Question

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 ' $a$ ' $(A P=B P=D P=P C=a=\sqrt{2} m )$, $a$ 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 $\frac{\Omega}{4 \pi \varepsilon_{0}} N$. The value of $\Omega$ is_________

Answer 1

Charges at $A, B, C$ and $D$ are placed at equilateral position of dipole. Hence force on each of them due to dipole
$F_{1}=\frac{Q p}{4 \pi \epsilon_{0}(a / \sqrt{2})^{3}} .$
This force is downward on charges. Hence force due to these charges on dipole is $4 F_{1}$ upwards.
Force on dipole due to charge at $P$ :
$F_{2}=Q . \frac{2 p Q}{4 \pi \epsilon_{0}(a / \sqrt{2})^{2}} \text { (upward) }$
Net force on dipole: $F=4 F_{1}+F_{2}=\frac{3 \sqrt{2} Q p}{\pi \epsilon_{0} a^{3}}$ upward
After substituting given values we get $F=\frac{6}{4 \pi \varepsilon_{0}} N$.