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JEE Main 202431 JanuaryMorning Shift
Exam
JEE Main
Year
2024
Date
31 January
Shift
Morning Shift
Subject
Physics
Chapter
Electrostatics

JEE Main 2024 · 31 January · Morning Shift · Physics · Electrostatics

Two charges q and 3q are separated by a distance ‘ r ’ in air. At a...

Two charges qq and 3q3q are separated by a distance ‘rr’ in air. At a distance xx from charge qq, the resultant electric field is zero. The value of xx is :Two charges q and 3q are separated by a distance ‘ r ’ in air. At a distance x from charge q , the resultant electric

Options

  1. A(1+3)r\dfrac{( 1 + \sqrt{3} )}{r}
  2. Br3(1+3)\dfrac{r}{3 ( 1 + \sqrt{3} )}
  3. Cr(1+3)\dfrac{r}{( 1 + \sqrt{3} )}
  4. Dr(1+3)r ( 1 + \sqrt{3} )

Correct answer

C. r(1+3)\dfrac{r}{( 1 + \sqrt{3} )}

Step-by-step solution

Two charges q and 3q are separated by a distance ‘ r ’ in air. At a distance x from charge q , the resultant electric

Given: (E→net)P=0{\left(\overset{\to}{E}net\right)}_{P} = 0

⇒kqx2=k×3q(r−x)2⇒ \dfrac{kq}{{x}^{2}} = \dfrac{k \times 3q}{{\left(r - x\right)}^{2}}

⇒(r−x)2=3x2⇒ {\left(r - x\right)}^{2} = 3{x}^{2}

⇒r−x=3x⇒ r - x = \sqrt{3}x

⇒x=r3+1⇒ x = \dfrac{r}{\sqrt{3} + 1}

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