P19S2017 – Charged Ring and Three Point Charges

To determine the electric force acting on charge $q$ at the center of the ring, we must first determine the net electric field at the center of the ring by superposition.

Electric field created by the charged ring:
By symmetry (best friend), the electric field created by the charged ring at the center of the ring is equal to zero. Indeed, the electric field created by any (red) portion of the ring on the right is canceled by the electric field created by the symmetry (blue) portion of the ring on the left (opposite) as shown below

Electric field created by the three equally spaced point charges:
We draw the electric field created by each individual point charge at the center after numbering the charges as shown below

By symmetry, the electric field at the center points to the right and its horizontal component is equal to

\begin{aligned}
E_{net\ x}&=E_{1x}+E_{2x}+E_{3x} \\
\\
&=E_1+E_2{\mathrm{cos} \left(\theta \right)\ }+E_3{\mathrm{cos} \left(\theta \right)\ } \\
\\
&=\frac{k\left|Q\right|}{a^2}+\frac{kQ}{a^2}{\mathrm{cos} \left(\theta \right)\ }+\frac{kQ}{a^2}{\mathrm{cos} \left(\theta \right)\ } \\
\\
&=\frac{k\left|Q\right|}{a^2}+\frac{2kQ}{a^2}{\mathrm{cos} \left(\theta \right)\ } \\
\\
&=\frac{k}{a^2}\left(\left|Q\right|+2Q{\mathrm{cos} \left(\theta \right)\ }\right) \\
\\
&=\frac{9\times {10}^9}{4}\left(2\times {10}^{-4}+2\cdot 2\times {10}^{-4}\cdot \frac{1}{2}\right) \\
\\
&=9\times {10}^5\ \ N/C
\end{aligned}

The net electric field at the center of the ring points right and therefore, since the charge $q$ is negative, the electric force acting on $q$ point lefts and has magnitude

\boxed{F_E=\left|q\right|E_{net\ x}=4\times {10}^{-5}\cdot 9\times {10}^5=36 \ \ N}

Note: by symmetry, since all the point charges 2 and 3 have the same charge, we conclude that $E_{2y}$ and $E_{3y}$ are equal in magnitude. Thus, $E_{net\ y}$ is zero as derived below

E_{net\ y}=E_{1y}+E_{2y}+E_{3y}=0-E_2{\mathrm{sin} \left(\theta \right)\ }+E_3{\mathrm{sin} \left(\theta \right)\ }=0