Question

For the electrostatic charge system as shown in Fig. 3.121, find
a. The net force on electric dipole, and
b. Electrostatic energy of the system.

Answer 1

a. Force exerted by the upper charge on dipole

$F_1=\frac{1}{2\pi {\epsilon }_0}\frac{pq}{a^3}$ (down)

Force exerted by the left charge on dipole

$F_2=\frac{1}{4\pi {\epsilon }_0}\frac{pq}{a^3}$ (up)

Force exerted by the right charge on dipole

$F_3=\frac{1}{4\pi {\epsilon }_0}\frac{pq}{a^3}$ (up)

Net force on the dipole due to all charges

$F=F_1+F_2+F_3=0$

Hence, net force on the dipole is zero. The total electric potential energy consists of interaction of all

the three charges among themselves and interaction of these three charges with dipole. So,

$U=2\left(\frac{1}{4\pi {\epsilon }_0}\frac{q^2}{\sqrt{2}a}\right)+\frac{1}{4\pi {\epsilon }_0}\frac{q^2}{2a}$

$-\overrightarrow{P}.{\overrightarrow{E}}_{up}-\overrightarrow{P}.{\overrightarrow{E}}_{left}-\overrightarrow{P}.{\overrightarrow{E}}_{right}$

$\overrightarrow{P}.{\overrightarrow{E}}_{left}=\overrightarrow{P}.{\overrightarrow{E}}_{right}=0$

(Because electric fields produced by left and right charges are perpendicular to P)

$-\overrightarrow{P}.{\overrightarrow{E}}_{up}=-P.\left(\frac{1}{4\pi {\epsilon }_0}\frac{q}{a^2}\right)cos\pi =\frac{1}{4\pi {\epsilon }_0}\frac{qp}{a^2}$

b. Putting the values, we get

$U=\frac{1}{4\pi {\epsilon }_0}\frac{q^2}{2a}[2\sqrt{2}+1]+\frac{1}{4\pi {\epsilon }_0}\frac{pq}{a^2}$