
\end{array}
\end{equation}
Since you have already chosen integral lower and upper bounds to be $\frac{3}{4}\pi$ and $\frac{3}{4}\pi$ then
$$
\begin{array}{rcl}
\vec{E} & = & \displaystyle \frac{1}{4\pi\epsilon_0} \frac{\lambda}{R} \int_{\frac34 \pi}^{\frac54 \pi} ( {\hat{x} \cos\theta + \hat{y} \sin\theta} ) d\theta \newline
& = & \displaystyle \frac{1}{4\pi\epsilon_0} \frac{\lambda}{R} \left[ {\hat{x} \sin\theta - \hat{y} \cos\theta}\right]_{\theta = \frac34 \pi}^{\frac54 \pi} \newline
& = & \displaystyle \frac{1}{4\pi\epsilon_0} \frac{\lambda}{R} \left[ \hat{x}(-\tfrac12\sqrt{2} - \tfrac12\sqrt{2} ) - \hat{y}(-\tfrac12\sqrt{2} + \tfrac12\sqrt{2}) \right] \newline
& = & \displaystyle \frac{1}{4\pi\epsilon_0} \frac{\lambda}{R} \sqrt{2} \hat{x}.
\end{array}
$$
Finally using your result $\lambda = 2Q/\pi R$ will lead us to
$$
\begin{array}{rcl}
\vec{E} & = & \displaystyle \frac{1}{4\pi\epsilon_0} \frac{2Q}{\pi R^2} \sqrt{2} \hat{x} \newline
& = & \displaystyle \frac{\sqrt{2}}{2} \frac{Q}{\pi^2 \epsilon_0 R^2} \hat{x}
\end{array}
$$
that you expect. Bt the way Eqn. \eqref{eqn:electric-field-arc} is more general and you can choose any initial and final angles you want.
