Two qual point charges of $8\mu$ C are fixed at points $(2m,0),(-2m,0)$. Another charge $q$ with a mass of $91mg$ is released at point \((0,0.1 m )\) at $t=0$. It is observed that $q$ oscillates in simple harmonic motion about the origin.

at t=0,  force experienced by \(q\) is $9\times {10}^{-3}N$.

Charge $q$ is

$\text{ Charge }q\text{ oscillates, so }q\text{ is negative. Force on }q\text{ is }$

So, net force on $q$ is

$$\begin{gathered} F_{\text{net }}=2F\cos \theta \\ =2\times \frac{1}{4\pi {\epsilon }_0}\times \frac{Qq}{y^2+a^2}\times \cos \theta \\ =2\times \frac{1}{4\pi {\epsilon }_0}\times \frac{Qq}{y^2+a^2}\times \frac{y}{{\left(y^2+a^2\right)}^{1/2}} \\ =\frac{1}{4\pi {\epsilon }_0}·\frac{2Qqy}{{\left(y^2+a^2\right)}^{3/2}} \end{gathered}$$

For $y<<a,F_{\text{net }}=\frac{1}{4\pi {\epsilon }_0}·\frac{2Qqy}{a^3}$

Now, as $a=2m$ and $y=0.1m$

We can use approximation

$$\begin{gathered} \therefore F=\frac{1}{4\pi {\epsilon }_0}\times \frac{2\times Qqy}{a^3} \\ \Rightarrow 9\times {10}^{-3}=\frac{9\times {10}^9\times 2\times 8\times {10}^{-6}\times q\times 0.1}{2^3} \\ \Rightarrow q=5\times {10}^{-6}C \\ =5\mu C \end{gathered}$$