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Electrostatics question

2013 · Q134
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Electrostatics question

2013 · Q134

NEETPhysicsElectrostaticsMCQ+4 / −1
A charge q is placed at the centre of the line joining two equal charges Q. The system of the three charges will be in equilibrium if q is equal to
  1. A
    −-−Q/4
  2. B
    Q/4
  3. C
    −-−Q/2
  4. D
    Q/2
View written solutionFree

Correct answer: A

Let the distance between given changes be 2x

NEET 2013 (Karnataka) Physics - Electrostatics Question 60 English Explanation

In equilibrium, ∣F→AB∣+∣F→AC∣=0\left| {{{\overrightarrow F }_{AB}}} \right| + \left| {{{\overrightarrow F }_{AC}}} \right| = 0​FAB​​+​FAC​​=0

So, ∣FAB∣=−∣FAC∣\left| {{F_{AB}}} \right| = - \left| {{F_{AC}}} \right|∣FAB​∣=−∣FAC​∣

14πε0Q24x2=−14πε0Qqx2{1 \over {4\pi {\varepsilon _0}}}{{{Q^2}} \over {4{x^2}}} = - {1 \over {4\pi {\varepsilon _0}}}{{Qq} \over {{x^2}}}4πε0​1​4x2Q2​=−4πε0​1​x2Qq​

⇒q=−Q/4 \Rightarrow q = - Q/4⇒q=−Q/4

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