What is the force on a dipole of dipole moment p placed as shown in fig.

Options

A. `(2qp)/(4 pi epsilon_(0)a^(2))`
B. `(2qp)/(4 pi epsilon_(0)a^(3))`
C. `(qp)/(4 pi epsilon_(0)a^(3))`
D. `(qp)/(4 pi epsilon_(0)a^(2))`

Answer

C

Explanation

Force on any q by dipole is
`F=qE_("dipole")=(q)/(4 pi epsilon_(0))(p)/(a^(3))` downward
So from Newton's third law, force on the dipole due to both chages is
`2F=(qp)/(2 pi epsilon_(0)a^(3))` upward.