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

Answer

i. zero ii. `(1)/(4 pi epsilon_0) (q^2)/(2 a)[2 sqrt(2) + 1] + (1)/(4 pi epsilon_0) (p q)/(a^2)`

Explanation

a. Force exerted by the upper charge on dipole
`F_1 = (1)/(2 pi epsilon_0) (p q)/(a^3)` (down)
force extered by the left charge on dipole
`F_2 = (1)/(4 pi epsilon_0) (p q)/(a^3)` (up)
Force extered by the right charge on dipole
`F_3 = (1)/(4 pi epsilon_0) (p q)/(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 ((1)/(4 pi epsilon_0)(q^2)/(sqrt(2 a))) + (1)/(4 pi epsilon_0)(q^2)/(2 a)`
`- vec P . vec E _(up) -vec P . vec E _("left") - vec P . vec E _("right")`
`vec P . vec E_("left") = vec P . vec E_("right") = 0`
(Because electric fields produced by left and right charges are perpendicular to P.)
`- vec P . vec E_(up) = - P. ((1)/(4 pi epsilon_0) q/(a^2)) cos pi = (1)/(4 pi epsilon_0) (q p)/(a^2)`
B Putting the values, we get
`U = (1)/(4 pi epsilon_0) (q^2)/(2a) [ 2 sqrt(2) + 1] + (1)/(4 pi epsilon_0) (p q)/(a^2)`.