Question

Four charge the particles each having charge Q are fixed at the comers of the base (at A, B, C, and D) of .a square pyramid with slant length a (AP = BP = DP = PC= a). A charge -Q is fixed at point P. A dipole with dipole moment P is placed at'the center of base and perpendicular to its plane as shown in Fig. 3.122.Find

a. the force on dipole due to charge particles, and

b. the potential energy of the system.

Answer 1

Charges at A, B, C, and D are placed at an equilateral position of dipole. Hence, the force on each of them due to dipole is

$F_1=\frac{Qp}{4\pi {\epsilon }_0(a/\sqrt{2}{)}^3}$

This force is downward on charges. Hence, force due to these charges on dipole is $4F_1$ (upward). Force

on dipole due to charge at P is

$F_2=\frac{2pQ}{4\pi {\epsilon }_0(a/\sqrt{2}{)}^3}$(upward)

Net force on dipole is

$F=4F_1+F_2=\frac{3\sqrt{2}QP}{\pi {\epsilon }_0{a}^3}$ (upward)

b.PE of the system is

U= (10 pairs of charged particles) + (5 pairs of dipole and charged particles)

As potential energy of the dipole with four charges at A, B, C,and D will be zero,

$U=4\left[\frac{1}{4\pi {\epsilon }_0}\frac{Q^2}{a}\right]+2\frac{1}{4\pi {\epsilon }_0}\frac{Q^2}{\sqrt{2}a}-4\frac{1}{4\pi {\epsilon }_0}\frac{Q^2}{a}$

$-\frac{1}{4\pi {\epsilon }_0}\frac{pQ}{(a/\sqrt{2}{)}^2}$

$U=\frac{Q^2}{2\sqrt{2}\pi {\epsilon }_{0}a}-\frac{pQ}{2\pi {\epsilon }_0{a}^2}$