Question

Four charge particles each having charge $Q=1C$ are fixed at corners of base (at $A,B,C$ and $D$ ) of a square pyramid with slant length ' $a$ ' $(AP=BP=DP=PC=a=\sqrt{2}m)$, $a$ charge $-Q$ is fixed at point $P$. A dipole with dipole moment $p=1C-m$ 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 $\frac{\Omega }{4\pi {\epsilon }_0}N$. The value of $\Omega$ is_________

Answer 1

Answer: 6

Charges at $A,B,C$ and $D$ are placed at equilateral position of dipole. Hence force on each of them due to dipole
$F_1=\frac{Qp}{4\pi {ϵ}_0(a/\sqrt{2}{)}^3}.$
This force is downward on charges. Hence force due to these charges on dipole is $4F_1$ upwards.
Force on dipole due to charge at $P$ :
$F_2=Q.\frac{2pQ}{4\pi {ϵ}_0(a/\sqrt{2}{)}^2}\text{ (upward) }$
Net force on dipole: $F=4F_1+F_2=\frac{3\sqrt{2}Qp}{\pi {ϵ}_0{a}^3}$ upward
After substituting given values we get $F=\frac{6}{4\pi {\epsilon }_0}N$.