Q 12-01-060JEE MainJEE Main 2024 (8 Apr, Shift 2)Medium
If the net electric field at point P along the Y axis is zero, then the ratio of $\left|\dfrac{q_2}{q_3}\right|$ is $\dfrac{8}{5\sqrt x}$, where $x =$ ______.
Numerical value type. Enter your answer.
Answer: 5
Distances from P: $r_2 = \sqrt{2^2 + 4^2} = \sqrt{20}\ \text{cm}$, $r_3 = \sqrt{3^2 + 4^2} = 5\ \text{cm}$.
The field of $+q_2$ at P has an upward $y$-component and the field of $-q_3$ (towards $q_3$) has a downward $y$-component. For zero net $y$-component:
$$\frac{kq_2}{r_2^2}\cdot\frac{4}{r_2} = \frac{kq_3}{r_3^2}\cdot\frac{4}{r_3} \;\Rightarrow\; \frac{q_2}{q_3} = \frac{r_2^3}{r_3^3} = \frac{20\sqrt{20}}{125} = \frac{8\sqrt5}{25} = \frac{8}{5\sqrt5}$$
So $x = 5$.
Solution by Sreeraj P, M.Sc Physics