question is based on the concept of electric charges
2007-02-02 02:25:51 · 3 answers · asked by dr.jasdeep 1 in Science & Mathematics ➔ Physics
above is wrong AS HE HAS NOT CONSIDERD THE TWO q CHARGES WICH IS ALSO IN SYSTEM
u know electric potential energy is
V=kq q /r
hence
k q^2/2a - (2kq^2/a) - (2kq^2/a)
= - 7k q^2/2a
2007-02-02 23:08:47 · answer #1 · answered by n nitant 3 · 0⤊ 0⤋
there is not any such factor obtainable. ideally in case of two opposite costs of diverse magnitude, we would place the third cost on the axis of the two costs, such that it lies on one area of the two costs. because of the fact the charges are equivalent, the rigidity exerted could be equivalent.yet because of the fact the area from the two costs would be diverse, this can not ensue
2016-11-24 19:08:10 · answer #2 · answered by Anonymous · 0⤊ 0⤋
kq(-2q)/a+kq(-2q)/a= -4kq2/a[2 is power of q]
2007-02-02 02:49:37 · answer #3 · answered by Anonymous · 0⤊ 2⤋