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 $$x4$$πε$$0N$$. Find the value of x?

Question

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  $$x4$$πε$$0N$$. Find the value of x?

Infinity Learn · Accepted Answer

Charges at A, B, C, and D are placed at equilateral position of dipole. Hence, force on each of them due to dipole is $$F1=Qp4$$πε$$0(a/2)3$$ 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 $$isF2=2Qp4$$πε$$0(a/2)3(upward)Net$$ force on dipole is $$F=4F1+F2=32$$  Qpπε$$0a3upwardAfter$$ substituting given values, we get $$F=64$$πε$$0N$$

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