In H.C.verma's con. of phy vol-2 pg.108 eg.29.3 Why is 3 multiplied with the total potential energy plsss.... reply yaar !!! Pls...help Pls i will approve the ans anyway help me aieee approaching n givin me the spine chills !!!

in initial condition p.e. of one particle is given by (10μC*10μC)/(4πε0*10 cm) so for calculating p.e of whole sys i.e. for 3 particles we multiply the p.e. of d single particle by 3(as p.e is just a scalar we will simply multiply by 3) and thus we get d energy of whole sys as U=(3*10μC*10μC)/(4πε0*10 cm)=27 J

tejaswi

charges are placed at three vertices of triangle ...

let A,B,C are the vertices of the triangle then potential energy of system is given by

potential energy of all the possible paires of charges ...

total potential energy = [ (PE bw charge at A & B) + (PE bw charge at A & C) + (PE bw charge at B & C)]    ..............1

since charges are same & also distance bw any two charges is same so PE bw any two charge = kq²/d

from eq 1

total PE = [ kq²/d + kq²/d + kq²/d ]

           = 3 [ kq²/d ]

k = columbs constant = 9*10⁹

general formula for finding total paires possible is n(n-1)/2

here in the above quest n = 3 so total  paires are 3(3-1)/2 = 3