JEE MAIN - Physics (2024 - 8th April Evening Shift - No. 22)
If the net electric field at point $$\mathrm{P}$$ along $$\mathrm{Y}$$ axis is zero, then the ratio of $$\left|\frac{q_2}{q_3}\right|$$ is $$\frac{8}{5 \sqrt{x}}$$, where $$x=$$ ________.
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Answer
5
Explanation
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$$\begin{aligned} \Rightarrow \quad & E_2 \cos \theta=E_3 \cos \phi \\ & \frac{q_2}{\left(2^2+4^2\right)} \frac{4}{\sqrt{2^2+4^2}}=\frac{q_3}{4^2+3^2} \frac{4}{\sqrt{4^2+3^2}} \\ & \frac{q_2}{q_3}=\frac{20}{25} \frac{\sqrt{20}}{5}=\frac{4}{5} \times \frac{2 \sqrt{5}}{5} \\ \Rightarrow \quad & n=5 \end{aligned}$$
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