Physics, asked by urmikr2811, 9 months ago

Two fixed point changes +4q and +2q are separated by a distance a cm in air where should the therd point charge q be placed from +4q to be equilibrium

Answers

Answered by amansharma264
9

EXPLANATION

  • Two fixed charges +4q and +2q
  • each seperate by a distance a cm
  • put +Q both the charges
  • so that charge is in equilibrium
  • put +Q charge towards the small +2q charges
  • F = kQ1Q1 / R^2
  • F1 = 4kqQ / (a - x) ^2
  • F2 = 2KqQ / x^2
  • F1 = F2
  • 4kqQ / ( a - x) ^2 = 2kqQ / x^2
  • a - x / x = √2
  • a - x = √2x
  • a = x (√2 + 1 )

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