Four charge particles each having charge Q=1 C are fixed at the corners of the base(A,B,C, and D) of a square pyramid with slant length `a(AP=BP=PC=a=sqrt(2)m)`, a charge -Q is fixed at point P. A dipole with dipole moment p=1 C-m is placed at the center of the bases and perpendicular to its plane as shown in fig. Force on the dipole due to the charge particles is `(square)/(4 pi epsilon_(0))N`.

Answer

6

Explanation

(i) charge at A, B, C, and D are placed at equilateral position of dipole. Hence, force on each of them due to dipole is
`F_(1)=(Qp)/(4 pi epsilon_(0)(a//sqrt(2))^(3))`
this force is downward on charges. Hence, force due to theses charges on dipole is `4F_(1)` upward. Force on dipole due to change at p is
`F_(2)=Q(2 Qp)/(4 pi epsilon_(0)(a//sqrt(2))^(2))` (upward)
Net force on dipole is `F=4F_(1)+F_(2)=(3sqrt(2)QP)/(pi epsilon_(0)a^(3))` upward
After substituting given values, we get `F=(6)/(4 pi epsilon_(0))N`.