If the net electric field at point P along Y axis is zero, then the ratio of |(q₂)/(q₃)| is (8)/(5√(x)), where x = ________. Numerical answer type. Work it out, then open the answer.

Answer: 5. Distances: r₂ = √(2² + 4²) = √(20) cm, r₃ = √(3² + 4²) = 5 cm. The y -component of +q₂ 's field (upward) must cancel that of -q₃ 's field (downward). (kq₂)/(20)·(4)/(√(20)) = (kq₃)/(25)·4/5 (q₂)/(q₃) = (20√(20))/(125) = (8√(5))/(25) = (8)/(5√(5)) So x = 5.

If the net electric field at point P along Y axis is zero, then the… | JEE Main 2024 (8 Apr, Shift 2)

Q18JEE Main 2024 · 8 APR · SHIFT 2Medium

Electric Field & Field Lines

If the net electric field at point $P$ along $Y$ axis is zero, then the ratio of $\left|\frac{q_2}{q_3}\right|$ is $\frac{8}{5\sqrt{x}}$ , where $x =$ ________.

Numerical answer type. Work it out, then open the answer.

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Answer: 5

Distances: $r_2 = \sqrt{2^2 + 4^2} = \sqrt{20}$ cm, $r_3 = \sqrt{3^2 + 4^2} = 5$ cm.
The $y$ -component of $+q_2$ 's field (upward) must cancel that of $-q_3$ 's field (downward).
$\dfrac{kq_2}{20}\cdot\dfrac{4}{\sqrt{20}} = \dfrac{kq_3}{25}\cdot\dfrac{4}{5}$
$\dfrac{q_2}{q_3} = \dfrac{20\sqrt{20}}{125} = \dfrac{8\sqrt{5}}{25} = \dfrac{8}{5\sqrt{5}}$
So $x = 5$ .

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