Verified Solution • 2025–26 Session moderate

QJEEM19Ph1f01: JEE Main 2019 Physics - Class 12 JEEM19Ph1f01 Solution: Three charges + Q, q, + Q are placed respectively, at distance, …

Three charges + Q, q, + Q are placed respectively, at distance, 0, d/2 and d from the origin, on the x-axis. If the net force experienced by + Q, placed at x = 0, is zero, then value of q is :

Multiple Choice Options

  • A −- Q4{Q \over 4}
  • B + Q2{Q \over 2}
  • C + Q4{Q \over 4}
  • D −- Q2{Q \over 2}

Step-by-Step Verified Solution

JEE Main 2019 (Online) 9th January Morning Slot Physics - Electrostatics Question 188 English Explanation

Force on + Q charge at x = 0 due to q charge, F1 = KQq(d2)2{{KQq} \over {{{\left( {{d \over 2}} \right)}^2}}}

Force on +Q charge at x = 0 due to + Q charge at x = d is,

      F2 = KQQd2{{KQQ} \over {{d^2}}}

According to the question,

F1 + F2 = 0

⇒\Rightarrow    F1 = −- F2

⇒\Rightarrow     KQq(d2)2{{KQq} \over {{{\left( {{d \over 2}} \right)}^2}}} = −- KQQd2{{KQQ} \over {{d^2}}}

⇒\Rightarrow    4q = −- Q

⇒\Rightarrow    q = −- Q4{Q \over 4}

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