# Four point charges,each +q, are fixed at the corners of a square of side a. Another point charge Q is placed at a height h vertically above the center of square,assuming the square to be in a horizontal plane. Magnitude of force experienced by Q is :

User
10 Points
5 years ago
Its "0" because Q is equidistant from all four point and it cancels out and we get "0" okay assume the square and Q as a charge on pillar on centre the distance of end of the pillar from points is "2 s power half a"then now we want distamce of top of pillar from points now we can see aright angle triangle of pillars hiegjt h and base "2 s power half a" and we have the hypotneuse equal from both points and then they all cancel out because this force is vector
Ojasvi Singh
20 Points
5 years ago
The Answer will be 4KQqh/(h^2 +(a^2)/2)^3/2 where K is constant. This is because of its summation of its components in upper direction.
Ojasvi Singh
20 Points
5 years ago
The Answer will be 4KQqh/(h^2+(a^2)/2)^3/2 where K is constant. This is because of the summation of the components of force in upper direction.