Physics · Electrostatics

JEE Main 2024 — 8 April, Shift 2 — Question 50

If the net electric field at point P along Y axis is zero, then the ratio of ∣q2q3∣\left|\frac{\mathrm{q}_{2}}{\mathrm{q}_{3}}\right| is 85x\frac{8}{5 \sqrt{\mathrm{x}}}, where x=\mathrm{x}= \qquad

Question figure

Answer: 5

Numerical answer — enter this value.

Step-by-step solution

Kq220cos⁡β=Kq325cos⁡θ\frac{\mathrm{Kq}_{2}}{20} \cos \beta=\frac{\mathrm{Kq}_{3}}{25} \cos \theta

Kq220420=Kq325425\frac{\mathrm{Kq}_{2}}{20} \frac{4}{\sqrt{20}}=\frac{\mathrm{Kq}_{3}}{25} \frac{4}{\sqrt{25}}

q2q3=20252025=85x\frac{\mathrm{q}_{2}}{\mathrm{q}_{3}}=\frac{20}{25} \sqrt{\frac{20}{25}}=\frac{8}{5 \sqrt{\mathrm{x}}}

⇒x=8×25255×2020\Rightarrow \sqrt{\mathrm{x}}=\frac{8 \times 25 \sqrt{25}}{5 \times 20 \sqrt{20}} x=5\mathrm{x}=5

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Subject
Physics
Chapter
Electrostatics
Topic
Electrostatic Field & motion of charge
If the net electric field at point P along Y axis is zero, then the… | JEE Main 2024 PYQ with Solution · DhiX AI