Two point charges `+20nC` and `-20nC` are separated by a distance of 0.10m. Calculate the electric field intensity at a perpendicular distance 0.10 m from the centre of a dipole.

Explanation

We know that
`E_(p)=(1/(4 pi epsilon_(0)))p/((r^(2)+a^(2))^(3/2))Vm^(-1)`
i.e. `E_(p)=(9xx10^(9)xx(0.1xx20xx10^(-9)))/((0.1^(2)+0.05^(2))^(3/2))`
`=18/([(10xx10^(-2))^(2)xx(5xx10^(-2))^(2)]^(3/2))`
`=18/(10^(-6)[125]^(3/2))`
`=0.0129xx10^(6)Vm^(-1)`
`E_(p)=1.29xx10^(4)Vm^(-1)` along PC (parallel to AB and opposite to the dipole moment)