Three particles have charges 20 mc each and they are fixed at the corners of an eq. triangle of side 0.5m The force on each of the particle has magnitude

Dear pallavi

force between two charge

F= 1/4∏ε (20 *10-6)²/(.5)²

  = 9*10⁹  * 4*10^-10 /.25

  =14.4 N

this force act along the side of triangle .

in one charge two force act along the direction of side and angel between then is 60

resultant force F'=[F² +F² +2FFcos60]^1/2

                              ⁼F√3

                     =24.94 N

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