QJEEM24Ph012a: JEE Main 2024 Physics - Class 12 JEEM24Ph012a Solution: If the net electric field at point along axis is zero, then the …

$\begin{aligned}\Rightarrow \quad & E_2 \cos \theta=E_3 \cos \phi \\& \frac{q_2}{\left(2^2+4^2\right)} \frac{4}{\sqrt{2^2+4^2}}=\frac{q_3}{4^2+3^2} \frac{4}{\sqrt{4^2+3^2}} \\& \frac{q_2}{q_3}=\frac{20}{25} \frac{\sqrt{20}}{5}=\frac{4}{5} \times \frac{2 \sqrt{5}}{5} \\\Rightarrow \quad & n=5\end{aligned}$