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JEE Main 202227 JuneEvening Shift
Exam
JEE Main
Year
2022
Date
27 June
Shift
Evening Shift
Subject
Physics
Chapter
Electrostatics

JEE Main 2022 · 27 June · Evening Shift · Physics · Electrostatics

Three identical charged balls each of charge 2C are suspended from...

Three identical charged balls each of charge 2C2C are suspended from a common point PP by silk threads of 2m2m each (as shown in figure). They form an equilateral triangle of side 1m1m.

The ratio of net force on a charged ball to the force between any two charged balls will be

Three identical charged balls each of charge 2C are suspended from a common point P by silk threads of 2m each (as

Options

  1. A3:1\sqrt{3}: 1
  2. B3:2\sqrt{3}: 2
  3. C1:11: 1
  4. D1:41: 4

Correct answer

A. 3:1\sqrt{3}: 1

Step-by-step solution

Three identical charged balls each of charge 2C are suspended from a common point P by silk threads of 2m each (as

We have force between any two chargeF1=F2=q24πε0l2=F{F}_{1} = {F}_{2} = \dfrac{{q}^{2}}{4\pi{ε}_{0}{l}^{2}} = F

Resultant electrostatics force FR=F12+F22+2F1F2cos⁡60°{F}_{R} = \sqrt{{F}_{1}^{2} + {F}_{2}^{2} + 2{F}_{1}{F}_{2}\cos 60 °}

=F2+F2+(2F2×12)=3F2= \sqrt{{F}^{2} + {F}^{2} + \left(2{F}^{2} \times \dfrac{1}{2}\right)} = \sqrt{3{F}^{2}}

So, the ratio of net force on a charged ball to the force between any charged balls is

⇒FRF=3⇒ \dfrac{{F}_{R}}{F} = \sqrt{3}

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