JEE Main 2019 — Electrostatics Question with Solution
From: JEE Main 2019 (Online) 8th April Morning Slot
Question
The bob of a simple pendulum has mass 2g and
a charge of 5.0 μC. It is at rest in a uniform
horizontal electric field of intensity 2000 V/m.
At equilibrium, the angle that the pendulum
makes with the vertical is : (take g = 10 m/s2)
Choose an option
Show full solutionCorrect option: C
Correct answer
Ctan–1(0.5)
Step-by-step explanation
Tcos = mg
Tsin = qE
tan =
tan =
Tsin = qE
tan =
tan =
Practice this on the real CBT interface
Solve this JEE Main question (and the rest of the Electrostatics chapter) on PrepSharp's TCS iON-style CBT player — with timer, bookmarks and session analytics.
Solve interactively →About this question
This is a previous-year question from JEE Main 2019, covering the Electrostatics chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.