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 $a(AP=BP=BP=PC=a=\sqrt{2}m),$ a charge -Q is fixed at point P. A dipole with dipole moment $p=1$ Cm is placed at the center of the base and perpendicular to its plane as shown in figure. Force on the dipole due to the charge particles is $\frac{x}{4\pi {\epsilon }_0}N$. Find $x$.

Answer 1

Here, $OA=OB=OC=OD=r=\sqrt{(a/2{)}^2+(a/2{)}^2}=a/\sqrt{2}=1m$ and $OP=\sqrt{a^2-(a/\sqrt{2}{)}^2}=a/\sqrt{2}=r$
The electric field at the each vertex of the square ABCD due to dipole is $E=\frac{p}{4\pi {ϵ}_0{r}^3}$
Hence, force on each of them due to dipole is $F_1=QE=\frac{Qp}{4\pi {ϵ}_0{r}^3}$
This force is downward on charges. Hence, force due to these charges on dipole is $4F_1$ upward.
Now the force on dipole due to charge at P is $F_2=\frac{Qp}{4\pi {ϵ}_0(OP{)}^3}=\frac{Qp}{4\pi {ϵ}_0{r}^3}$ upward.
Net force on dipole is $F=4F_1+F_2=\frac{4Qp}{4\pi {ϵ}_0{r}^3}+\frac{Qp}{4\pi {ϵ}_0{r}^3}=\frac{5}{4\pi {ϵ}_0}$ as $Q=1,p=1,r=1$
thus, $x=5$