Verified Solution • 2025–26 Session moderate

QJEEM23Ph775f: JEE Main 2023 Physics - Class 12 JEEM23Ph775f Solution: Two equal positive point charges are separated by a distance .…

Two equal positive point charges are separated by a distance 2a2 a . The distance of a point from the centre of the line joining two charges on the equatorial line (perpendicular bisector) at which force experienced by a test charge q0\mathrm{q}_{0} becomes maximum is ax\frac{a}{\sqrt{x}} . The value of xx is __________.

Step-by-Step Verified Solution

JEE Main 2023 (Online) 1st February Morning Shift Physics - Electrostatics Question 82 English Explanation

FP=q0Ep=q0kqz(a2+z2)3/2F_{P}=q_{0} E_{p}=q_{0} \frac{k q z}{\left(a^{2}+z^{2}\right)^{3 / 2}}

 or FP=kqq0z(a2+z2)3/2\text { or } F_{P}=\frac{k q q_{0} z}{\left(a^{2}+z^{2}\right)^{3 / 2}}

To maximize dFPdz=0\frac{d F_{P}}{d z}=0

or kqq0(a2+z2)3/2−z32×2z(a2+z2)12(a2+z2)3=0k q q_{0} \frac{\left(a^{2}+z^{2}\right)^{3 / 2}-z \frac{3}{2} \times 2 z\left(a^{2}+z^{2}\right)^{\frac{1}{2}}}{\left(a^{2}+z^{2}\right)^{3}}=0

⇒z=a2\Rightarrow z=\frac{a}{\sqrt{2}}

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring