Four charge particles each having charge Q=1 C are fixed at corners of base (at A, B, C and D) of a square pyramid with slant length 'a' (AP=BP=DP=PC=a=2m),a charge -Q is fixed at point P. A dipole with dipole moment p= 1 C-m is placed at centre of base and perpendicular to its plane as shown in figure. If the force on dipole due to charge particles is Ω4πε0N. The value of Ω is _________.
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answer is 6.
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Detailed Solution
Charges at A, B, C and D are placed at equilateral position of dipole. Hence force on each of them due to dipoleF1=Qp4π∈0(a/2)3This force is downward on charges. Hence force due to these charges on dipole is 4F1 upwards.Force on dipole due to charge at P:F2=Q⋅2pQ4π∈0(a/2)2(upward) Net force on dipole: F=4F1+F2=32Qpπ∈0a3 upward After substituting given values we get F=64πε0N.