Physics-
General
Easy

Question

The charges Q comma blank plus q blank a n d plus q are placed at the vertices of a right-angled isosceles triangle as shown. The net electrostatic energy of the configuration is zero if Q is equal to

  1. fraction numerator negative q over denominator 1 plus square root of 2 end fraction  
  2. fraction numerator negative 2 q over denominator 2 plus square root of 2 end fraction  
  3. negative 2 q  
  4. plus q  

The correct answer is: fraction numerator negative 2 q over denominator 2 plus square root of 2 end fraction


    Net electrostatic energy U equals fraction numerator k Q q over denominator a end fraction plus fraction numerator k q to the power of 2 end exponent over denominator a end fraction plus fraction numerator k Q q over denominator a square root of 2 end fraction equals 0
    rightwards double arrow fraction numerator k q over denominator a end fraction open parentheses Q plus q plus fraction numerator Q over denominator square root of 2 end fraction close parentheses equals 0 rightwards double arrow Q equals negative fraction numerator 2 q over denominator 2 plus square root of 2 end fraction

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