$$ \begin{aligned} &\text { Using the Pythagorean theorem, The distance between }-q \text { and the positive charges }\\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\sqrt{\frac{l^2}{4}+\frac{3 l^2}{4}}=l \end{aligned} $$

IAT (IISER) 2024 Physics - Electrostatics Question 8 English Explanation 1

Let,

$$ |F|=\left|\frac{q^2}{4 \pi \epsilon_0 l^2}\right| $$

Then, the forces can be redrawn as the following:

IAT (IISER) 2024 Physics - Electrostatics Question 8 English Explanation 2

$$ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text { Simplifying: } $$

IAT (IISER) 2024 Physics - Electrostatics Question 8 English Explanation 3

Vectorially this can be written as:

$$ \frac{1}{2} F \hat{\imath}-\frac{3 \sqrt{3}}{2} F \hat{\jmath}=\frac{F}{2}(\hat{\imath}-3 \sqrt{3} \hat{\jmath}) $$

On substituting value of $F$ we get,

$$ F_{n e t}=\frac{q^2}{8 \pi \epsilon_0 l^2}(\hat{\imath}-3 \sqrt{3} \hat{\jmath}) $$

Using $F=m a, a=\frac{F_{\text {net }}}{m}$

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