An electric dipole consists of charges $\pm 2.0 x {10}^{-8} C$separated by a distance of $2.0\times {10}^{-3} m$. It is placed  near a long line charge of linear charge density $4.0\times {10}^{-4} C m^{-1}$  as shown in the figure, such that the negative charge is at a distance of $2.0 cm$ from the line charge. The force acting on the dipole will be

 

The electric field at a distance $r$ from the line charge of linear density $L$ is given by

$\mathrm{E}=\frac{\mathrm{\lambda }}{2{\mathrm{\pi \epsilon }}_0\mathrm{r}}$

Hence, the field at the negative charge,

${\mathrm{E}}_1=\frac{(4.0\times {10}^{-4})(2\times 9\times {10}^9)}{0.02}=3.6\times {10}^8 {\mathrm{NC}}^{-1}$

The force on the negative charge,

${\mathrm{F}}_1 = (3.6 \mathrm{x} {10}^8) (2.0 \mathrm{x} {10}^{-8}) = 7.2 \mathrm{N}$ towards the line charge. 

Similarly, the field at the positive charge i.e. at $r = 0.022 m$

${\mathrm{E}}_2= 3.3 \mathrm{x} {10}^8 \mathrm{N} {\mathrm{C}}^{-1}$

The force on the positive charge,

${\mathrm{F}}_2 =(3.3 \mathrm{x} {10}^8) \mathrm{x} (2.0 \mathrm{x} {10}^{-8}) = 6.6 \mathrm{N}$

Net force on the dipole

${\mathrm{F}}_1-{\mathrm{F}}_2=0.6 \mathrm{N}$ towards the line charge