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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=PD=PC=a=2 m), a charge −Q is fixed at point P. A dipole with dipole moment p=1 Cm is placed at the centre of the base and perpendicular to its plane as shown in figure. Force on the dipole due to the charge particles is x4πε0N. Find the value of x?

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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 dipole is F1=Qp4πε0(a/2)3 This force is downward on charges. Hence, force due to these charges on dipole is 4F1 upward. Force on dipole due to charge at P isF2=2Qp4πε0(a/2)3(upward)Net force on dipole is F=4F1+F2=32 Qpπε0a3upwardAfter substituting given values, we get F=64πε0N

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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=PD=PC=a=2 m), a charge −Q is fixed at point P. A dipole with dipole moment p=1 Cm is placed at the centre of the base and perpendicular to its plane as shown in figure. Force on the dipole due to the charge particles is x4πε0N. Find the value of x?